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![© Art Traynor 2011
Mathematics
Definitions
Expression
Symbol Operation Relation
Expression – A Mathematical Sentence
Proposition Formula
VariablesConstants
Operands ( Terms )
Equation
A formula stating an
equivalency class relation
Linear Equation
An equation in which each term is either
a constant or the product of a constant
and (a) variable[s] of the first degree
Mathematics
Polynomial](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-6-320.jpg)
![© Art Traynor 2011
Mathematics
Linear Equation
Linear Equation
Equation
An Equation consisting of:
Operands that are either
Any Variables are restricted to the First Order n = 1
Linear Equation
An equation in which each term
is either a constant or the
product of a constant and (a)
variable[s] of the first order
Expression
Proposition
Equation
Formula
n Constant(s) or
n A product of Constant(s) and
one or more Variable(s)
The Linear character of the Equation derives from the
geometry of its graph which is a line in the R2 plane
As a Relation the Arity of a Linear Equation must be
at least two, or n ≥ 2 , or a Binomial or greater Polynomial
Polynomial](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-21-320.jpg)
![© Art Traynor 2011
Mathematics
Equation
Linear Equation
Linear Equation
An equation in which each term is either a constant or the product
of a constant and (a) variable[s] of the first order
Term ai represents a Coefficient
b = Σi= 1
n
ai xi = ai xi + ai+1 xi+1…+ an – 1 xn – 1 + an xn
Equation of a Line in n-variables
A linear equation in “ n ” variables, xi + xi+1 …+ xn-1 + xn
has the form:
n Coefficients are distributed over a defined field
( e.g. N , Z , Q , R , C )
Term xi represents a Variable ( e.g. x, y, z )
n Term a1 is defined as the Leading Coefficient
n Term x1 is defined as the Leading Variable
Section 1.1, (Pg. 2)
Section 1.1, (Pg. 2)
Section 1.1, (Pg. 2)
Section 1.1, (Pg. 2)
Coefficient = a multiplicative factor
(scalar) of fixed value (constant)
Section 1.1, (Pg. 2)](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-22-320.jpg)
![© Art Traynor 2011
Mathematics
Linear Equation
Equation
Standard Form ( Polynomial )
Ax + By = C
Ax1 + By1 = C
For the equation to describe a line ( no curvature )
the variable indices must equal one
ai xi + ai+1 xi+1 …+ an – 1 xn –1 + an xn = b
ai xi
1 + ai+1 x 1 …+ an – 1 x 1 + a1 x 1 = bi+1 n – 1 n n
ℝ
2
: a1 x + a2 y = b
ℝ
3
: a1 x + a2 y + a3 z = b
Blitzer, Section 3.2, (Pg. 226)
Section 1.1, (Pg. 2)
Test for Linearity
A Linear Equation can be expressed in Standard Form
As a species of Polynomial , a Linear Equation
can be expressed in Standard Form
Every Variable term must be of precise order n = 1
Linear Equation
An equation in which each term
is either a constant or the
product of a constant and (a)
variable[s] of the first order
Expression
Proposition
Equation
Formula
Polynomial](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-23-320.jpg)
![© Art Traynor 2011
Mathematics
Matrix Representation
Linear Algebra
Matrix Representation Methods
Uppercase Letter Designation
Section 1.2, (Pg. 40)
A , B , C
Bracket-Enclosed Representative Element
[ aij ] , [ bij ] , [ cij ]
a11
a21
a31
am1
.
.
.
a12
a22
a32
am2
.
.
.
a13
a23
a33
am3
.
.
.
. . .
. . .
. . .
. . .
.
.
.
a1n
a2n
a3n
amn
.
.
.
Rectangular Array
Brackets denote a Matrix ( i.e. not a specific element/real number)](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-47-320.jpg)
![© Art Traynor 2011
Mathematics
Matrix Equality
Linear Algebra
Matrix Equality Section 1.2, (Pg. 40)
A = [ aij ]
B = [ bij ]
are equal
when
Amxn = Bmxn
aij = bij
1 ≤ i ≤m
1 ≤ j ≤n
a1 a2 a3 . . . ana =
C1 C2 C3 . . . Cn
Row Matrix / Row Vector
A 1 x n (“ 1 by n ”) matrix is a single row
Column Matrix / Column Vector
b1
b2
b3
bm
.
.
.
C1
An m x 1 (“ m by 1 ”) matrix is a single column](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-48-320.jpg)
![© Art Traynor 2011
Mathematics
Matrix Operations
Linear Algebra
Matrix Summation Section 1.2, (Pg. 41)
A = [ aij ]
B = [ bij ]
is given
by
+ A + B = [ aij + bij ]
– 1
0
2
1
1
– 1
3
2
+ = ( – 1 + 1 ) =
( 0 + [ – 1] )
( 2 + 3 )
( 1 + 2 )
0
– 1
5
3
Scalar Multiplication
1
– 3
2
2
0
1
4
– 1
2
A = 3A =
3 ( 1 ) 3 ( 2 ) 3 ( 4 )
3 ( – 3 ) 3 ( 0 ) 3 ( – 1 )
3 ( 2 ) 3 ( 1 ) 3 ( 2 )
3
– 9
6
6
0
3
12
– 3
6
3A =](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-49-320.jpg)
![© Art Traynor 2011
Mathematics
Matrix Operations
Linear Algebra
Section 1.2, (Pg. 42)Matrix Multiplication
– 1
4
5
3
– 2
0
A =
A = [ aij ] Amx n
B = [ bij ] Bnx p
then AB = [ cij ] = Σk = 1
n
aik bkj = ai 1 b1 j + ai2 b2j +…+ ain –1 bn-1j + ain bnj
The entries of Row “ Aik” ( the i-th row ) are multiplied by the entries of “ Bkj” ( the j-th column )
and sequentially summed through Row “ Ain” and Column “ Bnj” to form the entry at [ cij ]
– 3
– 4
2
1
B =
c11 c12
C = c21 c22
c31 c32
a11b11 + a12b21 a11b12 + a12b22
= a21b11 + a22b21 a21b12 + a22b22
a31b11 + a32b21 a31b12 + a32b22
Product Summation Operand Count
For Each Element of AB (single entry)
Product Summation (Column-Row) Index
For the product of two matrices to be defined, the column count of the multiplicand matrix
must equal the row count of the multiplier matrix ( i.e. Ac = Br )
ABmx p](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-50-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Proofs
Let A = [ aij ], B = [ bij ]
Introduce/Define/Declare The Constituents to be Proven
This statement declares A & B to be Matrices
Specifies the row & column count index variables](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-55-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Properties Of Matrices – Proofs
A( BC ) = ( AB ) C
Order of Terms is preserved
Affects Order of Operations
Sequence – PEM-DAS
Associative
(Multiplication)
Σk = 1
n
T = [ tij ] = ( aik bkj )ckjΣk = 1
n
Σi = 1
n
( yj )Σj = 1
n
( xi )
Σi = 1
n
( xi ) ( y1 + y2 +…+ yn –1 + yn )
( x1 + x2 +…+ xn –1 + xn ) y1 + ( x1 + x2 +…+ xn –1 + xn ) y2 +…
( x1 + x2 +…+ xn –1 + xn ) yn –1 + ( x1 + x2 +…+ xn –1 + xn ) yn
x1 y1 + x2 y1 +…+ xn –1 y1 + xn y1 + x1 y2 + x2 y2 +…+ xn –1 y2 + xn y2 +…
x1 yn –1 + x2 yn –1 +…+ xn –1 yn –1 + xn yn –1 + x1 yn + x2 yn +…+ xn –1 yn + xn yn](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-58-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Properties Of Matrices – Proofs
A( BC ) = ( AB ) C
Order of Terms is preserved
Affects Order of Operations
Sequence – PEM-DAS
Associative
(Multiplication)
Σk = 1
n
T = [ tij ] = ( aik bkj )ckjΣk = 1
n
Σi = 1
n
( yj )Σj = 1
n
( xi )
Σi = 1
n
( xi ) ( y1 + y2 +…+ yn –1 + yn )
( x1 + x2 +…+ xn –1 + xn ) y1 + ( x1 + x2 +…+ xn –1 + xn ) y2 +…
( x1 + x2 +…+ xn –1 + xn ) yn –1 + ( x1 + x2 +…+ xn –1 + xn ) yn
x1 y1 + x2 y1 +…+ xn –1 y1 + xn y1 + x1 y2 + x2 y2 +…+ xn –1 y2 + xn y2 +…
x1 yn –1 + x2 yn –1 +…+ xn –1 yn –1 + xn yn –1 + x1 yn + x2 yn +…+ xn –1 yn + xn yn
Σi = 1
n
Σj = 1
n
xi yj](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-59-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Properties Of Matrices – Proofs
A = [ aij ] Amx n
B = [ bij ] Bnx p
then AB = [ cij ] = Σk = 1
n
aik bkj = ai 1 b1 j + ai2 b2j +…+ ain –1 bn-1j + ain bnj
Product Summation Operand Count
Product Summation (Column-Row) Index
ABmx p
The entries of Row “ Aik” ( the i-th row ) are multiplied by the entries of “ Bkj” ( the j-th column )
and sequentially summed through Row “ Ain” and Column “ Bnj” to form the entry at [ cij ]
ai,1 b1, j + ai,1 b1, j +1 ai,1 b1,n – 1 + ai,1 b1,n
ai+1,2 b2, j + ai+1,2 b2, j +1 ai+1,2 b2, n – 1 + ai+1,2 b2, n
.
.
.
.
.
.
.
.
.
+ . . .+
+ . . .+
.
.
.
.
.
.
an – 1,n – 1bn – 1, j + an – 1,n – 1bn – 1, j +1 an – 1,n – 1bn – 1, n – 1 + an – 1,n – 1bn
an,nbn, j + an,n bn, j +1 an,n bn,n – 1 + an,n bn,n
+ . . .+
+ . . .+
Section 1.2, (Pg. 42)
For Each Element of AB (single entry)](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-60-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Transpose Matrix
a11
a21
am1
.
.
.
a12
a22
am2
.
.
.
. . .
. . .
. . .
.
.
.
a1n
a2n
amn
.
.
.
A =
a11
a12
a1n
.
.
.
a21
a22
a2n
.
.
.
. . .
. . .
. . .
.
.
.
am1
am2
amn
.
.
.
AT =
1
2
0
2
1
0
0
0
1
C =
1
2
0
2
1
0
0
0
1
CT =
Symmetric Matrix: C = CT
If C = [ cij ] is a symmetric matrix, cij = cji for i ≠ j
C = [ cij ] is a symmetric matrix, Cmx n = CT
nx p for m = n = p](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-62-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
1
2
0
2
1
0
0
0
1
C =
1
2
0
2
1
0
0
0
1
CT =
If C = [ cij ] is a symmetric matrix, cij = cji , "i,j | i ≠ j
C = [ cij ] is a symmetric matrix, Cmx n = CT
nx p , "m,n, p | m = n = p
Symmetric Matrix
A Symmetric Matrix is a
Square Matrix that is
equal to it Transpose ( e.g. Cmx n = CT
mx n , "m,n | m = n)
](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-63-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Matrix Inverse
A matrix Anx n is Invertible or Non-Singular when
$ matrix Bnx n | AB = BA = In
In : Identity Matrix of Order n
Bnx n : The Multiplicative Inverse of A
A matrix that does not have an inverse is Non-Invertible or Singular
Non-square matrices do not have inverses
n For matrix products Amx n Bnx p where m ≠ n ≠ p,
AB ≠ BA as [ aij ≠ bij ] ??
](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-65-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Matrix Inverse
There are two methods for determining an inverse matrix A-1
of A ( if an inverse exists):
Solve Ax = In for X
Adjoin the Identity Matrix In (on RHS ) to A forming the doubly-
augmented matrix [ A In ] and perform EROs concluding in RREF to
produce an [ In A-1 ] solution
A test for determining whether an inverse matrix A-1 of A exists:
Demonstrate that either/or AB = In = BA
Section 2.3 (Pg. 64)](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-66-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Matrix Inverse – by Gauss-Jordan Elimination ( GJE )
An invertible coefficient Anx n matrix can be combined with its corresponding
xnx n unknown/variable matrix to form an Axnx n = In equation matrix
This equation matrix is composed itself of identical coefficient
column vectors
1 x11 + 4 x21 = 1
– 1 x21 + ( – 3 x21 ) = 0
1 x11 + 4 x21 = 0
– 1 x21 + ( – 3 x21 ) = 1
Ax = In Ax = In
Rather than solve the two column equation vectors separately,
they can be solved simultaneously by adjoining the identity
matrix to the shared coefficient matrix…
1
– 1
4
– 3
A
1
0
0
1
In
…then execute ERO’s to effect a GJ-Elimination of the
“ doubly augmented ” [ A I ] matrix the conclusion of
which will yield an [ I A-1 ] inverse matrix](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-71-320.jpg)
![© Art Traynor 2011
Mathematics
Algebra Of Matrices
Linear Algebra
Matrix Inverse – by Gauss-Jordan Elimination
An invertible coefficient Anx n matrix can be combined with its corresponding
xnx n unknown/variable matrix to form an Axnx n = In equation matrix
This equation matrix is composed itself of identical coefficient
column vectors
1 x11 + 4 x21 = 1
– 1 x21 + ( – 3 x21 ) = 0
1 x11 + 4 x21 = 0
– 1 x21 + ( – 3 x21 ) = 1
Ax = In Ax = In
The adjoined, “ doubly-augmented ” coefficient matrix , by means of ERO’s , is reduced by
GJ-Elimination to produce the [ I A-1 ] inverse matrix
1
– 1
4
– 3
A
1
0
0
1
In
– 3
1
– 4
1
A-1
1
0
0
1
In
Which is confirmed by verifying either of the following
n AA-1 = I
n AA-1 = A-1 A](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-72-320.jpg)
![© Art Traynor 2011
Mathematics
Linear Algebra
Definitions
EROs & Determinants (Properties)
Permutation:
[ A ] Pij [ B ]
det ( B ) = – det ( A )
|B | = – | A |
Multiplication by a Scalar:
[ A ] cRi [ B ]
det ( B ) = c det ( A )
|B | = c | A |
Addition to a Row Multiplied by a Scalar:
[ A ] Ri + cRj [ B ]
det ( B ) = det ( A )
|B | = | A |
There are three “effects” to a resultant
matrix which are unique to each of the
three EROs
Permutation
Scalar Multiplication
Row Addition](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-89-320.jpg)
![© Art Traynor 2011
Mathematics
Definition
Vectors
Vector ( Euclidean )
A geometric object (directed line segment)
describing a physical quantity and characterized by
Direction: depending on the coordinate system used to describe it; and
Magnitude: a scalar quantity (i.e. the “length” of the vector)
Aka: Geometric or Spatial Vector
originating at an initial point [ an ordered pair : ( 0, 0 ) ]
and concluding at a terminal point [ an ordered pair : ( a1 , a2 ) ]
Other mathematical objects
describing physical quantities and
coordinate system transforms
include: Pseudovectors and
Tensors
Not to be confused with elements of Vector Space (as in Linear Algebra)
Fixed-size, ordered collections
Aka: Inner Product Space
Also distinguished from statistical concept of a Random Vector
From the Latin Vehere (to carry)
or from Vectus…to carry some-
thing from the origin to the point
constituting the components of the vector 〈 a1 , a2 〉](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-142-320.jpg)
![© Art Traynor 2011
Mathematics
Vector Spaces
Vector Spaces – Classification
Polynomial Vector Spaces
P = set of all polynomials
Pn = set of all polynomials of degree ≤ n
Continuous Functions ( Calculus ) Vector Spaces
C ( – ∞ , ∞) = set of all continuous functions
defined on the real number line
C [ a, b ] = set of all continuous functions
defined on a closed interval [ a, b ]
Section 4.2, (Pg. 157)
Vector Space](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-150-320.jpg)
![© Art Traynor 2011
Mathematics
Vector Subspaces
W 1
Polynomial
functions
W 5
Functions
W 2
Differentiable
functions
W 3
Continuous
functions
W 4
Integrable
functions
W5 = Vector Space " f Defined on [ 0, 1 ]
W4 = Set " f Integrable on [ 0, 1 ]
W3 = Set " f Continuous on [ 0, 1 ]
W2 = Set " f Differentiable on [ 0, 1 ]
W2 = Set " Polynomials
Defined on [ 0, 1 ]
W1 W2 W3 W4 W5
W 1 – Every Polynomial function is Differentiable W1 W2
W 2 – Every Differentiable function is Continuous W2 W3
W 3 – Every Continuous function is Integrable W3 W4
W 4 – Every Integrable function
is a Function W4 W5
Function Space Section 4.3, (Pg. 164)
Vector Space](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-152-320.jpg)
![© Art Traynor 2011
Mathematics
Matrix Nullspace and Nullity
Nullspace
Section 4.6, (Pg. 194)
The solution set for the system forms a subspace of Rn designated as
the Nullspace of A and denoted as N ( A )
Vector Space
For a homogenous linear system Ax = 0 where A is an matrix of m-rows
and n-columns Amx n , x = [ xi , xi+1 ,…xn – 1 , xn ] T is a column
vector of unknowns 0 = [ 0 , 0 , … 0 , 0 ] T is the zero vector in Rm
N ( A ) = { x Rn | Ax = 0 }
The Dimension of the Nullspace of A is designated as its Nullity
dim ( N ( A ) )](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-175-320.jpg)
![© Art Traynor 2011
Mathematics
Solution Space Consistency
Solution Space
Section 4.6, (Pg. 198)
Vector Space
For a matrix A composed of m-rows and n-columns Amx n ,
of which Ax = b defines a particular solution xp to the nHLES
The nHLES is consistent if-and-only-if
The solution vector b is among the Column Space of A
The solution vector b = [ bi , bi+1 ,…bn – 1 , bn ] T
represents a Linear Combination of the columns of A
The solution vector b is among those populating the Subspace Rm
Spanned by the columns of A
There is an additional
implication that the respective
Ranks of the nHLES
coefficient and augmented
matrices are equivalent
Section 4.6, (Pg. 198)
I think what is meant by this
is that if b is adjoined to A
then reduced by GJE to an RREF
matrix B, so long as the column
representing b remains, the system is thus
demonstrated consistent](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-203-320.jpg)
![© Art Traynor 2011
Mathematics
Coordinate Representation Relative to a Basis
Basis
Section 4.7, (Pg. 202)
Vector Space
For an Ordered Basis set B = { vi , vi+1 ,…vn – 1 , vn },
within Vector Space V , there exists a vector x in V
that can be expressed as a Linear Combination, or sum of
scalar multiples of the constituent vectors of x such that:
xn = ci vi + ci+1 vi+1 +…+ cn –1 vn – 1 + cn vn
The coordinate matrix, or coordinate vector, of x relative to B is the
column matrix in Rn whose components are the coordinates of x
the scalars of which ci = [ ci , ci+1 ,…cn – 1 , cn ] are otherwise
referred to as the coordinates of x relative to the Basis B
c1
c2
cn
.
.
.
[ xn ]Bn
=
Chump Alert: Coordinate
matrix representation relative
to a Basis ( standard or
otherwise) is directly
analogous to Normalized ( ? )
Unit Vector representation](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-205-320.jpg)
![© Art Traynor 2011
Mathematics
y´
Vector Space
Basis
Coordinate Matrices and Bases Section 4.7 (Pg. 203),
Example 2
Example: Find the coordinate matrix of x relative to a non-standard basis
the ordered pair collection of vector scalar coefficients of the arbitrary
vector x is explicitly corresponded with the Basis vectors the
products of which form a resultant Linear Combination.
x
y
x´
O
O ´
1B 2B 3B 4B 5B
1B ⇌ ½B´
1B´ 2B´ 3B´
2B ⇌ 1B´
3B ⇌ 1½ B´
4B ⇌ 2B´
v2 = 2 u2
( 1 , 2 )
v1 = 1 u1
( 1 , 0 )
xn = c1 v1 + c2 v2
cn = ( c1 , c2 )
A scalar coefficient of
“one” is implicit
c1
c2
cn
.
.
.
ci
311
221
The ordered pair of
scalar vector
coefficients of x
Restated as a Column
Vector (scalar
coefficients of x)
B´ = { v1 , v2 }
The set of Basis
vectors in SNF
Juxtaposed as below,
it is clear that x is a
linear combination of
the Basis Vectors
B´ = { 1 v1 , 1 v2 }
ci ≠ 1 implies that the
arbitrary vector is a
Linear Combination of
the Basis Vector set
cn = [ x ]B´ =
An illuminating change of notation is here introduced [ x ]B´ whereby
1B´ 2B´ 3B´ 4B´ 5B´
5a](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-215-320.jpg)
![© Art Traynor 2011
Mathematics
y´
Vector Space
Basis
Coordinate Matrices and Bases Section 4.7 (Pg. 203),
Example 2
Example: Find the coordinate matrix of x relative to a non-standard basis
x
y
x´
O
O ´
1B 2B 3B 4B 5B
1B ⇌ ½B´
1B´ 2B´ 3B´
2B ⇌ 1B´
3B ⇌ 1½ B´
4B ⇌ 2B´
v2 = 2 u2
( 1 , 2 )
v1 = 1 u1
( 1 , 0 )
c1
c2
cn
.
.
.
ci
311
221
Stated as a Column
Vector (scalar
coefficients of x)
Juxtaposed as below,
it is clear that x is a
linear combination of
the Basis Vectors
cn = [ x ]B´ =
The novel notation for the set (ordered pair, or column vector form)
of vector scalar coefficients ci directs our attention to the specific
Basis by which the resulting coordinates are generated.
e.g.: x is expressed relative to Basis B
“ Relative to ”
“ Defined by ”
“ Within ”
xn = c1 v1 + c2 v2
cn = ( c1 , c2 )
B´ = { v1 , v2 }
B´ = { 1 v1 , 1 v2 }1B´ 2B´ 3B´ 4B´ 5B´
5b](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-216-320.jpg)
![© Art Traynor 2011
Mathematics
y´
Vector Space
Basis
Coordinate Matrices and Bases Section 4.7 (Pg. 203),
Example 2
Example: Find the coordinate matrix of x relative to a non-standard basis
①
x
y
x´
O
O ´
1B 2B 3B 4B 5B
1B ⇌ ½B´
1B´ 2B´ 3B´
2B ⇌ 1B´
3B ⇌ 1½ B´
4B ⇌ 2B´
v2 = 2 u2
( 1 , 2 )
v1 = 1 u1
( 1 , 0 )
c1
c2
cn
.
.
.
ci
311
221
Stated as a Column
Vector (scalar
coefficients of x)
Juxtaposed as below,
it is clear that x is a
linear combination of
the Basis Vectors
cn = [ x ]B´ =
Vector x therefore can be graphically represented by a Position
Vector demarcated in the B-Prime coordinate system [ i.e. three
increments of “ x ” from the alternative Basis ( aB ) origin and two
increments of “ y ” from the aB origin ].
“ Relative to ”
“ Defined by ”
“ Within ”
111 012
021 122
B
u1 u2
u2
u1
v2
v1
111 112
021 222
B ´
v1 v2
xi
yj
“Standard” Basis
Element
“Alternative” Basis
Element
1B
1B
1B´
2B ⇌ 1B´
xB = ( 3 , 2 )
1B´ 2B´ 3B´ 4B´ 5B´](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-217-320.jpg)
![© Art Traynor 2011
Mathematics
y´
Vector Space
Basis
Coordinate Matrices and Bases Section 4.7 (Pg. 203),
Example 2Example: Find the coordinate matrix of x relative to a non-standard basis
①
x
y
x´
O
O ´
1B 2B 3B 4B 5B
1B ⇌ ½B´
1B´ 2B´ 3B´
2B ⇌ 1B´
3B ⇌ 1½ B´
4B ⇌ 2B´
v2 = 2 u2
( 1 , 2 )
v1 = 1 u1
( 1 , 0 )
To restate the scalar set of the arbitrary vector
coefficients in terms of the standard basis ( sB ),
the aB vector components need simply be scaled
by the vector equation for x to yield the scalar
solution set in terms of the sB
xB = ( 3 , 2 )
xn = c1 v1 + c2 v2
B´ = { ( 1 , 0 ) , ( 1 , 2 ) }
B´ = { v1 , v2 }
xn = c1 ( 1 , 0 ) + c2 ( 1 , 2 )
cn = [ x ]B
= ( c1 , c2 )
cn = [ x ]B
= ( 3 , 2 )
xn = 3( 1 , 0 ) + 2 ( 1 , 2 )
x = ( 3·1 + 2·1 ) , ( 3·0 + 2·2 )
x = ( 3 + 2 ) , ( 0 + 4 )
x = ( 5 ) , ( 4 )
xn = 3 ( v1 ) + 2 ( v2 )
1B´ 2B´ 3B´ 4B´
5B´](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-218-320.jpg)
![© Art Traynor 2011
Mathematics
Vector Space
Basis
Coordinate Matrices and Bases Section 4.7 (Pg. 203),
Example 2Example: Find the coordinate matrix of x relative to a non-standard basis
① Note finally that, as B ′ ( B-Prime ) is the Identity
Matrix ( IM ), the B matrix is equivalent to the
Transition Matrix (from non-standard to standard
Basis) conventionally noted as P – 1
Section 4.7 (Pg. 208)
[ B′ In ]
Adjoin
[ In P – 1 ]
RREF
GJE
( EROs )
P – 1 = ( B′ ) – 1
′
Change of Basis
Non-Standard Standard
111 112
021 222
v1 v2
xi
yj
P – 1 = ( B′ ) = ′](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-219-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( aka Transformation )
Basis Transformation
Vector Space
Section 4.7 (Pg. 203),
Example 3
Example: Find the coordinate matrix of x relative to a non-standard basis
With this restatement, we next introduce the arbitrary vector
coordinates ( to which we are tasked to find the coordinate matrix
corresponding to the aB ) rendering it into a column vector form.
①
B´ = { u1 , u2 , u3 }
B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) }
1
0
1
0
– 1
2
– 2
– 3
– 5
aB Coefficient Matrix
B´ =
c1
c2
c3
u1 u2 u3
These given arbitrary
vector coordinates are
presumably introduced
relative to the Standard
Basis?
xn = ( 1 , 2 , – 1 )
c1
c2
cn
.
.
.
ci
cn = [ x ]B´ =
1
2
c1
c2
c3
xi
– 1“ Relative to ”
“ Defined by ”
“ Within ”
Keep in mind that the ‘given’ arbitrary vector is stated relative to the
standard Basis and thus represents a solution set by which the set of
scalar multiples applicable to the alternative Basis will be derived.
Alternative Basis
constituent vectors
expressed in Set
Notation Form SNF](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-221-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( aka Transformation )
Basis Transformation
Vector Space
Section 4.7 (Pg. 203),
Example 3Example: Find the coordinate matrix of x relative to a non-standard basis
We next introduce a column vector of scalars the solution to which
will represent the coordinate matrix of the arbitrary vector with
respect to the alternative Basis ( aB )
①
B´ = { u1 , u2 , u3 }
B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) }
1
0
1
0
– 1
2
– 2
– 3
– 5
aB Coefficient Matrix
u1 u2 u3
Alternative Basis
constituent vectors
expressed in Set
Notation Form SNF
These given arbitrary
vector coordinates are
presumably introduced
relative to the Standard
Basis?
1
2
– 1
c1
c2
c3
xi
xn = ( 1 , 2 , – 1 )
Keep in mind that the ‘given’
arbitrary vector is stated relative
to the standard Basis and thus
represents a solution set by
which the set of scalar multiples
associated with the alternative
Basis will be determined
c1
c2
ci
– c3
=
cn = [ x ]B
= ( c1 , c2 , c3 )](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-222-320.jpg)
![© Art Traynor 2011
Mathematics
Vector Space
Basis Transformation
Change of Basis ( aka Transformation )
Section 4.7 (Pg. 203),
Example 3Example: Find the coordinate matrix of x relative to a non-standard basis
No parameterization is necessary as the RREF form gives explicit
values for each unknown
1 0 – 0 5
0 – 1 – 0 – 8
– 0 0 – 1 – 2
1d1 + 0x2 + 2c3 = 5s
3x1 – 1d2 + 3c3 = – 8s
1c1 + 2c2 – 5d3 = – 2s
u1 u2 u3 b
10
5
– 2
d1
d2
d3
xi
– 1
d1
d2
dn
.
.
.
di
dn = [ x ]B´ =
“ Relative to ”
“ Defined by ”
“ Within ”
5
– 2
d1
d2
d3
xi
– 1
d1
d2
d3
x = d1u1 + d2u2 + d3u3 + xp
System Solution(s)
x = xh + xp
xh = d1u1 + d2u2 + d3u3
Ax = b
Ax = 0 ( HLES )
( nHLES )
RREF ( Basis Set ) for nHLES](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-233-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( aka Transformation )
Basis Transformation
Vector Space
Section 4.7 (Pg. 203),
Example 3
Example: Find the coordinate matrix of x relative to a non-standard basis
Continued…13
d1
d2
dn
.
.
.
di
dn = [ x ]B´ =
“ Relative to ”
“ Defined by ”
“ Within ”
5
– 2
d1
d2
d3
xi
– 1
xn = ci · ui + ci+1 · ui+1 +…+ cn –1 vn – 1 + cn vn
Arbitrary Vector, generally
xnn = c1 · u1 + c2 · u2 + c3 · v3
Arbitrary Vector, particular
xn = 51 ( 1 , 0, 1 ) + ( – 82 ) ( 0 , – 1, 2 ) + ( – 22 )( 2 , 3, – 5 )
Arbitrary Vector, as nHLES
linear combination relative
to aB](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-236-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( via Transition Matrix)
Basis Transformation
Vector Space
Section 4.7 (Pg. 204)
For an arbitrary vector x in vector space V described by coordinates relative to
a Standard Basis B , an ancillary description – in coordinate terms relative
to an Alternate Basis B′ ( B-Prime ) – can be determined by operation
of a Transition Matrix P ( the entries of which are populated by the
components of x ) .
The matrix product of P and a column vector [ x ]B′ of scalars
ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix
of x relative to the alternate basis B′ ( B-Prime ) yield the
column vector [ x ]B′ , the “ root ” basis.
The “ Change of Basis ” is thus the solution to the unknown column
vector [ x ]B′ of scalars ci = [ ci , ci+1 ,…cn – 1 , cn ] .
](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-237-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( via Transition Matrix)
Basis Transformation
Vector Space
Section 4.7 (Pg. 204)
The matrix product of P and a column vector [ x ]B′ of scalars
ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix
of x relative to the alternate basis B′ ( B-Prime ) yield the
column vector [ x ]B′ .
P = B´ = { u1 , u2 , u3 }
P = B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) }
xn = ci ui + ci+1 ui+1 +…+ cn –1 un – 1 + cn un
Note the convention wherein
blue font is assigned to those
elements of the B-Basis set,
whose component vectors are
assigned to “ u ”
Red font accordingly is
assigned to those elements of
the B-Prime ( B′ ) Basis set,
whose component vectors are
assigned to “ v ”
P = B´ =
Coefficient Matrix
u1 u2 u3
d1
d2
d3
xi = [ x ]B′
=
p11
Alternative Basis
constituent vectors
expressed in Set
Notation Form SNF
Note that Matrix P = B′ is
expressed in Transition
Matrix Form ( TMF )
wherein the columns are
populated by the individual
Basis vector components
with the rows collecting like
“ unknown ” terms.
xi = [ x ]B′
c1
c2
c3
Root BasisAlternate Basis
p21
p31
p12
p22
p32
p13
p23
p33](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-238-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( via Transition Matrix)
Basis Transformation
Vector Space
Section 4.7 (Pg. 204)
The matrix product of P and a column vector [ x ]B′ of scalars
ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix
of x relative to the alternate basis B′ ( B-Prime ) yield the
column vector [ x ]B′ .
P = B´ = { u1 , u2 , u3 }
P = B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) }
xn = ci ui + ci+1 ui+1 +…+ cn –1 un – 1 + cn un
P = B´ =
1
0
1
0
– 1
2
– 2
– 3
– 5
Coefficient Matrix
u1 u2 u3
d1
d2
d3
xi = [ x ]B′
=
xi = [ x ]B′
Root BasisAlternate Basis
– 1
– 2
– 1](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-239-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( via Transition Matrix)
Basis Transformation
Vector Space
Section 4.7 (Pg. 204)
The matrix product of P and a column vector [ x ]B′ of scalars
ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix
of x relative to the alternate basis B′ ( B-Prime ) yield the
column vector [ x ]B′ .
P [ x ]B′ = [ x ]B′
P [ x ]B′ = [ x ]B′
P2 P2
P [ x ]B′ = [ x ]B′
P2 P2
P [ x ]B′ = P – 1 [ x ]B′
Whereas the product of the
Transition Matrix and the
Alternate Basis ( aB ) yields
the Root Basis, it is
equivalent to state that the
product of the Transition
Matrix Inverse and the Root
Basis will yield the Alternate
Basis.Change of Basis
B′ B
Change of Basis
B B′
We recall that to find an
inverse matrix we adjoin the
Identity Matrix In (on RHS ) to
P forming matrix [ A In ]
and perform EROs by GJE to
arrive at an RREF which will
then result in matrix [ In A –1 ]
with the inverse occupying the
RHS of the resultant matrix](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-240-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( via Transition Matrix)
Basis Transformation
Vector Space
Section 4.7 (Pg. 204)
P = B´ =
Transition Matrix
u1 u2 u3
d1
d2
d3
xi = [ x ]B′
=
p11
xi = [ x ]B′
c1
c2
c3
Root BasisAlternate Basis
p21
p31
p12
p22
p32
p13
p23
p33
P – 1 = B´
Transition Matrix Inverse
d1
d2
d3
xi = [ x ]B′
=
p – 1
xi = [ x ]B′
c1
c2
c3
Root BasisAlternate Basis
p12
p22
p32
p13
p23
p33
v1 v2 v3
11
p – 1
21
p – 1
31](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-241-320.jpg)
![© Art Traynor 2011
Mathematics
Change of Basis ( via Transition Matrix)
Basis Transformation
Vector Space
Section 4.7 (Pg. 206)
[ B′ B ]
The Transition Matrix ( TM ) from a Root Basis ( RB = B ) to
an Alternate Basis ( aB = B′ , B-Prime ) is found, as is the case for
similar matrix inverses, by adjoining the Root Basis matrix with the
aB matrix ( on the LHS ) and applying EROs via GJE to arrive at
a RREF reduced matrix, the result of which will form an adjoined
matrix composed of the Identity Matrix ( IM – on the LHS ) and the
P – 1 Transition Matrix ( from B to B′ – on the RHS ).
Adjoin
[ In P – 1 ]
RREF
GJE
( EROs ) ](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-242-320.jpg)
![© Art Traynor 2011
Mathematics
Transition Matrix
Basis Transformation
Vector Space
Section 4.7 (Pg. 208)
When the Root Basis ( RB = B ) is equivalent to the Identity Matrix
( IM ) the process of finding a basis-changing Transition Matrix
( TM ) is simplified to a symmetric operation whereby the Root-
Identity Matrix is adjoined ( on the RHS ) to an Alternate Basis
Matrix ( ABM = B′ , B-Prime, on the LHS ) and applying EROs
via GJE to arrive at a RREF reduced matrix, the result of which will
form an adjoined matrix composed of the Root-Identity Matrix ( RIM
– on the LHS ) and the P – 1 Transition Matrix ( from B to B′ –
on the RHS ).
[ B′ In ]
Adjoin
[ In P – 1 ]
RREF
GJE
( EROs )
P – 1 = ( B′ ) – 1 ′ Change of Basis
Standard Non-Standard](https://image.slidesharecdn.com/2d826fdb-8c8e-4ed6-a046-0eb4c5ad7212-160620064415/85/LinearAlgebra_160423_01-243-320.jpg)
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- 10. © Art Traynor2011 Mathematics Arity Arity ( Cardinality of Expression Variables ) Expression A relation can not be defined for Expressions of Arity less than two: n < 2 Nullary Unary n = 0 n = 1 Binary n = 2 Ternary n = 3 1-ary 2-ary 3-ary Quaternary n = 4 4-ary Quinary n = 5 5-ary Senary n = 6 6-ary Septenary n = 7 7-ary Octary n = 8 8-ary Nonary n = 9 9-ary n-ary VariablesConstants Operands Expression Polynomial 0-ary
- 11. © Art Traynor2011 Mathematics Operand Parity – Property of Operands Parity n is even if $ k | n = 2k n is odd if $ k | n = 2k+1 Even Even Integer Parity Same Parity Even Odd Opposite Parity
- 12. © Art Traynor2011 Mathematics Polynomial Expression A well-formed symbolic representation of operands, of discrete arity, upon which one or more operations can structure a Relation Expression Polynomial Expression A Mathematical Expression , the Terms ( Operands ) of which are a compound composition of : Polynomial Constants – referred to as Coefficients Variables – also referred to as Unknowns And structured by the Polynomial Structure Criteria ( PSC ) arithmetic Laws of Composition ( LOC’s ) including : Addition / Subtraction Multiplication / Non-Negative Exponentiation LOC ( Pn ) = { + , – , x bn ∀ n ≥ 0 } Wiki: “ Polynomial ” An excluded equation by Polynomial Structure Criteria ( PSC ) Σ an xi n i = 0 P( x ) = an xn + an – 1 xn – 1 +…+ ak+1 xk+1 + ak xk +…+ a1 x1 + a0 x0 Variable Coefficient Polynomial Term From the Greek Poly meaning many, and the Latin Nomen for name
- 13. © Art Traynor2011 Mathematics Degree Expression Polynomial Degree of a Polynomial Polynomial Wiki: “ Degree of a Polynomial ” The Degree of a Polynomial Expression ( PE ) is supplied by that of its Terms ( Operands ) featuring the greatest Exponentiation For a multivariate term PE , the Degree of the PE is supplied by that Term featuring the greatest summation of Variable exponents P = Variable Cardinality & Variable Product Exponent Summation & Term Cardinality Arity Latin “ Distributive ” Number suffix of “ – ary ” Degree Latin “ Ordinal ” Number suffix of “ – ic ” Latin “ Distributive ” Number suffix of “ – nomial ” 0 = 1 = 2 = 3 = Nullary Unary Binary Tenary Constant Linear Quadratic Cubic Monomial Binomial Trinomial An Expression composed of Constants ( Coefficients ) and Variables ( Unknowns) with an LOC of Addition, Subtraction, Multiplication and Non- Negative Exponentiation
- 14. © Art Traynor2011 Mathematics Degree Polynomial Degree of a Polynomial Nullary Unary p = 0 p = 1 Linear Binaryp = 2 Quadratic Ternaryp = 3 Cubic 1-ary 2-ary 3-ary Quaternaryp = 4 Quartic4-ary Quinaryp = 5 5-ary Senaryp = 6 6-ary Septenaryp = 7 7-ary Octaryp = 8 8-ary Nonaryp = 9 9-ary “ n ”-ary Arity Degree Monomial Binomial Trinomial Quadranomial Terms Constant Quintic P Wiki: “ Degree of a Polynomial ” Septic Octic Nonic Decic Sextic aka: Heptic aka: Hexic
- 15. © Art Traynor2011 Mathematics Degree Expression Polynomial Degree of a Polynomial Polynomial Wiki: “ Degree of a Polynomial ” An Expression composed of Constants ( Coefficients ) and Variables ( Unknowns) with an LOC of Addition, Subtraction, Multiplication and Non- Negative Exponentiation The Degree of a Polynomial Expression ( PE ) is supplied by that of its Terms ( Operands ) featuring the greatest Exponentiation For a PE with multivariate term(s) , the Degree of the PE is supplied by that Term featuring the greatest summation of individual Variable exponents P( x ) = ai xi 0 Nullary Constant Monomial P( x ) = ai xi 1 Unary Linear Monomial P( x ) = ai xi 2 Unary Quadratic Monomial ai xi 1 yi 1P( x , y ) = Binary Quadratic Monomial Univariate Bivariate
- 16. © Art Traynor2011 Mathematics Degree Expression Polynomial Degree of a Polynomial Polynomial Wiki: “ Degree of a Polynomial ” The Degree of a Polynomial Expression ( PE ) is supplied by that of its Terms ( Operands ) featuring the greatest Exponentiation For a multivariate term PE , the Degree of the PE is supplied by that Term featuring the greatest summation of Variable exponents P( x ) = ai xi 0 Nullary Constant Monomial P( x ) = ai xi 1 Unary Linear Monomial P( x ) = ai xi 2 Unary Quadratic Monomial ai xi 1 yi 1P( x , y ) = Binary Quadratic Monomial ai xi 1 yi 1zi 1P( x , y , z ) = Ternary Cubic Monomial Univariate Bivariate Trivariate Multivariate
- 17. © Art Traynor2011 Mathematics Quadratic Expression Polynomial Quadratic Polynomial Polynomial Wiki: “ Degree of a Polynomial ” A Unary or greater Polynomial composed of at least one Term and : Degree precisely equal to two Quadratic ai xi n ∀ n = 2 ai xi n yj m ∀ n , m n + m = 2|: Etymology From the Latin “ quadrātum ” or “ square ” referring specifically to the four sides of the geometric figure Wiki: “ Quadratic Function ” Arity ≥ 1 ai xi n ± ai + 1 xi + 1 n ∀ n = 2 Unary Quadratic Monomial Binary Quadratic Monomial Unary Quadratic Binomial ai xi n yj m ± ai + 1 xi + 1 n ∀ n + m = 2 Binary Quadratic Binomial
- 18. © Art Traynor2011 Mathematics Equation Equation Expression An Equation is a statement or Proposition ( aka Formula ) purporting to express an equivalency relation between two Expressions : Expression Proposition A declarative expression asserting a fact whose truth value can be ascertained Equation A symbolic formula, in the form of a proposition, expressing an equality relationship Formula A concise symbolic expression positing a relationship between quantities VariablesConstants Operands Symbols Operations The Equation is composed of Operand terms and one or more discrete Transformations ( Operations ) which can render the statement true ( i.e. a Solution ) Polynomial
- 19. © Art Traynor2011 Mathematics Equation Solution Solution and Solution Sets Free Variable: A symbol within an expression specifying where a substitution may be made Contrasted with a Bound Variable which can only assume a specific value or range of values Solution: A value when substituted for a free variable which renders an equation true Analogous to independent & dependent variables Unique Solution: only one solution can render the equation true (quantified by $! ) General Solution: constants are undetermined General Solution: constants are value-specified (bound?) Unique Solution Particular Solution General Solution Solution Set n A family (set) of all solutions – can be represented by a parameter (i.e. parametric representation) Equivalent Equations: Two (or more) systems of equations sharing the same solution set Section 1.1, (Pg. 3) Section 1.1, (Pg. 3) Section 1.1, (Pg. 6) Any of which could include a Trivial Solution Section 1.2, (Pg. 21)
- 20. © Art Traynor2011 Mathematics Equation Solution Solution and Solution Sets Solution: A value when substituted for a free variable which renders an equation true Unique Solution: only one solution can render the equation true (quantified by $! ) General Solution: constants are undetermined General Solution: constants are value-specified (bound?) Solution Set n For some function f with parameter c such that f(xi , xi+1 ,…xn – 1 , xn ) = c the family (set) of all solutions is defined to include all members of the inverse image set such that f(x) = c f -1(c) = x f -1(c) = {(ai , ai+1 ,…an-1 , an ) Ti· Ti+1 ·…· Tn-1· Tn | f(ai , ai+1 ,…an-1 , an ) = c } where Ti· Ti+1 ·…· Tn-1· Tn is the domain of the function f o f -1(c) = { }, or empty set ( no solution exists ) o f -1(c) = 1, exactly one solution exists ( Unique Solution, Singleton) o f -1(c) = { cn } , a finite set of solutions exist o f -1(c) = {∞ } , an infinite set of solutions exists Inconsistent Consistent Section 1.1, (Pg. 5)
- 21. © Art Traynor2011 Mathematics Linear Equation Linear Equation Equation An Equation consisting of: Operands that are either Any Variables are restricted to the First Order n = 1 Linear Equation An equation in which each term is either a constant or the product of a constant and (a) variable[s] of the first order Expression Proposition Equation Formula n Constant(s) or n A product of Constant(s) and one or more Variable(s) The Linear character of the Equation derives from the geometry of its graph which is a line in the R2 plane As a Relation the Arity of a Linear Equation must be at least two, or n ≥ 2 , or a Binomial or greater Polynomial Polynomial
- 22. © Art Traynor2011 Mathematics Equation Linear Equation Linear Equation An equation in which each term is either a constant or the product of a constant and (a) variable[s] of the first order Term ai represents a Coefficient b = Σi= 1 n ai xi = ai xi + ai+1 xi+1…+ an – 1 xn – 1 + an xn Equation of a Line in n-variables A linear equation in “ n ” variables, xi + xi+1 …+ xn-1 + xn has the form: n Coefficients are distributed over a defined field ( e.g. N , Z , Q , R , C ) Term xi represents a Variable ( e.g. x, y, z ) n Term a1 is defined as the Leading Coefficient n Term x1 is defined as the Leading Variable Section 1.1, (Pg. 2) Section 1.1, (Pg. 2) Section 1.1, (Pg. 2) Section 1.1, (Pg. 2) Coefficient = a multiplicative factor (scalar) of fixed value (constant) Section 1.1, (Pg. 2)
- 23. © Art Traynor2011 Mathematics Linear Equation Equation Standard Form ( Polynomial ) Ax + By = C Ax1 + By1 = C For the equation to describe a line ( no curvature ) the variable indices must equal one ai xi + ai+1 xi+1 …+ an – 1 xn –1 + an xn = b ai xi 1 + ai+1 x 1 …+ an – 1 x 1 + a1 x 1 = bi+1 n – 1 n n ℝ 2 : a1 x + a2 y = b ℝ 3 : a1 x + a2 y + a3 z = b Blitzer, Section 3.2, (Pg. 226) Section 1.1, (Pg. 2) Test for Linearity A Linear Equation can be expressed in Standard Form As a species of Polynomial , a Linear Equation can be expressed in Standard Form Every Variable term must be of precise order n = 1 Linear Equation An equation in which each term is either a constant or the product of a constant and (a) variable[s] of the first order Expression Proposition Equation Formula Polynomial
- 24. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Solution Consistency Solution: A value when substituted for a free variable which renders an equation true Unique Solution Particular Solution General Solution Solution Set n A family (set) of all solutions – can be represented by a parameter No Solution - Inconsistent 1 0 0 2 1 0 – 1 0 0 4 3 – 2 Represents “ 0 = – 2 ” , a contradiction, and thus no solution { } to the LE system for which the augmented matrix stands 1x1 + 0x2 – 3x3 = – 1 System 0x1 + 1x2 – 1x3 = 0 x1 – 3x3 = – 1 System x2 = x3 Section 1.1, (Pg. 8)
- 25. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Solution Consistency Solution: A value when substituted for a free variable which renders an equation true Solution Set - Consistent n A family (set) of all solutions – can be represented by a parameter 1x1 + 0x2 – 3x3 = – 1 System 0x1 + 1x2 – 1x3 = 0 x1 = 3x3 – 1 System x2 = x3 Note that x3 can be parameterized ( as a composite function f ○ g → ( f ○ g )( x ) = f ( g ( x )) with y = f ( u ) and u = g ( x3 ) ) x2 = x3 = u Tautology/Identity* x1 = 3u – 1 The solution set for “ f(u) ” can thus be indexed by/over Z+ representing a countably infinte solution set * Section 1.1, (Pg. 3)
- 26. © Art Traynor2011 Mathematics ‘ f(x) ’ ‘– f ’ ‘ x ’ ‘ f(x) ’ ‘ +f -1 ’ ‘ x ’ Linear Algebra Solution Linear Equation – Solution Set Solution: A value when substituted for a free variable which renders an equation true Solution Set n For some function f with parameter c such that f ( xi , xi+1 ,…xn – 1 , xn ) = c the family ( set ) of all solutions is defined to include all members of the inverse image set such that f ( x ) = c f -1( c ) = x
- 27. © Art Traynor2011 Mathematics Linear Algebra Solution aij xi + aij+1 xi+1 + . . . + ain – 1 xn – 1 + ain xn = bi ai+1j xi + ai+1j+1 xi+1 + . . . + ai+1n – 1 xn – 1 + ai+1n xn = bi+1 am – 1j xi + am – 1j+1 xi+1 + . . . + am – 1n – 1 xn – 1 + am – 1n xn = bm – 1 . . . . . . . . . . . . . . . amj xi + amj+1 xi+1 + . . . + amn xn – 1 + amn xn = bm Linear Equation – System A system of m linear equations in n variables is a set of m equations , each of which is linear in the same n variables Linear Equation System Solution Set The set S = { si , si+1 ,…sn-1 , sn } which renders each of the equations in the system true Section 1.1, (Pg. 4) Section 1.1, (Pg. 4)
- 28. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Back Substitution x – 2y = 5 y = – 2 System x – 2 y = 5 x – 2 (– 2 ) = 5 x + 4 = 5 x = 5 – 4 x = 1 Solution Set – Singleton, Unique Solution, ( exactly one solution ) S = { 1, – 2 } Section 1.1, (Pg. 6)
- 29. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation Equivalence Equivalent Linear Equations: Two (or more) systems of linear equations sharing the same solution set Gaussian Elimination Operations Producing Linear Equation Equivalent Systems Permutation/Interchange – of two equations Multiply – an equation by a non-zero constant Add – a multiple of an equation to another equation Section 1.1, (Pg. 6) Section 1.1, (Pg. 7) Otherwise known as Elementary Row Operations Section 1.2, (Pg. 14) n ERO’s should always proceed with an Augend/Multiplicand of lesser rank and Summand/Multiplier of greater rank ( Aij < Amn ) yielding a Sum/Product substituted for the second Operand n Multiplication by a scalar ( non-zero constant ) need not affect any change in rank for the resultant row
- 30. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Row Echelon Form (REF) 1 0 0 2 1 0 – 1 0 1 4 3 – 2 A matrix in Row-Echelon Form ( REF ) has three distinguishing characteristics Any rows consisting entirely of zeros is positioned at the bottom of the matrix For each row that does not consist entirely of zeros, the first non-zero entry is a “ 1 ” ( called the Leading One, aka Pivot ) 1 0 0 2 1 0 – 1 0 1 4 3 – 2 Section 1.1, (Pg. 6) Section 1.2, (Pg. 15) Section 1.2, (Pg. 15) 0 0 0 0 0 0 0 0 Section 1.2, (Pg. 16) Every matrix is row equivalent to a matrix in Row-Echelon Form ( REF )
- 31. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Row Echelon Form (REF) A matrix in Row-Echelon Form ( REF ) has three distinguishing characteristics For two successive (non-zero) rows, the leading one in the higher row is farther to the left than the leading one ( Pivot ) in the lower row 1 0 0 2 1 0 – 1 0 1 4 3 – 2 Section 1.1, (Pg. 6) Section 1.2, (Pg. 15) 0 0 0 0 A matrix in Reduced Row-Echelon Form ( RREF ) has one additional characteristic Every column that has a leading one ( Pivot ) has zeros in every position above and below its leading one ( Pivot ) 1 0 0 0 1 0 0 0 1 4 3 – 2 Section 1.2, (Pg. 15) 0 0 0 0 Every matrix is row equivalent to a matrix in Row-Echelon Form ( REF )
- 32. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Reduced Row Echelon Form (RREF) 1 – 1 2 – 2 3 – 5 3 0 5 9 – 4 17 x – 2y + 3z = 9 – x + 3y = – 4 2x – 5y + 5z = 17 System Augmented Matrix R2 : R2 + R1 R2´ – 1 3 0 – 4 + R1 : 1 – 2 3 9 = R2´ : 0 1 3 5 1 0 2 – 2 1 – 5 3 3 5 9 5 17 R3 – 2R1 R3´
- 33. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Reduced Row Echelon Form (RREF) 1 2 – 2 – 5 3 5 9 17 Augmented Matrix R3 : – 2R1 : – 2 4 – 6 – 18 = R3´ : 0 – 1 – 1 – 1 1 0 0 – 2 1 – 1 3 3 – 1 9 5 – 1 R3 + R2 R3´´ 0 1 3 5 R3 – 2R1 R3´ 2 – 5 5 17
- 34. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Reduced Row Echelon Form (RREF) Augmented Matrix R3 : + R2 : = R3´´ : 0 0 2 4 1 0 – 2 1 3 3 9 5 R3 + R2 R3´´ 1 – 2 3 9 0 1 3 5 0 – 1 – 1 – 1 0 – 1 – 1 – 1 0 1 3 5 0 0 2 4 R3 R3´´ ´1 2
- 35. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Reduced Row Echelon Form (RREF) Augmented Matrix R3 : 1 – 2 3 9 0 1 3 5 R3 R3´´ ´1 2 1 – 2 3 9 0 1 3 5 0 0 2 4 R3 : 1 2 R3´´ ´ : 0 0 2 4 0 0 1 2 0 0 1 2 0 0 1 2 Matrix is in Row-Echelon Form ( REF ) Proceeding to Reduced Row-Echelon Form ( RREF ) R1 + 2R2 R1´
- 36. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Reduced Row Echelon Form (RREF) Augmented Matrix R1 : + 2R2 : R1´ : 0 2 6 10 1 0 9 19 1 0 9 19 0 1 3 5 0 0 1 2 1 – 2 3 9 0 1 3 5 0 0 1 2 R1 + 2R2 R1´ 1 – 2 3 9 R2 – 3R3 R2´´
- 37. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Reduced Row Echelon Form (RREF) Augmented Matrix R2 : – 3R3 : R2´´ : 0 0 – 3 – 6 0 1 0 – 1 1 0 9 19 0 1 0 – 1 0 0 1 2 R1 + 9R3 R1´´ 1 0 9 19 0 1 3 5 0 0 1 2 R2 – 3R3 R2´´ 0 1 3 5
- 38. © Art Traynor2011 Mathematics Linear Algebra Solution Linear Equation – Reduced Row Echelon Form (RREF) Augmented Matrix R1 : – 9R3 : = R1´´ : 0 0 – 9 – 18 1 0 0 1 0 1 0 – 1 0 0 1 2 R1 – 9R3 R1´´ 1 0 9 19 0 1 0 – 1 0 0 1 2 1 0 0 1 1 0 9 19 Matrix is in Reduced Row-Echelon Form ( RREF )
- 39. © Art Traynor2011 Mathematics Matrices Matrix For positive integers m and n, an m x n (“m by n”) matrix is a rectangular array populated by entries aij , located at the i-th row and the j-th column: Linear Algebra M = N: the matrix is a square of order n The a11 , a22 , a33 , amn , sequence of entries is the main diagonal ( ↘ ) of the matrix M = # of Rows i = Row Number Index N = # of Columns j = Column Number Index Row 1 a11 Row 2 Row 3 Row m . . . a21 a31 am1 . . . a12 a22 a32 am2 . . . a13 a23 a33 am3 . . . . . . . . . . . . . . . . . . a1n a2n a3n amn . . . mi mi+1 mi+2 mm nj nj+1 nj+2 nn C1 C2 C3 . . . C4 Section 1.2, (Pg. 13) Section 1.2, (Pg. 13) Section 1.2, (Pg. 13)
- 40. © Art Traynor2011 Mathematics Matrices Matrix For positive integers m and n, an m x n (“m by n”) matrix is a rectangular array Row 1 populated by entries aij , located at the i-th row and the j-th column: Linear Algebra a11 Row 2 Row 3 Row m . . . a21 a31 am1 . . . a12 a22 a32 am2 . . . a13 a23 a33 am3 . . . . . . . . . . . . . . . . . . a1n a2n a3n amn . . . mi mi+1 mi+2 mm nj nj+1 nj+2 nn M = # of Rows i = Row Number Index N = # of Columns j = Column Number Index C1 C2 C3 . . . Cn Section 1.2, (Pg. 13) “ i ” is the row subscript “ j ” is the column subscript
- 41. © Art Traynor2011 Mathematics Linear Algebra Definitions Diagonal Matrix A matrix Anx n is said to be diagonal when all of the entries outside the main diagonal ( ↘ ) are zero Section 2.1, (Pg. 50) The matrix Dnx n is diagonal if : dij = 0 if i ≠ j "ij { dii , di+1,i+1 ,…, dn–1,n–1 , dnn } Any square diagonal matrix is also a Symmetric Matrix A diagonal matrix is also both Upper-Triangular and Lower-Triangular The Identity Matrix In is a diagonal matrix Any square Zero Matrix is a diagonal matrix d11 d21 d31 dm1 . . . d12 d22 d32 dm2 . . . d13 d23 d33 dm3 . . . . . . . . . . . . . . . . . . d1n d2n d3n dmn . . . Dnx n =
- 42. © Art Traynor2011 Mathematics Linear Algebra Definitions Tr ( A ) = Σi = 1 n aii = a11 + a22 +…+ an –1n – 1 + ann Matrix Trace The trace of a matrix Anx n is the sum of the main diagonal entries Section 2.1, (Pg. 50) a11 a21 a31 am1 . . . a12 a22 a32 am2 . . . a13 a23 a33 am3 . . . . . . . . . . . . . . . . . . a1n a2n a3n amn . . . A
- 43. © Art Traynor2011 Mathematics Matrices Linear Algebra 1 – 1 2 – 4 3 0 3 – 1 – 4 5 – 3 6 x – 4y + 3z = 5 – x + 3y – z = – 3 2x – 4z = – 6 System Augmented Matrix 1 – 1 2 – 4 3 0 3 – 1 – 4 Coefficient Matrix M = # of Rows i = Row Number Index N = # of Columns j = Column Number Index Augmented Matrix A matrix representing a system of linear equations including both the coefficient and constant terms Coefficient = a multiplicative factor (scalar) of fixed value (constant) Section 1.2, (Pg. 13) Coefficient Matrix A augmented matrix excluding any constant terms and populated only by the variable coefficients Section 1.2, (Pg. 13)
- 44. © Art Traynor2011 Mathematics Linear Algebra Solution Gaussian Elimination With Back-Substitution Express the system of linear equations as an Augmented Matrix Interchange – of two equations Multiply – an equation by a non-zero constant Add – a multiple of an equation to another equation Section 1.2, (Pg. 16) Every matrix is row equivalent to a matrix in Row-Echelon Form ( REF ) Section 1.2, (Pg. 16) Apply ERO’s to restate the matrix in Row Echelon Form (REF) Section 1.2, (Pg. 13) Section 1.2, (Pg. 14) Section 1.1, (Pg. 6) Section 1.2, (Pg. 15) Use Back Substitution to solve for unknown variables Section 1.1, (Pg. 6) Order Matters! Operate from left-to-right Multiply – an equation by a non-zero constant
- 45. © Art Traynor2011 Mathematics Linear Algebra Solution Gauss-Jordan Elimination Follow steps 1 & 2 of Gaussian Elimination Section 1.2, (Pg. 19) Every matrix is row equivalent to a matrix in Row-Echelon Form ( REF ) Section 1.2, (Pg. 16) Apply ERO’s to restate the matrix in Row Echelon Form (REF) Section 1.2, (Pg. 13) Section 1.2, (Pg. 14) Section 1.1, (Pg. 6) Section 1.2, (Pg. 15) Use Back Substitution to solve for unknown variables Section 1.1, (Pg. 6) Order Matters! Operate from left-to-right Multiply – an equation by a non-zero constant Express the system of linear equations as an Augmented Matrix n Interchange – of two equations n Multiply – an equation by a non-zero constant n Add – a multiple of an equation to another equation Keep Going! Continue to apply ERO’s until matrix assumes Reduced Row Echelon Form ( RREF ) Section 1.2, (Pg. 15)
- 46. © Art Traynor2011 Mathematics Linear Algebra Homogeneity Homogenous Systems of Linear Equations A linear equation system in which each of the constant terms is zero Section 1.2, (Pg. 21) aij xi + aij+1 xi+1 + . . . + ain – 1 xn – 1 + ain xn = 0 ai+1j xi + ai+1j+1 xi+1 + . . . + ai+1n – 1 xn – 1 + ai+1n xn = 0 am – 1j xi + am – 1j+1 xi+1 + . . . + am – 1n – 1 xn – 1 + am – 1n xn = 0 . . . . . . . . . . . . . . . amj xi + amj+1 xi+1 + . . . + amn xn – 1 + amn xn = 0 A homogenous LE system Must have At Least One Solution Section 1.2, (Pg. 21) Every homogenous LE system is Consistent # Equations < # Variables Infinitely Many Solutions
- 47. © Art Traynor2011 Mathematics Matrix Representation Linear Algebra Matrix Representation Methods Uppercase Letter Designation Section 1.2, (Pg. 40) A , B , C Bracket-Enclosed Representative Element [ aij ] , [ bij ] , [ cij ] a11 a21 a31 am1 . . . a12 a22 a32 am2 . . . a13 a23 a33 am3 . . . . . . . . . . . . . . . . . . a1n a2n a3n amn . . . Rectangular Array Brackets denote a Matrix ( i.e. not a specific element/real number)
- 48. © Art Traynor2011 Mathematics Matrix Equality Linear Algebra Matrix Equality Section 1.2, (Pg. 40) A = [ aij ] B = [ bij ] are equal when Amxn = Bmxn aij = bij 1 ≤ i ≤m 1 ≤ j ≤n a1 a2 a3 . . . ana = C1 C2 C3 . . . Cn Row Matrix / Row Vector A 1 x n (“ 1 by n ”) matrix is a single row Column Matrix / Column Vector b1 b2 b3 bm . . . C1 An m x 1 (“ m by 1 ”) matrix is a single column
- 49. © Art Traynor2011 Mathematics Matrix Operations Linear Algebra Matrix Summation Section 1.2, (Pg. 41) A = [ aij ] B = [ bij ] is given by + A + B = [ aij + bij ] – 1 0 2 1 1 – 1 3 2 + = ( – 1 + 1 ) = ( 0 + [ – 1] ) ( 2 + 3 ) ( 1 + 2 ) 0 – 1 5 3 Scalar Multiplication 1 – 3 2 2 0 1 4 – 1 2 A = 3A = 3 ( 1 ) 3 ( 2 ) 3 ( 4 ) 3 ( – 3 ) 3 ( 0 ) 3 ( – 1 ) 3 ( 2 ) 3 ( 1 ) 3 ( 2 ) 3 – 9 6 6 0 3 12 – 3 6 3A =
- 50. © Art Traynor2011 Mathematics Matrix Operations Linear Algebra Section 1.2, (Pg. 42)Matrix Multiplication – 1 4 5 3 – 2 0 A = A = [ aij ] Amx n B = [ bij ] Bnx p then AB = [ cij ] = Σk = 1 n aik bkj = ai 1 b1 j + ai2 b2j +…+ ain –1 bn-1j + ain bnj The entries of Row “ Aik” ( the i-th row ) are multiplied by the entries of “ Bkj” ( the j-th column ) and sequentially summed through Row “ Ain” and Column “ Bnj” to form the entry at [ cij ] – 3 – 4 2 1 B = c11 c12 C = c21 c22 c31 c32 a11b11 + a12b21 a11b12 + a12b22 = a21b11 + a22b21 a21b12 + a22b22 a31b11 + a32b21 a31b12 + a32b22 Product Summation Operand Count For Each Element of AB (single entry) Product Summation (Column-Row) Index For the product of two matrices to be defined, the column count of the multiplicand matrix must equal the row count of the multiplier matrix ( i.e. Ac = Br ) ABmx p
- 51. © Art Traynor2011 Mathematics Systems Of Linear Equations Linear Algebra Linear Equation System a11 x1 + a12 x2 + a13 x3 = b1 a21 x1 + a22 x2 + a23 x3 = b2 a31 x1 + a32 x2 + a33 x3 = b3 Matrix-Vector Notation a11 a13 Ax = b a21 a23 a31 a33 = a12 a22 a32 A x1 x2 x3 x b1 b2 b3 b
- 52. © Art Traynor2011 Mathematics Systems Of Linear Equations Linear Algebra Partitioned Matrix Form (PMF) Ax = b = A x b a11 a21 am1 . . . a12 a22 am2 . . . . . . . . . . . . . . . a1n a2n amn . . . x1 x2 xn . . . Ax = b = ai1 b a11 a21 am1 . . . x1 ai2 a12 a22 am2 . . . + x2 + . . . + xn ain a1n a2n amn . . . Ax = b = Ax b a11 x1 a21 x1 am1 x1 . . . + a12 x2 + + a22 x2 + + am2 x2 + . . . . . . . . . . . . . . . + a1n xn + a2n xn + amn xn . . . Ax = x1 a1 + x2 a2 + . . . + xn an = b
- 53. © Art Traynor2011 Mathematics Section 2.1 Review Linear Algebra Section 2.1 Review Introduce Three Basic Martrix Operations Matrix Addition Scalar Multiplication Matrix Multiplication
- 54. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Matrices – Addition & Scalar Multiplication Commutative (Addition) Associative (Addition)A +( B + C ) = ( A +B ) + C Changes Order of Operations as per “PEM-DAS”, Parentheses are the principal or first operation A +B = B +A Re-Orders Terms Does Not Change Order of Operations – PEM-DAS Associative (Multiplication)( cd ) A = c ( dA ) Distributive ( Scalar Over Matrix Addition )c ( A + B ) = c A + cB Distributive ( Scalar Addition Over Matrix Addition ) ( c + d ) A = c A + dA
- 55. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Proofs Let A = [ aij ], B = [ bij ] Introduce/Define/Declare The Constituents to be Proven This statement declares A & B to be Matrices Specifies the row & column count index variables
- 56. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Matrices – Identities & Zero Matrices Multiplicative Identity Multiplicative Zero Identity 1A = A Additive IdentityA + 0mx n = A Additive InverseA + ( – A ) = 0mx n c A = 0mx n if c = 0 or A = 0mx n
- 57. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Matrices – Matrix Multiplication Distributive ( LHS ) A( BC ) = ( AB ) C Order of Terms is preserved Affects Order of Operations Sequence – PEM-DAS Distributive ( RHS ) Associative ( Scalar Over Matrix Multiplication ) A( B + C ) = AB + AC ( A + B ) C = AC + BC c ( AB ) = ( c A )B = A ( c B ) Associative (Multiplication) Order of Terms is preserved Order of Terms is preserved Order of Terms is preserved Order of Terms is preserved AC = BC CA = CB ( C is invertible ) Right Cancellation Property A = B if then Left Cancellation Property
- 58. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Matrices – Proofs A( BC ) = ( AB ) C Order of Terms is preserved Affects Order of Operations Sequence – PEM-DAS Associative (Multiplication) Σk = 1 n T = [ tij ] = ( aik bkj )ckjΣk = 1 n Σi = 1 n ( yj )Σj = 1 n ( xi ) Σi = 1 n ( xi ) ( y1 + y2 +…+ yn –1 + yn ) ( x1 + x2 +…+ xn –1 + xn ) y1 + ( x1 + x2 +…+ xn –1 + xn ) y2 +… ( x1 + x2 +…+ xn –1 + xn ) yn –1 + ( x1 + x2 +…+ xn –1 + xn ) yn x1 y1 + x2 y1 +…+ xn –1 y1 + xn y1 + x1 y2 + x2 y2 +…+ xn –1 y2 + xn y2 +… x1 yn –1 + x2 yn –1 +…+ xn –1 yn –1 + xn yn –1 + x1 yn + x2 yn +…+ xn –1 yn + xn yn
- 59. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Matrices – Proofs A( BC ) = ( AB ) C Order of Terms is preserved Affects Order of Operations Sequence – PEM-DAS Associative (Multiplication) Σk = 1 n T = [ tij ] = ( aik bkj )ckjΣk = 1 n Σi = 1 n ( yj )Σj = 1 n ( xi ) Σi = 1 n ( xi ) ( y1 + y2 +…+ yn –1 + yn ) ( x1 + x2 +…+ xn –1 + xn ) y1 + ( x1 + x2 +…+ xn –1 + xn ) y2 +… ( x1 + x2 +…+ xn –1 + xn ) yn –1 + ( x1 + x2 +…+ xn –1 + xn ) yn x1 y1 + x2 y1 +…+ xn –1 y1 + xn y1 + x1 y2 + x2 y2 +…+ xn –1 y2 + xn y2 +… x1 yn –1 + x2 yn –1 +…+ xn –1 yn –1 + xn yn –1 + x1 yn + x2 yn +…+ xn –1 yn + xn yn Σi = 1 n Σj = 1 n xi yj
- 60. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Matrices – Proofs A = [ aij ] Amx n B = [ bij ] Bnx p then AB = [ cij ] = Σk = 1 n aik bkj = ai 1 b1 j + ai2 b2j +…+ ain –1 bn-1j + ain bnj Product Summation Operand Count Product Summation (Column-Row) Index ABmx p The entries of Row “ Aik” ( the i-th row ) are multiplied by the entries of “ Bkj” ( the j-th column ) and sequentially summed through Row “ Ain” and Column “ Bnj” to form the entry at [ cij ] ai,1 b1, j + ai,1 b1, j +1 ai,1 b1,n – 1 + ai,1 b1,n ai+1,2 b2, j + ai+1,2 b2, j +1 ai+1,2 b2, n – 1 + ai+1,2 b2, n . . . . . . . . . + . . .+ + . . .+ . . . . . . an – 1,n – 1bn – 1, j + an – 1,n – 1bn – 1, j +1 an – 1,n – 1bn – 1, n – 1 + an – 1,n – 1bn an,nbn, j + an,n bn, j +1 an,n bn,n – 1 + an,n bn,n + . . .+ + . . .+ Section 1.2, (Pg. 42) For Each Element of AB (single entry)
- 61. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Identity Matrix For Amx n A In = A A Im = A Matrix Exponentiation For Ak = AA…A K factors
- 62. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Transpose Matrix a11 a21 am1 . . . a12 a22 am2 . . . . . . . . . . . . . . . a1n a2n amn . . . A = a11 a12 a1n . . . a21 a22 a2n . . . . . . . . . . . . . . . am1 am2 amn . . . AT = 1 2 0 2 1 0 0 0 1 C = 1 2 0 2 1 0 0 0 1 CT = Symmetric Matrix: C = CT If C = [ cij ] is a symmetric matrix, cij = cji for i ≠ j C = [ cij ] is a symmetric matrix, Cmx n = CT nx p for m = n = p
- 63. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra 1 2 0 2 1 0 0 0 1 C = 1 2 0 2 1 0 0 0 1 CT = If C = [ cij ] is a symmetric matrix, cij = cji , "i,j | i ≠ j C = [ cij ] is a symmetric matrix, Cmx n = CT nx p , "m,n, p | m = n = p Symmetric Matrix A Symmetric Matrix is a Square Matrix that is equal to it Transpose ( e.g. Cmx n = CT mx n , "m,n | m = n)
- 64. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Matrices – Transposes ( AT ) T = A Transpose of a Scalar Multiple ( A + B ) T = A T + B T ( c A ) T = c ( A T ) Transpose of a Transpose Transpose of Sum ( AB ) T = B T A T Transpose of a Product Reverse Order of Terms ( interchange multiplicand & multiplier terms in the product expression ) Symmetry of A Matrix & The Product of Its Transpose AAT = ( AAT ) T ATA is also symmetric
- 65. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse A matrix Anx n is Invertible or Non-Singular when $ matrix Bnx n | AB = BA = In In : Identity Matrix of Order n Bnx n : The Multiplicative Inverse of A A matrix that does not have an inverse is Non-Invertible or Singular Non-square matrices do not have inverses n For matrix products Amx n Bnx p where m ≠ n ≠ p, AB ≠ BA as [ aij ≠ bij ] ??
- 66. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse There are two methods for determining an inverse matrix A-1 of A ( if an inverse exists): Solve Ax = In for X Adjoin the Identity Matrix In (on RHS ) to A forming the doubly- augmented matrix [ A In ] and perform EROs concluding in RREF to produce an [ In A-1 ] solution A test for determining whether an inverse matrix A-1 of A exists: Demonstrate that either/or AB = In = BA Section 2.3 (Pg. 64)
- 67. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse Uniqueness Property If A is invertible, then its inverse is Unique Notation: The inverse of A is denoted as A-1 If A is invertible, then the LE system represented by Ax = b has a Unique Solution given by x = A– 1 b
- 68. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse – by Matrix Equation x11 x12 x21 x22 1 – 1 4 – 3 + = A x 1 0 0 1 In For a coefficient matrix Anx n the A-1 nx n matrix is that whose product yields a solution matrix to the corresponding In identity matrix 1x11 + – 1x21 + 4x21 ( – 3x21 ) = Ax 1x11 + – 1x21 + 4x21 ( – 3x21 ) 1 0 0 1 In 1x11 + 4x21 = 1 – 1x21 + ( – 3x21 ) = 0 1x11 + 4x21 = 0 – 1x21 + ( – 3x21 ) = 1 – 3 1 – 4 1 A-1Ax = In Ax = In
- 69. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse – by Gauss-Jordan Elimination x11 x12 x21 x22 1 – 1 4 – 3 + = A x 1 0 0 1 In An invertible coefficient Anx n matrix can be combined with its corresponding xnx n unknown/variable matrix to form an Axnx n = In equation matrix This equation matrix is composed itself of identical coefficient column vectors 1 x11 + – 1 x21 + 4 x21 ( – 3 x21 ) = Ax 1 x11 + – 1 x21 + 4 x21 ( – 3 x21 ) 1 0 0 1 In
- 70. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse – by Gauss-Jordan Elimination An invertible coefficient Anx n matrix can be combined with its corresponding xnx n unknown/variable matrix to form an Axnx n = In equation matrix This equation matrix is composed itself of identical coefficient column vectors 1 x11 + – 1 x21 + 4 x21 ( – 3 x21 ) = Ax 1 x11 + – 1 x21 + 4 x21 ( – 3 x21 ) 1 0 0 1 In 1x11 + 4x21 = 1 – 1x21 + ( – 3x21 ) = 0 1x11 + 4x21 = 0 – 1x21 + ( – 3x21 ) = 1 Ax = In Ax = In Rather than solve the two column equation vectors separately, they can be solved simultaneously by adjoining the identity matrix to the shared coefficient matrix
- 71. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse – by Gauss-Jordan Elimination ( GJE ) An invertible coefficient Anx n matrix can be combined with its corresponding xnx n unknown/variable matrix to form an Axnx n = In equation matrix This equation matrix is composed itself of identical coefficient column vectors 1 x11 + 4 x21 = 1 – 1 x21 + ( – 3 x21 ) = 0 1 x11 + 4 x21 = 0 – 1 x21 + ( – 3 x21 ) = 1 Ax = In Ax = In Rather than solve the two column equation vectors separately, they can be solved simultaneously by adjoining the identity matrix to the shared coefficient matrix… 1 – 1 4 – 3 A 1 0 0 1 In …then execute ERO’s to effect a GJ-Elimination of the “ doubly augmented ” [ A I ] matrix the conclusion of which will yield an [ I A-1 ] inverse matrix
- 72. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse – by Gauss-Jordan Elimination An invertible coefficient Anx n matrix can be combined with its corresponding xnx n unknown/variable matrix to form an Axnx n = In equation matrix This equation matrix is composed itself of identical coefficient column vectors 1 x11 + 4 x21 = 1 – 1 x21 + ( – 3 x21 ) = 0 1 x11 + 4 x21 = 0 – 1 x21 + ( – 3 x21 ) = 1 Ax = In Ax = In The adjoined, “ doubly-augmented ” coefficient matrix , by means of ERO’s , is reduced by GJ-Elimination to produce the [ I A-1 ] inverse matrix 1 – 1 4 – 3 A 1 0 0 1 In – 3 1 – 4 1 A-1 1 0 0 1 In Which is confirmed by verifying either of the following n AA-1 = I n AA-1 = A-1 A
- 73. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Matrix Inverse – 2x2 Matrix ( Special Case ) For a square matrix A2x 2 , given by: The inverse A-1 of the root matrix A2x 2 is given by the following product: a c b d = ad – cb d – c – b a The difference of the diagonal products forms the multiplicand denominator of the matrix whose product yields the inverse of the root matrix 1 ad – cb A-1 = NegateSwitcheroo Abstract Algebra, Lecture 2 @ 18:30 The scalar multiple is the inverse of the root matrix Determinant! The “multiplier” matrix is a half-negated permutation of the root matrix!
- 74. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties Of Inverse Matrices A : An Invertible Matrix k : A positive integer , Z+ c : A non-zero scalar, c ≠ 0 A– 1 Ak c A AT are Invertible and the following are true: ( A– 1 ) – 1 = A ( Ak ) – 1 = A– 1 A– 1 … A– 1 = ( A– 1 ) k K factors ( cA ) – 1 = A– 1 1 c ( AT ) – 1 = ( A– 1 ) T Aj Ak = Aj+k ( Aj ) k = Ajk ( AB ) – 1 = B– 1 A– 1 ( B is also invertible )
- 75. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Properties of Matrix Exponentiation Aj Ak = Aj+k ( Aj ) k = Ajk
- 76. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Elementary Matrices An Elementary Matrix, Anx n is: A square matrix ( n x n ) Obtained from a corresponding Identity Matrix In Results from a single Elementary Row Operation ( ERO ) If E is an Elementary Matrix, then: E is obtained from an ERO on a corresponding Identity Matrix Im EA is the product of the same ERO performed on an Am x n matrix Matrices Amx n & Bmx n are Row Equivalent when: $ a finite set of Elementary Matrices E1 , E2 ,… , Ek such that B = Ek Ek – 1 … E2 E1 A
- 77. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Elementary Matrices - Properties If E is an Elementary Matrix then: E– 1 exists E– 1 is an Elementary Matrix A square matrix A is invertible if-and-only-if: A can be expressed as the product of elementary matrices Every Elementary Matrix has an inverse Matrix Equivalency conditions, for Anx n matrix: “ A ” is invertible Ax = b has a unique solution for every n x 1 column matrix b Ax = 0 has only the trivial solution “ A ” is row-equivalent to In “ A ” can be written as the product of elementary matrices
- 78. © Art Traynor2011 Mathematics Linear Algebra Definitions Upper & Lower Triangular Matrices a11 a21 a31 am1 . . . 0 a22 a32 am2 . . . 0 0 a33 am3 . . . . . . . . . . . . . . . . . . 0 0 0 amn . . . L For an Anx n square matrix: “ L ” is a lower triangular matrix where all entries above the Main Diagonal are zero, and only the lower half is populated with non-zero entries. a11 0 0 0 . . . a12 a22 0 0 . . . a13 a23 a33 0 . . . . . . . . . . . . . . . . . . a1n a2n a3n amn . . . U “ U ” is a lower triangular matrix where all entries below the Main Diagonal are zero, and only the upper half is populated with non-zero entries.
- 79. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra LU Factorization A square matrix Anx n can be written as a product A = LU if: “ L ” is a lower triangular matrix where all entries above the Main Diagonal are zero, and only the lower half is populated with non-zero entries, and … “ U ” is a lower triangular matrix where all entries below the Main Diagonal are zero, and only the upper half is populated with non-zero entries
- 80. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Determinants Every square matrix Anx n can be associated with a real number defined as its Determinant Notation: det ( A ) = |a | a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 Example: 2-LE System with (2) unknowns yields solutions with common denominators b1 a22 – b2 a12 a11 a22 – a21 a12 x1 = b2 a11 – b1 a21 a11 a22 – a21 a12 x2 = Determinant of a 2 x 2 Matrix a11 a12 a21 a22 A = = det ( A ) = |A | = a11 a22 – a21 a12 a11 a21 a12 a22 = a11a22 – a21a12 Determinant is the difference of the product of the diagonals The Determinant is a polynomial of Order “ n ”
- 81. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Determinants Every square matrix Anx n can be associated with a real number defined as its Determinant Notation: det ( A ) = |a | a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 Example: 2-LE System with (2) unknowns yields solutions with common denominators b1 a22 – b2 a12 a11 a22 – a21 a12 x1 = b2 a11 – b1 a21 a11 a22 – a21 a12 x2 = Determinant of a 2 x 2 Matrix Determinant is the difference of the product of the diagonals The Determinant is a polynomial of Order “ n ” a c b d = ad – cb The Determinant is the Area (an n-Manifold ) of the parallelogram suggested by the addition of the vectors represented by the matrix
- 82. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Minors & Cofactors For a square matrix Anx n The Minor Mij of the entry aij is the determinant of the matrix obtained by deleting the ith row and jth column of A The Cofactor Cij of the entry aij is Cij = ( – 1 )i+j Mij Example: a11 a13 a21 a23 a31 a33 a12 a22 a32 Minor of a21 a11 a13 a21 a23 a31 a33 a12 a22 a32 Minor of a22 a12 a32 a13 a33 , M21 = a11 a31 a13 a33 , M22 = A Minor IS A DETERMINANT!!
- 83. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Minors & Cofactors For a square matrix Anx n The Cofactor Cij of the entry aij is Cij = ( – 1 )i+j Mij Example: a11 a13 a21 a23 a31 a33 a12 a22 a32 Minor of a21 a11 a13 a21 a23 a31 a33 a12 a22 a32 Minor of a22 a12 a32 a13 a33 , M21 = a11 a31 a13 a33 , M22 = Cofactor of a21 Cofactor of a22 C21 = ( – 1 )2+1 M21 = – M21 C22 = ( – 1 )2+2 M22 = M22
- 84. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Determinant of a Square Matrix For a square matrix Anx n of order n ≥ 2 , then: The Determinant of A is the sum of the entries in the first row of A multiplied by their respective Cofactors det ( A ) = |A | = a1j C1j = a11 C11 + a12 C12 +…+ a1, n –1 Cn, n –1 + a1n C1nΣj = 1 n The process of determining this sum is Expanding The Cofactors ( in the first row ) Section 3.1, (Pg. 106)
- 85. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Expansion By Cofactors For a square matrix Anx n of order n , the determinant of A is given by: An ith row expansion det ( A ) = |A | = aij Cij = ai1 Ci1 + ai2 Ci2 +…+ ai, n –1 Ci, n –1 + ain CinΣj = 1 n A jth column expansion det ( A ) = |A | = aij Cij = a1j C1j + a2j C2j +…+ an –1, j Cn –1, j + anj CnjΣj = 1 n Section 3.1, (Pg. 107)
- 86. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Determinants – 3x3 Matrix ( Special Case ) For a square matrix A3x 3 , the determinant of A is given by: The first two columns are adjoined to the RHS of the matrix a11 a13 a21 a23 a31 a33 a12 a22 a32 a11 a21 a31 a12 a22 a32 Product sums are formed by first multiplying along the main diagonal proceeding to the right a11 a13 a21 a23 a31 a33 a12 a22 a32 a11 a21 a31 a12 a22 a32 ➀ ➁ ➂ = a11a22 a33 + a12a23 a31 + a13a21 a32 = UD Upper Diagonal
- 87. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Determinants – 3x3 Matrix ( Special Case ) For a square matrix A3x 3 , the determinant of A is given by: Remaining product differences are then formed by multiplying along the LHS bottom diagonal proceeding to the right a11 a13 a21 a23 a31 a33 a12 a22 a32 a11 a21 a31 a12 a22 a32 ➃ ➄ ➅ = UD – a31a22 a13 – a32a23 a11 – a33a21 a12 = UD – LD Upper Diagonal minus Lower Diagonal
- 88. © Art Traynor2011 Mathematics Linear Algebra Definitions Diagonal Matrix A matrix Anx n is said to be diagonal when all of the entries outside the main diagonal ( ↘ ) are zero Section 2.1, (Pg. 50) The matrix Dnx n is diagonal if : dij = 0 if i ≠ j "ij { dii , di+1,i+1 ,…, dn–1,n–1 , dnn } d11 d21 d31 dm1 . . . d12 d22 d32 dm2 . . . d13 d23 d33 dm3 . . . . . . . . . . . . . . . . . . d1n d2n d3n dmn . . . Dnx n = A matrix Anx n that is both upper AND lower triangular is said to be diagonal The determinant of a triangular matrix Dnx n is the product of its main diagonal elements det ( D ) = |D | = aii = a11 a22 … a n –1, n –1 ain Πi = 1 n
- 89. © Art Traynor2011 Mathematics Linear Algebra Definitions EROs & Determinants (Properties) Permutation: [ A ] Pij [ B ] det ( B ) = – det ( A ) |B | = – | A | Multiplication by a Scalar: [ A ] cRi [ B ] det ( B ) = c det ( A ) |B | = c | A | Addition to a Row Multiplied by a Scalar: [ A ] Ri + cRj [ B ] det ( B ) = det ( A ) |B | = | A | There are three “effects” to a resultant matrix which are unique to each of the three EROs Permutation Scalar Multiplication Row Addition
- 90. © Art Traynor2011 Mathematics Linear Algebra Definitions Zero Determinants A matrix Anx n will feature a determinant of zero det ( A ) = 0 |A | = 0 if any of the following pertain One row/column of “ A ” consists of all zeros Two rows/columns of “ A ” are equal One row/column of “ A ” is a multiple of another
- 91. © Art Traynor2011 Mathematics Linear Algebra Definitions Determinant of a Matrix Product For matrices Anx n & Bnx n , of order “ n ” det ( AB ) = det ( A ) det ( B ) |AB | = | A | | B | Determinant of a Scalar Multiple of a Matrix For matrix Anx n of order “ n ” , and Scalar “ c ” det ( cA ) = cn det ( A ) |A | = cn | A | Determinant of an Invertible Matrix For matrix Anx n A is invertible if-and-only-if det ( A ) ≠ 0 |A | ≠ 0 Factors are not row-column specific (for whatever reason??) An invertible matrix must have a non- zero determinant, elsewise one would be dividing by zero to obtain the inverse of the matrix (undefined)
- 92. © Art Traynor2011 Mathematics Linear Algebra Definitions Determinant of an Inverse Matrix For matrix Anx n A is invertible if-and-only-if det ( A– 1 ) = |A | = 1 det ( A ) 1 |A | Determinant of a Transpose For matrix Anx n det ( A ) = det ( AT ) |A | = |AT | An invertible matrix must have a non- zero determinant, elsewise one would be dividing by zero to obtain the inverse of the matrix (undefined)
- 93. © Art Traynor2011 Mathematics Linear Algebra Definitions Equivalent Conditions For A Non-Singular Matrix For matrix Anx n , the following statements are equivalent “ A ” is invertible Ax = b has a unique solution for every n x 1 column matrix b Ax = 0 has only the trivial solution “ A ” is row-equivalent to In “ A ” can be written as the product of elementary matrices det ( A ) ≠ 0 ; |A | ≠ 0
- 94. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Adjoint of a Matrix For a square matrix Anx n The Cofactor Cij of the entry aij is Cij = ( – 1 )i+j Mij C11 C21 Cn1 . . . C12 C22 Cn2 . . . . . . . . . . . . . . . C1n C2n Cnn . . . Cofactor Matrix of A C11 C12 C1n . . . C21 C22 C2n . . . . . . . . . . . . . . . Cn1 Cn2 Cnn . . . adj ( A ) = Adjoint Matrix of A The transpose of the Cofactor Matrix Cij of “ A ” is Cij T
- 95. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Adjoint Equivalence with Matrix Inverse For invertible matrix Anx n , A– 1 is defined by A– 1 = adj ( A ) A– 1 = adj ( A ) 1 det ( A ) 1 |A |
- 96. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Cramer’s Rule Given a square matrix Anx n in “ n ” equations ( i.e. LE count = Airty ) a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 2-LE System with (2) unknowns yields solutions with common denominators b1 a22 – b2 a12 a11 a22 – a21 a12 x1 = b2 a11 – b1 a21 a11 a22 – a21 a12 x2 = which denominator forms the Determinant of the matrix “ A ” b1 b2 a12 a22 x1 = a11 a21 a12 a22 a11 a21 b1 b2 , x2 = a11 a21 a12 a22 , a11 a22 – a21 a12 ≠ 0 |A1 | = b1 b2 a12 a22 |A2 | = a11 a21 b1 b2 |A1 | |A | x1 = |A2 | |A | x2 =
- 97. © Art Traynor2011 Mathematics Algebra Of Matrices Linear Algebra Cramer’s Rule Given a system of “ n ” linear equations in “ n ” variables ( i.e. LE count = Airty ) with coefficient matrix “ A ” and non-zero determinant |A | the solution of the system is given as: |A1 | |A | x1 = , x2 = , … , xn = where the ith column of Ai is the “ constant ” vector in the LE system |A2 | |A | |An | |A | Linear Equation System a11 x1 + a12 x2 + a13 x3 = b1 a21 x1 + a22 x2 + a23 x3 = b2 a31 x1 + a32 x2 + a33 x3 = b3 Matrix-Vector Notation Example: a11 b1 a21 b2 a31 b3 a12 a22 a32 a11 a13 a21 a23 a31 a33 a12 a22 a32 |A3 | |A | x3 = =
- 98. © Art Traynor2011 Mathematics Topological Spaces Space Mathematical Space Mathematical Space A Mathematical Space is a Mathematical Object that is regarded as a species of Set characterized by: Structure Heirarchy and Inner Product Spaces Normed Vector Spaces Vector Spaces Metric Spaces Subordinate Spaces ( Subspaces ) inherit the properties of Parent Spaces such that subordinate Subspaces are said to Induce their properties onto the parent spaces in a recursive fashion e.g. an Algebra or Algebraic Structure
- 99. © Art Traynor2011 Mathematics ProjectiveEuclidean Mathematical Space Distance between two points is defined Distance is Undefined Space Mathematical Space Heirarchy Upper Level Classification Second Level Classification Non-EuclideanEuclidean Finite Dimensional Infinite Dimensional Compact Non-Compact Second Level Classification N n , Z n , Q n , R n , C n , E n This slide is very slippery It really needs a deeper dive to achieve necessary cogency
- 100. © Art Traynor2011 Mathematics MapFunction Morphism A Relation between a Set of inputs and a Set of permissible outputs whereby each input is assigned to exactly one output A Relation as a Function but endowed with a specific property of salience to a particular Mathematical Space A Relation as a Map with the additional property of Structure preservation as between the sets of its operation Structure A Set attribute by which several species of Mathematical Object are permitted to attach or relate to the Set which expand the enrichment of the Set Space Mathematical Space Measure The manner by which a Number or Set Element is assigned to a Subset Algebraic Structure A Carrier Set defined by one or more Finitary Operations Field A non-zero Commutative Ring with Multiplicative Inverses for all non-zero elements (an Abelian Group under Multiplication)
- 101. © Art Traynor2011 Mathematics FMM A unique Relation between Sets Structure A Set attribute by which several species of Mathematical Object are permitted to attach or relate to the Set which expand the enrichment of the Set Space Mathematical Space Measure The manner by which a Number or Set Element is assigned to a Subset Algebraic Structure A Carrier Set defined by one or more Finitary Operations Field A non-zero Commutative Ring with Multiplicative Inverses for all non-zero elements (an Abelian Group under Multiplication) Satisfies Group Axioms plus Commutativity Arithmetic Operations are defined ( +, – , x ,÷ ) Salient to a Mathematical Space Preserving of Structure FMM = Function~Map~Morphism Akin to the Holy Trinity Topology Those properties of a Mathematical Object which are invariant under Transformation or Equivalence Metric Space A Set for which distance between all Elements of the Set are defined The Triangle Inequality constitutes the principle Axiom from which three subsidiary axioms are derived F ≡ C R Q Z N
- 102. © Art Traynor2011 Mathematics Topology Structure A Set attribute by which several species of Mathematical Object are permitted to attach or relate to the Set which expand the enrichment of the Set Space Mathematical Space Manifold A Topologic Space resembling a Euclidean Space whose features may be charted to Euclidean Space by Map Projection Metric Space A Carrier Set defined by one or more Finitary Operations Riemann Manifold Order A Binary Set Relation exhibiting the Reflexive, Antisymmetric, and Transitive properties Equivalence Class Those properties of a Mathematical Object which are invariant under Transformation or Equivalence Surface of a Sphere is not a Euclidean Space! A Real Manifold enriched with an inner product on the Tangent Space varying smoothly at each point Geometry A Complete, Locally Homogenous, Reimann Manifold Scale Invariant - Exhibits Multiplicative Scaling Convergent A Binary Set Relation exhibiting the Reflexive, Symmetric, and Transitive properties
- 103. © Art Traynor2011 Mathematics Topology Structure A Set attribute by which several species of Mathematical Object are permitted to attach or relate to the Set which expand the enrichment of the Set Space Mathematical Space Manifold A Topologic Space resembling a Euclidean Space whose features may be charted to Euclidean Space by Map Projection Order A Binary Set Relation exhibiting the Reflexive, Antisymmetric, and Transitive properties Equivalence Class Those properties of a Mathematical Object which are invariant under Transformation or Equivalence Surface of a Sphere is not a Euclidean Space! A Binary Set Relation exhibiting the Reflexive, Symmetric, and Transitive properties Differential Structures A Structure on a Set rendering the Set into a Differential Manifold with n-dimensional Continuity defined by a CK Atlas of Bijection/Charts Categories Comprised of Object and Morphism Classes and Morphisms relating the Objects admitting Composition and satisfying the Associativity and Identity Axioms
- 104. © Art Traynor2011 Mathematics FMM A unique Relation between Sets Structure A Set attribute by which several species of Mathematical Object are permitted to attach or relate to the Set which expand the enrichment of the Set Space Mathematical Space Measure The manner by which a Number or Set Element is assigned to a Subset Salient to a Mathematical Space Preserving of Structure FMM = Function~Map~Morphism Akin to the Holy Trinity SurjectionInjective Functions One-to-One Onto Bijection Inversive One-to-One & Onto f : X Y A Function which returns a CoDomain equivalent to the Domain of another Function returning that same CoDomain aka: Automorphism
- 105. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M “① Example:
- 106. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M ”① Example: Set Theory entails that at least two improper subspaces are constituent of the Space: the Empty Set and “ M ” itself Proof: •Let M be a Space over some field F. •Every Space must contain at least two elements: the empty set { } , and itself Ms P ( M ) Ms
- 107. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M ”① Example: Set Theory entails that at least two improper subspaces are constituent of the Space: the Empty Set and “ M ” itself Proof: •Let M be a Space over some field F. •Every Space must contain at least two elements: the empty set { } , and itself Ms P ( M ) Ms
- 108. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M ”① Example: Set Theory entails that at least two improper subspaces are constituent of the Space: the Empty Set and “ M ” itself Proof: •Let M be a Space over some field F. •Every Space must contain at least two elements: the empty set { } , and itself Ms P ( M ) Ms
- 109. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M ”① Example: Set Theory entails that at least two improper subspaces are constituent of the Space: the Empty Set and “ M ” itself Proof: •Let M be a Space over some field F. •Every Space must contain at least two elements: the empty set { } , and itself Ms P ( M ) Ms
- 110. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M ”① Example: Set Theory entails that at least two improper subspaces are constituent of the Space: the Empty Set and “ M ” itself Proof: •Let M be a Space over some field F. •Every Space must contain at least two elements: the empty set { } , and itself Ms P ( M ) Ms
- 111. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M ”① Example: Set Theory entails that at least two improper subspaces are constituent of the Space: the Empty Set and “ M ” itself Proof: •Let M be a Space over some field F. •Every Space must contain at least two elements: the empty set { } , and itself Ms P ( M ) Ms
- 112. © Art Traynor2011 Mathematics Subspace Subspace ( General ) Mathematical Space Somewhat trivially, a mathematical Subspace is a Subset of a parent Mathematical Space which inherits and enriches the Structure of the superordinating Mathematical Space Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Mathematical Space M Mathematical Space Given Mathematical Space “ M ”① Example: Set Theory entails that at least two improper subspaces are constituent of the Space: the Empty Set and “ M ” itself P ( M ) Ss ② We next introduce into this Power Set / Spanning space a well defined non-zero element “ S ” and at least one additional Structuring operation (∗) such that Si ∗ Si Fi Ss Si ∗
- 113. © Art Traynor2011 Mathematics Vector Space Vector Space ( General ) Mathematical Space Vector Spaces Vector Spaces Metric Spaces Topological Spaces A Vector Space V is a species of Set over a Field F of scalars (e.g. R or C ) whose constituent point elements can be uniquely characterized by an ordered tuple of n-dimension ( Vectors ) Structured the Superposition Principle ( and its derivative Linear Operations ): Addition ( aka Additivity Property ) A function that assigns to the combination of any two or more elements of the space a resultant unique n-tuple ( Vector ) composed of the sum of the respective operand vector components. f ( 〈 a , b 〉 ) = 〈 an + bn 〉 = 〈 rn 〉 = r ( e.g. rn R n ) f : V + V V = a V f ( a ) V Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements The simplest Vector Space is the space populated by only the Field itself, known as a Coordinate Space Vector Addition corresponds to the Motion of Translation
- 114. © Art Traynor2011 Mathematics Addition Vectors Vector (General ) x y O initial point terminal point Free Vector r A Sum of Vectors – Vector Addition (Tail –to–Tip) B a ( ax, ay ) b ( bx, by ) ║a ║ ║b ║ Any two (or more) vectors can be summed by positioning the operand vector (or its corresponding-equivalent vector) tail at the tip of the augend vector. The summation (resultant) vector is then extended from (tail) the origin (tail) of the augend vector to the terminal point (tip) of the operand vector (tip-to-tip/head-to-head). ry rx r ( rx , ry ) θ “ Tail-to-Tip ” “ Tip-to-Tip ” Same procedure, sequence of operations whether for vector addition (summation) or vector subtraction (difference) Resultant is always tip-to-tip Operands are oriented “ tip-to-tail ” resultant vector is oriented “ tip-to- tip ” The resultant vector in a summation always originates at the displacement origin and terminates coincident at the terminus of the final displacement vector (e.g. tip-to-tip) Chump Alert: A vector summation is a species of Linear Comination
- 115. © Art Traynor2011 Mathematics Vector Space Vector Space ( General ) Mathematical Space Vector Spaces Vector Spaces Metric Spaces Topological Spaces A Vector Space V is a species of Set over a Field F of scalars (e.g. R or C ) whose constituent point elements can be uniquely characterized by an ordered tuple of n-dimension ( Vectors ) Structured by the following Linear Operations: Addition f ( c 〈 an 〉 ) = 〈 can 〉 = 〈 rn 〉 = r ( e.g. rn R n ) Scalar Multiplication A function that assigns to the combination of any element of the multiplicand field and any multiplier vector space element a resultant unique n-tuple ( Vector ) composed of the product of the respective multiplicand scalar and the constituent multiplier vector n-components supra. f : F x V V = a V f ( a ) V Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements
- 116. © Art Traynor2011 Mathematics Vectors Vector (General ) PQ x y Position Vector O initial point Free Vector C Vector Scalar Multiple – a species of Transformation where the CoDomain Set if positive effects an Expansion, or if negative effects a Dilation. O C ( cax , cay ) A ( ax , ay ) c OA = OC terminal point Example: F = ma Vector Scalar Multiple Operands are oriented “ tip-to-tail ” with the multiplicand ( vector to be scaled ) “ scaled ” by the multiplier-scalar. The result constitutes a vector addition of the product of the scalar and the multiplicand normalized unit vector (NUV) thus preserving multiplicand orientation in the result c 〈 ax , ay 〉 = 〈 cax , cay 〉 Chump Alert: A vector scalar is a species of Linear Comination
- 117. © Art Traynor2011 Mathematics Vector Space Vector Space ( General ) Mathematical Space Vector Spaces Vector Spaces Metric Spaces Topological Spaces A Vector Space V is a species of Set over a Field F of scalars (e.g. R or C ) whose constituent point elements can be uniquely characterized by an ordered tuple of n-dimension ( Vectors ) Structured by the following Linear Operations: Addition Scalar Multiplication Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Vector/Linear (VL ) Spaces are said to be “Algebraic” VL Space operations define figures (subspaces?) such as lines and planes The Dimension of a VL Space is determined by the maximal number of Linear Independent variables (identical to the minimal number of vectors that Span the space) Additional Structure apart from that characterizing general Vector Space is needed to define Nearness, Angles, or Distance A Vector Space V is a species of Set over a Field F
- 118. © Art Traynor2011 Mathematics Vector Space Vector Space ( General ) Mathematical Space Vector Spaces Vector Spaces Metric Spaces Topological Spaces A Vector Space V is a species of Set over a Field F of scalars (e.g. R or C ) whose constituent point elements can be uniquely characterized by an ordered tuple of n-dimension ( Vectors ) Structured by the following Linear Operations: Addition Scalar Multiplication Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements Vector Spaces are said to be “Linear” spaces (as distinct from Topological Spaces) Vector/Linear (VL ) Spaces are said to be “Algebraic” VL Space operations define figures (subspaces?) such as lines and planes The Dimension of a VL Space is determined by the maximal number of Linear Independent variables (identical to the minimal number of vectors that Span the space) Additional Structure apart from that characterizing general Vector Space is needed to define Nearness, Angles, or Distance
- 119. © Art Traynor2011 Mathematics Vector Space Vector Space ( General ) Mathematical Space Vector Spaces Vector Spaces Metric Spaces Topological Spaces A Vector Space V is a species of Set over a Field F of scalars (e.g. R or C ) whose constituent point elements can be uniquely characterized by an ordered tuple of n-dimension ( Vectors ) Structured by the following Linear Operations: Addition Scalar Multiplication Closed under the operation of Addition and Scalar Multiplication And which satisfy the ten axioms governing vector space elements The essential Structure of a Vector Space enables transformations of its elements that correspond to classes of Motion Vector Addition (as well as Scalar Multiplication, which is by extension repeated vector addition) corresponds to Translation Translation is classed as one of three species of Rigid Motion the other two Rotation and Reflection require additional Structure A Vector is understood to represent a difference (Displacement ) between the respective values of its constituent ordered tuples
- 120. © Art Traynor2011 Mathematics Vectors Vector Properties Multiplicative Inverse If c( v ) = 0 Zero Vector Scalar Identity – 1( v ) = – v Properties of Scalar Multiplication If v represents any element of a vector space V ( v V ) and c represents any scalar, then the following properties pertain: Zero Vector Multiplicative Identity0( v ) = 0 Scalar Zero Vector Multiplicative Identity c( 0 ) = 0 then c = 0 or v = 0
- 121. © Art Traynor2011 Mathematics Vector Space Axioms – Addition Abstraction Mathematical Space Vector Space A vector space is comprised of four elements: a set of vectors, a set of scalars, and two operations: u + v ( is in V ) Closure Under Addition ( u + v ) + w = u + ( v + w ) Changes Order of Operations as per “PEM-DAS”, Parentheses are the principal or first operation u + v = v + u Commutative Property of Addition Re-Orders Terms Does Not Change Order of Operations – PEM-DAS Associative Property of Addition u + 0 = u Additive Identity u + ( – u ) = 0 Additive Inverse If V is a vector space then $ 0 | "u V " u V , $ – u | "u V Operations: Addition & Scalar Mult. Section 4.2, (Pg. 155) Represents “ 0 = – 2 ” , a contradiction, and thus no solution { } to the LE system for which the augmented matrix stands Note that there is nothing in these axioms that entails Length/Distance or Magnitude of Vectors, nor corresponding attributes such as Angle or Nearness
- 122. © Art Traynor2011 Mathematics Multiplicative Identity cu ( is in V ) Closure Under Scalar Multiplication c ( u + v ) = cu + cv Distributive ( c + d )u = cu + du Distributive c( du ) = ( cd )u Associative Property of Multiplication 1( u ) = u Vector Space Axioms – Scalar Multiplication Abstraction A vector space is comprised of four elements: a set of vectors, a set of scalars, and two operations: Operations: Addition & Scalar Mult. Section 4.2, (Pg. 155) Let c = 0 and you therefore don’t need to state a separate scalar multiplicative zero element Mathematical Space Vector Space Note that there is nothing in these axioms that entails Length/Distance or Magnitude of Vectors, nor corresponding attributes such as Angle or Nearness
- 123. © Art Traynor2011 Mathematics Normed Vector Space Normed Vector Space Mathematical Space “Length” (aka Distance/Magnitude) is one type of norm The vector norm ║X║ is more formally defined as the ℓ2-norm Somewhat trivially, a Normed Vector Space is a Vector Space Structured by an Norm A norm is defined as a Mathematical Structure of the species of a Measure. Inner Product Spaces Normed Vector Spaces Vector Spaces There are several species of Norm A Norm on a vector space V is a function that maps a vector a in V to an element r of a Field F f : V R = a V f ( a ) F = rn ( e.g rn R n ) 2 2 ax + ay║a ║ = ║〈 ax , ay 〉║ =Magnitude p-Norm aka Euclidean Norm, L2 Norm, or ℓ2Norm ( or “ Length ” ) ║a ║p = ( Σ | ai |p )i = 1 n 1 p
- 124. © Art Traynor2011 Mathematics Magnitude Vectors Vector ( General ) Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) x y O θ A ( ax , ay ) Magnitude a Position Vector PVF: Position Vector Form xO θ a (adj ) b (opp ) r = c (hyp ) M A (1, 0) P ( cos θ, sin θ ) 1 tan θ cos θ Q sin θ y UCF: Unit Circle Form ay ax Unit Circle In PVF the magnitude of a vector a = 〈 ax , ay 〉 is equivalent to the hypotenuse ( c = ║a ║ ) of a right triangle whose adjacent side ( a ) is given by the coordinate a1 , and whose opposite side ( b ) is given by the coordinate a2 : 2 2 ax + ay║a ║ = ║ 〈 ax , ay 〉 ║ = Pythagorean Theorem derived
- 125. © Art Traynor2011 Mathematics Inner Product Space Inner Product Space ( IPS ) An Inner Product Space is a Vector Space over a Field of Scalars (e.g. R or C ) Structured by an Inner Product Inner Product Spaces Normed Vector Spaces For a Euclidean Space the Inner Product is defined as the Dot Product Positive-Definite Symmetric Bilinear Form Length/Distance or Magnitude of Vectors This Structure moreover defines IPS salients such as: Vector Subtended Angle Orthogonality of Vectors These spaces have a well-ordered semantic construction of the form: A Vector Space with an Inner Product “on” it… “ The IPS of conventional multiplication over the field of R ” “ The IPS of the dot product over the field of R ” Mathematical Space
- 126. © Art Traynor2011 Mathematics Inner Product Space ( IPS ) Axioms Inner Product Spaces Normed Vector Spaces A summation of the Scalar Product of the vector components, a = 〈 ax , ay 〉 , b = 〈 bx , by 〉 For a Euclidean Space the Inner Product is defined as the Dot Product Note here the possibility of describing the dot product as an equivalence class for its alternate expressions to every n-tuple of vectors a and b in V, a scalar in F Let V be a vector space over F , a Field of Scalars (e.g. R or C ) . An Inner Product on V is a function that assigns, f ( 〈 a , b 〉 ) = ( an bn + an bn ) = rn ( e.g. rn R n ) Inner Product Space Mathematical Space
- 127. © Art Traynor2011 Mathematics Inner Product Axioms Given vectors u , v , and w in Rn , and scalars c , the following axioms pertain: 〈 u , v 〉 = 〈 v , u 〉 Symmetry 〈 u , v + w 〉 = 〈 u , v 〉 + 〈 u , w 〉 Additive Linearity Positive Definiteness Inner Product Space Mathematical Space c 〈 u , v 〉 = 〈 cu , v 〉 Multiplicative Linearity 〈 v , v 〉 ≥ 0 〈 v , u 〉 〈 v , v 〉 = 0 if and only if v = 0 Section 5.2, (Pg. 237)
- 128. © Art Traynor2011 Mathematics Vectors x y Position Vector O initial point terminal point Free Vector A B O A ( ax , ay ) B ( bx , by ) ║a ║ ║b ║ Vector ( Euclidean ) Dot Product The dot product of two vectors is the scalar summation of the product Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) PVF: Position Vector Form UCF: Unit Circle Form a · b = ax bx + ay by of their components, a = < ax , ay > , b = < bx , by > Also referred to as the Scalar Product or Inner Product Pythagorean Theorem derived Inner (Dot) Product
- 129. © Art Traynor2011 Mathematics Vectors x y Position Vector O initial point terminal point Free Vector A B O A ( ax , ay ) B ( bx , by ) ║a ║ ║b ║ Vector ( Euclidean ) Dot Product & Angle Between Vectors For any two non-zero vectors sharing a common initial point the dot product of the two vectors is equivalent to the product of their magnitudes and the cosine of the angle between Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) Inner (Dot) Product θ θ a · b = ax bx + ay by a · b = ║b ║║a ║ cosθ cosθ = ║a ║║b ║ a · b You will be asked to find the angle between two vectors sharing a common initial point (origin)…a lot θ = cos– 1 ║a ║║b ║ a · b
- 130. © Art Traynor2011 Mathematics Vectors Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) x y Position Vector O Free Vector Physical Quantities represented by vectors include: Displacement, Velocity, Acceleration, Momentum, Gravity, etc. O A ( ax , ay ) B ( bx , by ) a b c A B θ θ ║a ║ cos θ Dot Product & Angle Between Vectors For any two non-zero vectors sharing a common initial point the dot product of the two vectors is equivalent to the product of their magnitudes and the cosine of the angle between Vector ( Euclidean ) a · b = ax bx + ay by a · b = ║b ║║a ║ cosθ = a · b OB – OA = AB Inner (Dot) Product
- 131. © Art Traynor2011 Mathematics Vectors Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) x y Position Vector O Free Vector Physical Quantities represented by vectors include: Displacement, Velocity, Acceleration, Momentum, Gravity, etc. O A ( a1, a2 ) B ( b1x , by ) a b c A B θ θ ║a ║ cos θ ║b ║ Area = ║b ║║a ║ cosθ = a · b Dot Product & Angle Between Vectors For any two non-zero vectors sharing a common initial point the dot product of the two vectors is equivalent to the product of their magnitudes and the cosine of the angle between Vector ( Euclidean ) a · b = ax bx + ay by OB – OA = AB Inner (Dot) Product
- 132. © Art Traynor2011 Mathematics Vectors x y Position Vector O A ( ax , ay ) B ( bx , by ) a b θ Vector Component as Projection Vector ( Euclidean ) Inner (Dot) Product ② ③ ④ The intersection of any two vectors with common origin will feature a shared angle ( the “Angle Between” ). ① ② In Position Vector Form (PVF), the vector system can be aligned so that the vector common origin coincides with a coordinate system origin and one of the vectors (the Multiplier vector “ b ”) can then be aligned along the x-coordinate axis ③ In this orientation the Multiplicand vector “ a ” (if the angle between is acute) will terminate in the first quadrant of the coordinate system. O A B θ Free Vector ① ④ Note that the X-component of a (i.e. ax) is geometrically equivalent to a vertical projection from a onto the X-axis and b ax
- 133. © Art Traynor2011 Mathematics Vectors x y Position Vector O A ( ax , ay ) B ( bx , by ) a b θ Vector Component as Projection Vector ( Euclidean ) Inner (Dot) Product ⑤ Recalling the trigonometric relationships of the Unit Circle, it can be further noted that the X-component of a (i.e. ax) – previously noted to be geometrically equivalent to a vertical projection from a onto the X-axis and b – is also geometrically equivalent to the product of the length of a (its Magnitude) and the cosine of the angle formed with the x-axis ax 2 2 ax + ay║a ║ = ║〈 ax , ay 〉║ = ax = ║a ║ cosθ ║b ║ 1 compb a = a · b O U ! ! cos θ r = 1 = c r = c (hyp ) tan θ sin θ θ x y a (adj ) b (opp ) Unit Circle
- 134. © Art Traynor2011 Mathematics Vectors x y Position Vector O A ( ax , ay ) B ( bx , by ) a b θ Vector Component as Projection Vector ( Euclidean ) Inner (Dot) Product ⑤ Recalling the trigonometric relationships of the Unit Circle, it can be further noted that the X-component of a (i.e. ax) – previously noted to be geometrically equivalent to a vertical projection from a onto the X-axis and b – is also geometrically equivalent to the product of the length of a (its Magnitude) and the cosine of the angle formed with the x-axis ax 2 2 ax + ay║a ║ = ║〈 ax , ay 〉║ = Another way to express this geometrical equivalence is to note the inherent relationship between the lengths of two vectors in composition sharing a common origin and the angle between the two supplied by the inner (dot) product relationship xO θ a (adj ) b (opp ) r = c (hyp ) M A (1, 0) P ( cos θ, sin θ ) 1 tan θ (r|c|1) cos θ Q sin θ y Unit Circle a OA = OP = AP = t = θ = 1
- 135. © Art Traynor2011 Mathematics Vectors Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) x y Position Vector O Physical Quantities represented by vectors include: Displacement, Velocity, Acceleration, Momentum, Gravity, etc. A ( a1, a2 ) B ( b1x , by ) a b c θ ║a ║ cos θ ║b ║ Area = ║b ║║a ║ cosθ Vector Component as Projection For any two non-zero vectors sharing a common initial point the dot product of the two vectors is equivalent to the product of their magnitudes and the cosine of the angle between Vector ( Euclidean ) Inner (Dot) Product a · b = ║b ║║a ║ cosθ
- 136. © Art Traynor2011 Mathematics Vectors Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) xO The notion of “component along” is a direct consequence of the definition of an inner (dot) product – relating the two “sides” of a vector “triangle” via a ratio given by the cosine of the “angle between” O A ( ax , ay ) B ( b1, b2 ) a b c A Bθ θ ║a ║ cos θ Vector Component Along an Adjoining Vector Vector ( Euclidean ) y Position Vector The component of OA along OB that has the same direction as OB ║b ║ 1 compb a = a · b x y Position Vector ║b ║ b compb a = a · Compb a = a ·║b ║ b is the dot product of OA with the unit vector 1 ║u ║ u ║a ║ û = u =( ( Dot Product
- 137. © Art Traynor2011 Mathematics Vectors Col. 1 Col. 2 Col. 3 . . . Col. n a1 a2 a3 . . . ana · b = aTb = Col. 1 b1 b2 b3 bm . . . Vector ( Euclidean ) Dot Product ( Determinant Form ) The dot product of two vectors, is the matrix product of the 1 x n transpose of the multiplicand vector and the m x 1 multiplier vector Col. 1 b1 b = b2 b3 bm . . . Col. 1 a1 a = a2 a3 an . . . a · b = aTb = a1b1 + a2b2 + a3b3 …+ anbm Inner (Dot) Product
- 138. © Art Traynor2011 Mathematics Vector Spaces – Rn Vectors Tuple Properties An “ n-tuple” is characterized by the following: A sequence An ordered list Comprising “ n ” elements ( n is a non-negative integer) Canonical “ n-tuples ” 0-tuple: null tuple 1-tuple: singleton 2-tuple: ordered pair 3-tuple: triplet
- 139. © Art Traynor2011 Mathematics Vector Spaces – Rn Vectors Vector “ n-tuple ” Representation An ordered n-tuple represents a vector in n-space Section 4.1, (Pg. 149) n Of the form ( ai , ai+1 ,…an – 1 , an ) n The Set of all n-tuples is n-space, denoted by Rn n An n-tuple can be rendered as a point in Rn whose coordinates describe a unique vector a n n-tuples delineate Rn such that all points in Rn can be represented by a unique n-tuple Tuple Properties
- 140. © Art Traynor2011 Mathematics Vector Spaces – Rn Vectors “ n-tuple ” distinguished from a set n-tuples delineate Rn space such that tuples of disparate n-order: ( 1, 2, 3, 2 ) ≠ ( 1, 2, 3 ) are not equal as the same sequence expressed as elements of a set { 1, 2, 3, 2 } = { 1, 2, 3 } Tuple elements are ordered: ( 1, 2, 3 ) ≠ ( 3, 2, 1 ) whereas for a set { 1, 2, 3 } = { 3, 2, 1 } A tuple is composed of a finite population of elements whereas a set may contain infinitely many elements Tuple: Sequence Matters Set: Sequence Does Not Matter Set: Order Matters Set: Order Does Not Matter Tuple Properties
- 141. © Art Traynor2011 Mathematics Vector Spaces – Rn Vectors Tuples as Functions An n-tuple can be rendered as a function “ F ” the domain of which is represented by the tuple’s element index/indices or “ X ” the codomain of which is represented by the tuple’s elements or “ Y ” X = { i , i + 1 ,…, n – 1 , n } ( ai , ai+1 ,…an – 1 , an ) = ( X , Y , F ) ( a1 , a2 ,…, an – 1 , an ) = ( X , Y , F ) X = { 1 , 2 ,…, n – 1 , n } or or Y = { a1 , a2 ,…, an –1 , an } F = { ( 1, a1 ) , ( 2, a2 ) ,…, ( n – 1, an – 1 ) , ( n, an ) } Tuple Properties
- 142. © Art Traynor2011 Mathematics Definition Vectors Vector ( Euclidean ) A geometric object (directed line segment) describing a physical quantity and characterized by Direction: depending on the coordinate system used to describe it; and Magnitude: a scalar quantity (i.e. the “length” of the vector) Aka: Geometric or Spatial Vector originating at an initial point [ an ordered pair : ( 0, 0 ) ] and concluding at a terminal point [ an ordered pair : ( a1 , a2 ) ] Other mathematical objects describing physical quantities and coordinate system transforms include: Pseudovectors and Tensors Not to be confused with elements of Vector Space (as in Linear Algebra) Fixed-size, ordered collections Aka: Inner Product Space Also distinguished from statistical concept of a Random Vector From the Latin Vehere (to carry) or from Vectus…to carry some- thing from the origin to the point constituting the components of the vector 〈 a1 , a2 〉
- 143. © Art Traynor2011 Mathematics Vectors Vector ( Euclidean ) Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) Vector – Properties ( PVF Form) Each (position) vector determines a unique Ordered Pair ( a1 , a2 ) The coordinates a1 and a2 form the Components of vector 〈 a1 , a2 〉 x y Position Vector ║a ║ O θ A ( a1, a2 ) initial point terminal point a a1 a2 Position Vector A vector represented in PVF is Unique n There is precisely one free-vector equivalent in PVF: a = OA n The unique ordered pair describing the vector is a unique n-tuple in Rn
- 144. © Art Traynor2011 Mathematics Vector Standard Operations in Rn Vectors Sum of Vectors: u + v u + v = ( u1+ v1 , u2+ v2 , … , un –1 + vn –1 , un + vn ) Given u = ( ui , ui+1 ,…un – 1 , un ) and v = ( vi , vi+1 ,…vn – 1 , vn ) Scalar Multiple of Vectors: cu cu = ( cu1 , cu2 , … , cun –1 , cun ) Vector Operations
- 145. © Art Traynor2011 Mathematics Addition Vectors Vector ( Euclidean ) x y O initial point terminal point Free Vector r A Sum of Vectors – Vector Addition (Tail –to–Tip) B a ( ax, ay ) b ( bx, by ) ║a ║ ║b ║ Any two (or more) vectors can be summed by positioning the operand vector (or its corresponding-equivalent vector) tail at the tip of the augend vector. The summation (resultant) vector is then extended from (tail) the origin (tail) of the augend vector to the terminal point (tip) of the operand vector (tip-to-tip/head-to-head). ry rx r ( rx , ry ) θ “ Tail-to-Tip ” “ Tip-to-Tip ” Same procedure, sequence of operations whether for vector addition (summation) or vector subtraction (difference) Resultant is always tip-to-tip Operands are oriented “ tip-to-tail ” resultant vector is oriented “ tip-to- tip ” The resultant vector in a summation always originates at the displacement origin and terminates coincident at the terminus of the final displacement vector (e.g. tip-to-tip) Chump Alert: A vector summation is a species of Linear Comination
- 146. © Art Traynor2011 Mathematics Vector Standard Operations in Rn Vectors Vector Operations Difference of Vectors: u – v u – v = ( u1 – v1 , u2 – v2 , … , un –1 – vn –1 , un – vn ) Given u = ( ui , ui+1 ,…un – 1 , un ) and v = ( vi , vi+1 ,…vn – 1 , vn ) Scalar Multiplicative Inverse: – cu ( c = 1) – u = ( – u1 , – u2 , … , – un –1 , – un )
- 147. © Art Traynor2011 Mathematics Subtraction Vectors Vector ( Euclidean ) x y O ry rx θ initial point terminal point Free Vector r = a + bcorr a b O “ Tail-to-Tip ” “ Tip-to-Tip ” ( Addition ) bcorr – bcorr “ Tip-to-Tip ” ( Difference ) Position Vector r = a – bcorr Difference of Vectors – Vector Subtraction ( Tail –to–Tip ) Any two (or more) vectors can be subtracted by positioning the tail of a corresponding-equivalent subtrahend vector (initial point) at the tip (terminal point) of the minuend vector. The difference (resultant) vector is then extended from the tail (initial point ) of the minuend vector (tail-to-tail) to the terminal point (tip) of the subtrahend vector (tip-to-tip). minuend subtrahend Same procedure, sequence of operations whether for vector addition (summation) or vector subtraction (difference) Resultant is always tip-to-tip r r = a + bcorr bcorr – bcorr a b Operands are oriented “ tip-to-tail ” resultant vector is oriented “ tip-to- tip ” Chump Alert: A vector difference is a species of Linear Comination
- 148. © Art Traynor2011 Mathematics Vector Properties – Additive Identity & Additive Inverse Vectors Vector Properties Given vectors u , v , and w in Rn , and scalars c and d, the following properties pertain 0v = 0 Scalar Zero Element If u + v = v then u = 0 Additive Identity is Unique If v + u = 0 then u = – v Additive Inverse is Unique c0 = 0 Scalar Multiplicative Identity of Zero Vector If cv = 0 then c = 0 or v = 0 Zero Vector Product Equivalence – ( – v ) = v Negation Identity Section 4.1, (Pg. 151)
- 149. © Art Traynor2011 Mathematics Vector Spaces – Classification Real Number Vector Spaces R = set of all real numbers R2 = set of all ordered pairs R3 = set of all ordered triplets Rn = set of all n-tuple Matrix Vector Spaces Mm,n = set of all m x n matrices Mn,n = set of all n x n square matrices Section 4.2, (Pg. 157) Vector Spaces Vector Space
- 150. © Art Traynor2011 Mathematics Vector Spaces Vector Spaces – Classification Polynomial Vector Spaces P = set of all polynomials Pn = set of all polynomials of degree ≤ n Continuous Functions ( Calculus ) Vector Spaces C ( – ∞ , ∞) = set of all continuous functions defined on the real number line C [ a, b ] = set of all continuous functions defined on a closed interval [ a, b ] Section 4.2, (Pg. 157) Vector Space
- 151. © Art Traynor2011 Mathematics Vector Subspaces Subspace Definition A non-empty subset W ( W ≠ ) of a vector space V is a subspace of V when the following conditions pertain: W is a vector space under addition in V W is a vector space under scalar multiplication in V Subspace Test For a non-empty subset W ( W ≠ ) of a vector space V, W is a subspace of V if-and-only if the following pertain: If u and v are in W, then u + v is in W If u is in W and c is any scalar, then cu is in W Zero Subspace W = { 0 } Section 4.3, (Pg. 162) Section 4.3, (Pg. 162) Section 4.3, (Pg. 163) Vector Space
- 152. © Art Traynor2011 Mathematics Vector Subspaces W 1 Polynomial functions W 5 Functions W 2 Differentiable functions W 3 Continuous functions W 4 Integrable functions W5 = Vector Space " f Defined on [ 0, 1 ] W4 = Set " f Integrable on [ 0, 1 ] W3 = Set " f Continuous on [ 0, 1 ] W2 = Set " f Differentiable on [ 0, 1 ] W2 = Set " Polynomials Defined on [ 0, 1 ] W1 W2 W3 W4 W5 W 1 – Every Polynomial function is Differentiable W1 W2 W 2 – Every Differentiable function is Continuous W2 W3 W 3 – Every Continuous function is Integrable W3 W4 W 4 – Every Integrable function is a Function W4 W5 Function Space Section 4.3, (Pg. 164) Vector Space
- 153. © Art Traynor2011 Mathematics Vector Subspaces U V W V W Properties of Scalar Multiplication If V & W are both subspaces of a vector space U, then the intersection of V & W ( V W ) is also a subspace of U Vector Space
- 154. © Art Traynor2011 Mathematics Linear Combination of Vectors ( Definition ) A vector v in a vector space V with scalars c = ( ci , ci+1 ,…cn – 1 , cn ) is a Linear Combination of the vectors ( ui , ui+1 ,…un – 1 , un ) expressed as: v = ci ui + ci+1 ui+1 +…+ cn –1 un – 1 + cn un Section 4.1, (Pg. 152) Section 4.4, (Pg. 169) Linear Combination Example: S = { ( 1 , 3 , 1 ) , ( 0 , 1 , 2 ) , ( 1 , 0 , – 5 ) } v1 v2 v3 v1 = 3v2 + v3 v1 = 3( 0 , 1 , 2 ) + (1 , 0 , – 5 ) v1 = ( 0 , 3 , 6 ) + (1 , 0 , – 5 ) v1 = ( 0 + 1 ) , ( 3 + 0 ) , ( 6 – 5 ) = ( 1 , 3 , 1 ) V1 can be expressed as a combination of components of the other two vectors in the set S $ c F | ( cv2 v3 ) v1 $ c F | ( cv2 v3 ) v1 Chump Alert: Each of the Vector Space operations (e.g. summation, difference, scalar multiplications) is a species of Linear Combination Vector Space
- 155. © Art Traynor2011 Mathematics Spanning Set of a Vector Space Given S = { vi , vi+1 ,…vk – 1 , vk }, a subset of vector space V the set S is a Spanning Set of V when every vector in V can be written as a Linear Combination of vectors in S, “ S spans V ” x = ci vi + ci+1 vi+1 +…+ cn –1 vn – 1 + cn vn Section 4.1, (Pg. 152) Section 4.4, (Pg. 169) Section 4.4, (Pg. 171) Span Sounds analogous to the power set of a vector equaton? The set of unit vectors S = { i, j, k } are the minimal spanning set (Basis) for Rn Vector Space P ( S ) = { , S, cn û } Vector Space
- 156. © Art Traynor2011 Mathematics Vectors Vector ( Euclidean ) Aka: Geometric or Spatial Vector From the Latin Vehere (to carry) x y Position Vector O θ A ( ax , ay ) Unit Vector ( Components ) Any vector in PVF can be expressed as a scalar product of the vector sum of its unit ( multiplicative scalar identity ) components î = 〈 1, 0 〉 , ĵ = 〈 1, 0 〉 a ĵ î x y Position Vector O θ A ( ax , ay ) ay ĵ ax î a a = ax î + ay ĵ PVF: Position Vector Form c ( î ) = 〈 c1, c0 〉 , c ( ĵ ) = 〈 c 0, c 1 〉 ax ( î ) = 〈 ax 1, ax 0 〉 , ay ( ĵ ) = 〈 ay 0, ay 1 〉 Unit Vector The set of unit vectors S = { i, j, k } are the minimal spanning set (Basis) for Rn Vector Space
- 157. © Art Traynor2011 Mathematics Vectors Vector ( Euclidean ) Aka: Geometric or Spatial Vector Aka: Versor (Cartesian) x y Position Vector O θ Normalized Unit Vector A normalized unit vector ( NUV ) is the vector of unitary magnitude corresponding to the set of all vectors which share its direction ĵ î x y Position Vector O θ A ( ax, ay ) a2 ĵ a1 î A ( ax , ay ) û ( ax , ay )║a ║ 1 ║a ║ 1 Any vector can be specified by the scalar product of its corresponding normalized unit vector and its magnitude ( identity ) a = ax î + ay ĵ 1 ║a ║ a ║a ║ û = a = The NUV of a vector is the scalar product of the reciprocal of its magnitude ĵ î û ûa Normalized Unit Vector
- 158. © Art Traynor2011 Mathematics Span of a Set Given S = { vi , vi+1 ,…vk – 1 , vk }, is a set of vectors in a vector space V with scalars c = ( ci , ci+1 ,…cn – 1 , cn ) then the span of S is the set of all Linear Combinations of the vectors in S, span ( S ) = { ci vi + ci+1 vi+1 +…+ ck –1 vik– 1 + ck vk } Section 4.4, (Pg. 172) The span of S is denoted: span ( S ) or span { vi , vi+1 ,…vk – 1 , vk } When span ( S ) = V, it is said that: V is spanned by { vi , vi+1 ,…vk – 1 , vk } or S spans V Span P ( S ) = { , S, cn û } The set of unit vectors S = { i, j, k } are the minimal spanning set (Basis) for Rn Vector Space Vector Space
- 159. © Art Traynor2011 Mathematics Span ( S ) as Subspace of V Span Given S = { vi , vi+1 ,…vk – 1 , vk }, is a set of vectors in a vector space V then the span of S span ( S ) or span { vi , vi+1 ,…vk – 1 , vk } is a Subspace of V Section 4.4, (Pg. 172) The span of S denoted span ( S ) or span { vi , vi+1 ,…vk – 1 , vk } is the smallest Subspace of V containing Ssuch that every other Subspace of V containing S Must also contain span ( S ) It is not sufficiently obvious from this that the minimal cardinality of Spanning Set corresponds precisely to the dimension of the space The set of unit vectors S = { i, j, k } are the minimal spanning set (Basis) for Rn Vector Space P ( S ) = { , S, cn û } W = span ( S ) = { Σ ci vi | k N , vi S , ci F }i = 1 k V Sk – i SiSi W Si Si F = {ci , ci+1…ck – 1 , ck } Vector Space
- 160. © Art Traynor2011 Mathematics Linear Independence Linear Independence Section 4.4, (Pg. 173) Given S = { vi , vi+1 ,…vk – 1 , vk }, within vector space V over a field of scalars c = ( ci , ci+1 ,…cn – 1 , cn ) a vector equation of the form ci vi + ci+1 vi+1 +…+ ck –1 vik– 1 + ck vk = 0 expresses Linear Dependence if the solution set includes at least one non-zero solution and if the vector equation features only the trial solution 0 = ( ci , ci+1 ,…cn – 1 , cn ) it is said to express Linear Independence By setting it equal to the zero vector, are we looking for solutions to a homogenous system? Example: S = { ( 1 , 3 , 1 ) , ( 0 , 1 , 2 ) , ( 1 , 0 , – 5 ) } v1 v2 v3 0 = c1v1 + c2v2 + c3v3 0 = c1 ( 1 , 3 , 1 ) + c2 ( 0 , 1 , 2 ) + c3 ( 1 , 0 , – 5 ) ci = { 1 , – 3 , – 1 } 0 = ( 1 , 3 , 1 ) + ( 0 , – 3 , – 6 ) + ( – 1 , 0 , 5 ) 0 = xi ( 1 + 0 – 1 ) , yi ( 3 – 3 + 0 ) , zi ( 1 – 6 + 5 ) 0 = xi ( 0 ) , yi ( 0 ) , zi ( 0 ) It is not sufficiently obvious from this that the cardinality of the maximum set of linearly independent vectors corresponds precisely to the dimension of the space xi yi zi = x1 , y1, z1 x2 , y2, z2 x3 , y3, z3 Vector Space
- 161. © Art Traynor2011 Mathematics Linear Independence Linear Independence Section 4.4, (Pg. 173) Given S = { vi , vi+1 ,…vk – 1 , vk }, within vector space V over a field of scalars c = ( ci , ci+1 ,…cn – 1 , cn ) a vector equation of the form ci vi + ci+1 vi+1 +…+ ck –1 vik– 1 + ck vk = 0 expresses Linear Dependence if the solution set includes at least one non- zero solution and if the vector equation features only the trial solution By setting it equal to the zero vector, are we looking for solutions to a homogenous system? Example: S = { ( 1 , 3 , 1 ) , ( 0 , 1 , 2 ) , ( 1 , 0 , – 5 ) } v1 v2 v3 0 = c1v1 + c2v2 + c3v3 0 = c1 ( 1 , 3 , 1 ) + c2 ( 0 , 1 , 2 ) + c3 ( 1 , 0 , – 5 ) ci = { 1 , – 3 , – 1 } 0 = ( 1 , 3 , 1 ) + ( 0 , – 3 , – 6 ) + ( – 1 , 0 , 5 ) 0 = xi ( 1 + 0 – 1 ) , yi ( 3 – 3 + 0 ) , zi ( 1 – 6 + 5 ) 0 = xi ( 0 ) , yi ( 0 ) , zi ( 0 ) It is not sufficiently obvious from this that the cardinality of the maximum set of linearly independent vectors corresponds precisely to the dimension of the space 0 = 1 v1 + – 3 v2 + – 1 v3 xi yi zi = x1 , y1, z1 x2 , y2, z2 x3 , y3, z3 Vector Space
- 162. © Art Traynor2011 Mathematics Linear Independence Linear Independence Section 4.4, (Pg. 173) Given S = { vi , vi+1 ,…vk – 1 , vk }, within vector space V over a field of scalars c = ( ci , ci+1 ,…cn – 1 , cn ) a vector equation of the form ci vi + ci+1 vi+1 +…+ ck –1 vik– 1 + ck vk = 0 expresses Linear Dependence if the solution set includes at least one non-zero solution and if the vector equation features only the trial solution 0 = ( ci , ci+1 ,…cn – 1 , cn ) it is said to express Linear Independence Example: A Visitor to New York City asks directions to Carnegie Hall. He is instructed to proceed 3-blocks North then 4-blocks East. you are here 3N 4E 5NE These two directions are sufficient to allow him to reach his destination The system of vectors is Linearly Independent and corresponds to the dimension of the space (which would be elsewise if he needed to go to the 6th Floor) Adding that the destination is 5-blocks Northeast renders the system of vectors Linearly Dependent (as one of the vectors can be expressed as a Linear Combination of the other two). Vector Space
- 163. © Art Traynor2011 Mathematics Linear Independence Linear Independence Example: A Visitor to New York City asks directions to Carnegie Hall. He is instructed to proceed 3-blocks North then 4-blocks East. you are here 3N 4E 5NE These two directions are sufficient to allow him to reach his destination The system of vectors is Linearly Independent and corresponds to the dimension of the space (which would be elsewise if he needed to go to the 6th Floor) Adding that the destination is 5-blocks Northeast renders the system of vectors Linearly Dependent (as one of the vectors can be expressed as a Linear Combination of the other two). Vector Space Ban the Hypotenuse! It is not necessary. Any triangle can be described completely by a & b. C is not necessary
- 164. © Art Traynor2011 Mathematics Linear Independence Test Linear Independence Section 4.4, (Pg. 174) For S = { vi , vi+1 ,…vk – 1 , vk }, constituting a set of vectors within vector space V , to determine the linear independence of S: ci vi + ci+1 vi+1 +…+ ck –1 vik– 1 + ck vk = 0 State the set as a homogenous equation equivalent to a sum of vector products ( of a scalar coefficient and the respective constituent vector components) Perform GJE (Gauss-Jordan Elimination) to determine if the system has a Unique Solution Where only the trivial solution ci = 0 with satisfy the system, the Set S is demonstrated to be Linear Independent where the system is also satisfied by one or more non-trivial solutions, the Set S is demonstrated to be Linear Independent Vector Space
- 165. © Art Traynor2011 Mathematics Linear Dependence & Linear Combination Linear Independence Section 4.4, (Pg. 176) For S = { vi , vi+1 ,…vk – 1 , vk }, k ≥ 2 constituting a set of vectors within vector space V , the set can be demonstrated to be Linear Dependent if-and-only-if At least one element vector of the set can be expressed as a linear combination of any of the other vectors in the set For S = { v , u } constituting a set of vectors within vector space V , the set can be demonstrated to be Linear Dependent if-and-only-if One element vector of the set can be expressed as a scalar multiple of the other Vector Space
- 166. © Art Traynor2011 Mathematics Basis Criteria Basis Section 4.5, (Pg. 180) For S = { vi , vi+1 ,…vk – 1 , vk }, constituting a set of vectors within vector space V , the set can be demonstrated to form a Basis for the vector set if S spans V S is Linear Independent Infinite Dimensional examples: Vector Space P ( all polynomials ) Vector Space C ( all continuous functions) Finite Dimensional examples: Zero Vector { 0 } Standard Basis for an n x n matrix features a diagonal populated with ones with all other entries occupied by zeros Vector Space
- 167. © Art Traynor2011 Mathematics Basis Representation - Uniqueness Basis Section 4.5, (Pg. 182) For S = { vi , vi+1 ,…vk – 1 , vk }, constituting a set of vectors within vector space V , and forming a basis for that vector space, then Every element vector can only be expressed as a unique Linear Combination of the constituent vectors If there were more than one, their difference would yield the zero vector which would violate the Basis criteria requiring Linear Independence Vector Space
- 168. © Art Traynor2011 Mathematics Basis Cardinality Basis Section 4.5, (Pg. 184) For S = { vi , vi+1 ,…vn – 1 , vn }, constituting a set of vectors within vector space V , and forming a basis for that vector space with precisely “ n ” vectors, then Every basis for V will include precisely “ n ” vectors Basis Cardinality - Maximum For S = { vi , vi+1 ,…vn – 1 , vn }, constituting a set of vectors within vector space V , and forming a basis for that vector space, then any set in Rn with k vectors, where k > n will be Linear Dependent Section 4.5, (Pg. 183) Vector Space Easier way to think of it… Refer to the “Standard Basis” for the vector space (e.g. R3) Any alternative Basis must then include precisely that many vectors
- 169. © Art Traynor2011 Mathematics Vector Space Dimension Vector Space Dimension Section 4.5, (Pg. 185) For S = { vi , vi+1 ,…vk – 1 , vk }, constituting a set of vectors within vector space V , and forming a basis for that vector space with precisely “ n ” vectors, then The number “ n ” is denoted as the Dimension of V , or dim ( V ) dim ( Rn ) = n dim ( Pn ) = n + 1 dim ( Mm,n ) = m · n To determine the Dimension of a Subspace W of vector space V Identify a set S of Linear Independent vectors that Span subspace W The set S thus identified is a Basis for the subset W The dimension of subspace W is thus the count ( or cardinality ) of the vectors in the Basis
- 170. © Art Traynor2011 Mathematics Vector Space Dimension Vector Space Dimension Section 4.5, (Pg. 185) To determine the Dimension of a Subspace W of vector space V Identify a set S of Linear Independent vectors that Span subspace W The set S thus identified is a Basis for the subset W The dimension of subspace W is thus the count ( or cardinality ) of the vectors in the Basis Example: W = { ( d , c – d , c ) } Inspection reveals that this subspace has precisely two factors: c and d Wc = { c ( 0c · d , 1c – d , 1c ) } Wc = { c ( 0 , 1, 1 ) } Wd = { d ( c · 0d , c – 1d , c · 0d ) } Wd = { d ( 0 , – 1, 0 ) } Wcd = { c ( 0 , 1, 1 ) + d ( 0 , – 1, 0 ) } The Dimension of the subspace is thus two because we have precisely two vectors spanned by the set S = { (0,1,1), (1, –1, 0) }, which can be shown to be Linear Independent and thus a Basis for W
- 171. © Art Traynor2011 Mathematics Basis Test for an n-Dimensional Space Basis Section 4.5, (Pg. 186) For a vector space V of Dimension n if S = { vi , vi+1 ,…vk – 1 , vk }, constitutes a set of Linear Independent vectors in V, then: S is a Basis for V Vector Space If S spans V then S is a basis for V
- 172. © Art Traynor2011 Mathematics Subspace Spans of Vector Representations Vector Representations Section 4.6, (Pg. 189) Matrices A = A a11 a21 am1 . . . a12 a22 am2 . . . . . . . . . . . . . . . a1n a2n amn . . . ( a11 , a12 , … a1n ) Row Vectors of A ( a21 , a22 , … a2n ) . . . ( a21 , a22 , … a2n ) A = A a11 a21 am1 . . . a12 a22 am2 . . . . . . . . . . . . . . . a1n a2n amn . . . a11 a21 am1 . . . a12 a22 am2 . . . . . . . . . . . . . . . a1n a2n amn . . . Column Vectors of A For a matrix A of m-rows and n-columns Amx n The Row Space ( a subspace of Rn ) is spanned by the row vectors of A The Column Space ( a subspace of Rn ) is spanned by the column vectors of A
- 173. © Art Traynor2011 Mathematics Row Space for Row Equivalent Matrices Matrix Sub Spaces Section 4.6, (Pg. 190) Another Row-Equivalent m-by-n matrix Bm x n will share the same Row Space with Amx n Vector Space Row Space Basis Section 4.6, (Pg. 190) For a matrix A of m-rows and n-columns Amx n Another Row-Equivalent m-by-n matrix Bm x n expressed in Row Echelon Form ( REF ) will feature non-zero Row Vectors constituting a Basis for Amx n For a matrix A of m-rows and n-columns Amx n Row/Column Space Dimensional Equivalence Section 4.6, (Pg. 192) Both Row Space and Column Space share the same value for their respective Dimensions For a matrix A of m-rows and n-columns Amx n
- 174. © Art Traynor2011 Mathematics Matrix Rank Matrix Rank Section 4.6, (Pg. 193) The Dimension of a Row or Column Space defines the Rank of the Matrix, denoted rank ( A ) Vector Space For a matrix A of m-rows and n-columns Amx n
- 175. © Art Traynor2011 Mathematics Matrix Nullspace and Nullity Nullspace Section 4.6, (Pg. 194) The solution set for the system forms a subspace of Rn designated as the Nullspace of A and denoted as N ( A ) Vector Space For a homogenous linear system Ax = 0 where A is an matrix of m-rows and n-columns Amx n , x = [ xi , xi+1 ,…xn – 1 , xn ] T is a column vector of unknowns 0 = [ 0 , 0 , … 0 , 0 ] T is the zero vector in Rm N ( A ) = { x Rn | Ax = 0 } The Dimension of the Nullspace of A is designated as its Nullity dim ( N ( A ) )
- 176. © Art Traynor2011 Mathematics 1 3 1 2 6 2 – 2 – 5 0 1 4 3 Standard Coefficient Matrix Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix A = Standard Matrix Form (SMF) is that arrangement of matrix elements in which constituent rows are populated with individual expressions (equations) constituting the linear system (of which the matrix is a representation), and in which the columns are arrayed such that each is populated by a distinct “unknown” (variable) the entries of which are populated by their individual coefficients. Section 6.3, (Pg. 314) Section 4.6, (Pg. 195), Example 7 Designate each row with an Uppercase Alpha Character…this will allow the Elementary Row Operations (EROs) to be performed to be described in a summarized algebraic fashion. ① 1 3 1 2 6 2 – 2 – 5 0 1 4 3 A1 B1 C1
- 177. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Utilize Permutation to manipulate the simplest (most easily reduced) rows into the primary and higher positions in the matrix 1 3 1 2 6 2 – 2 – 5 0 1 4 3 A1 B1 C1 ② 3 1 6 2 – 5 0 4 3 B1 C1 The book prefers Row Three to be in the first position, so we’ll begin by permuting Rows One and Three A ⇌ C1 1 2 – 2 1A1 Once “fixed” at a value of one (1) circle this entry as an established pivot
- 178. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Continued…② 3 1 6 2 – 5 0 4 3 B1 C1 1 2 – 2 1A1 A ⇌ B1 3 1 6 2 – 5 0 4 3 B1 C1 1 2 – 2 1A1 Row Three appears to present a very simple reduction (by simply scaling it by a factor of negative one), so it too should be permuted with Row Two for clarity and ease of succeeding EROs operations. A ⇌ B1
- 179. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 From this established “Pivot” move down to the next row and render the entry there into a “zero” using EROs 1 ③ So now we perform our first real operation on the system. It can be noted “by inspection” that Row Two scaled by a factor of – 1 can make quick work of yielding zeros just where we want them – 1A1 = A1 3 6 2 – 5 0 4 3 B1 C1 1 2 – 2 1A1 – 1 – 2 2 – 1– 1A1 1 2 0 3C1 0 0 2 2A1 Adding the scaled Row Two to Row One we get a cleaned up replacement for Row Two in the next evolution of the GJE reduction C2 – 1A1 = A1
- 180. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Continued… 1 ③ So now we perform our first real operation on the system. It can be noted “by inspection” that Row Two scaled by a factor of – 1 can make quick work of yielding zeros just where we want them – 1A1 = A1 3 6 2 – 5 0 4 3 B1 C1 1 2 – 2 1A1 – 1 – 2 2 – 1– 1A1 1 2 0 3C1 0 0 2 2A1
- 181. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Continued…③ 1 3 6 2 – 5 0 4 3 B1 C1 0 0 2 2A1 Inspection suggests that Row one scaled by a factor of – 3 would allow for a handy reduction of Row Three into the desired zero element in the first column position B – 3C1 = B1 – 3 – 6 0 – 9– 3C1 0 0 – 5 – 5B1 3 6 – 5 4B1
- 182. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Having “zeroed-out” the first column, we proceed to the right… 1 2 0 3C1 0 0 2 2A1 0 0 – 5 – 5B1 Rows Two and Three can be dispatched with a scaling by their product 5A1 = A2 2B1 = B2 0 0 10 10A2 0 0 – 10 – 10B2 A simple summation of Rows Two and Three will thereby reduce this matrix to a very tame state A2 + B2 = B3 0 0 0 0B3 ④
- 183. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Continued… 1 2 0 3C1 0 0 0 0B3 0 0 10 10A2 Finally we note that Row 2 can be reduced by a common factor ( 10 ) to yield a maximally simplified row A2 = A3 1 10 1 2 0 3C1 0 0 0 0B3 0 0 1 1A3 Arrived!! ④
- 184. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Now we look to parameterizing the system. We look for coefficients ≠ 1 to prefer for selection so as to maximally simplify substitution back into the system. 1 2 0 3 0 0 0 0 0 0 1 1 x1 + 2x2 + 2x3 + 3x4 = 0 System of Linear Equations x1 + 2x2 + 1x3 + 1x4 = 0 ⑤ x1 + 2s2 + 2x3 + 3t4 = 0 x1 + 2x2 + 1x3 + 1t4 = 0 x1 + 2x2 + 1x3 + 1x2 = s x1 + 2x2 + 1x3 + 1x4 = t B =
- 185. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Continued… 1 2 0 3 0 0 0 0 0 0 1 1 System of Linear Equations⑤ x1 + 2s2 + 2x3 + 3t4 = 0 x1 + 2x2 + 1x3 3 + 1t4 = 0 x1 + 2x2 + 1x3 + 1x2 = s x1 + 2x2 + 1x3 + 1x4 = t x1 + 2s2 + 2x3 + 3t4 = – 2s – 3t x1 + 2s2 + 2x3 + x1 = – 2s – 3t x1 + 2s2 + 2x3 + x3 = – t B =
- 186. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 Now we re-configure the system into Partitioned Matrix Form (PMF) arraying the “unknowns” (variables) and the parameterized solutions into respective column vectors 1 2 0 3 0 0 0 0 0 0 1 1 System of Linear Equations x1 + 2s2 + 2x3 + 3t4 = 0 x1 + 2x2 + 1x3 3 + 1t4 = 0 x1 + 2x2 + 1x3 + 1x2 = s x1 + 2x2 + 1x3 + 1x4 = t x1 + 2s2 + 2x3 + 3t4 = – 2s – 3t x1 + 2s2 + 2x3 + x1 = – 2s – 3t x1 + 2s2 + 2x3 + x3 = – t ⑥ – 2s 1s 0 – 3t 0t – 1t 0 1t x = – 2 1 0 – 3 0 – 1 0 1 = s + t B =
- 187. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 1 2 0 3 0 0 0 0 0 0 1 1 – 2s 1s 0 – 3t 0t – 1t 0 1t x = – 2 1 0 – 3 0 – 1 0 1 = s + t B = N ( A ) = { ( – 2 , 1, 0, 0 ) , ( – 3 , 0, – 1, 1 ) } The Nullspace of the matrix thus forms a basis, synonymous with the Solution Space of the homogenous system Ax = 0 , ALL solutions of which are Linear Combinations of these two vectors 1 3 1 2 6 2 – 2 – 5 0 1 4 3 Standard Coefficient Matrix A = “ Row Equivalent” ( REF Basis Matrix ) v1 v2 v3 w1 w2
- 188. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 – 2s 1s 0 – 3t 0t – 1t 0 1t x = – 2 1 0 – 3 0 – 1 0 1 = s + t dim ( N ( A ) ) = 2 ; i.e. { w1 , w2 } The Dimension of the Nullspace is equivalent to the cardinality of the non-zero “Row Equivalent” REF Basis matrix row constituents, which is otherwise known as the Nullity of the Matrix 1 3 1 2 6 2 – 2 – 5 0 1 4 3 Standard Coefficient Matrix A = v1 v2 v3 1 2 0 3 0 0 0 0 0 0 1 1B = “ Row Equivalent” ( REF Basis Matrix ) w1 w2
- 189. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 – 2s 1s 0 – 3t 0t – 1t 0 1t x = – 2 1 0 – 3 0 – 1 0 1 = s + t The Rank and Nullity of the Solution Space “Row Equivalent” REF Basis matrix can be determined by the number of columns featuring leading “ ones ” ( i.e. two, the Rank , columns 1 & 3 ) and those columns which correspond to the free variable column vectors ( i.e. two, the Nullity , columns 2 & 4 ) . 1 3 1 2 6 2 – 2 – 5 0 1 4 3 Standard Coefficient Matrix A = v1 v2 v3 1 2 0 3 0 0 0 0 0 0 1 1B = “ Row Equivalent” ( REF Basis Matrix ) w1 w2
- 190. © Art Traynor2011 Mathematics Nullspace Vector Space Matrix Nullspace, Rank, and Nullity Example: find the nullspace of the matrix Section 4.6, (Pg. 195), Example 7 – 2s 1s 0 – 3t 0t – 1t 0 1t x = – 2 1 0 – 3 0 – 1 0 1 = s + t For a matrix A of m-rows and n-columns Amx n then, the total number of columns ( n ) is the sum of the Rank and Nullity of the Solution Space “Row Equivalent” REF Basis matrix ( i.e. Rank two, columns 1 & 3 and Nullity two, columns 2 & 4 ) . 1 3 1 2 6 2 – 2 – 5 0 1 4 3 Standard Coefficient Matrix A = v1 v2 v3 1 2 0 3 0 0 0 0 0 0 1 1B = “ Row Equivalent” ( REF Basis Matrix ) w1 w2
- 191. © Art Traynor2011 Mathematics Solution Space Dimension – Matrix Metrics Solution Space Section 4.6, (Pg. 196) Vector Space For a matrix A composed of m-rows and n-columns Amx n , of Rank “ r ” , the Dimension of the Solution Space Ax = 0 is The difference of the matrix column cardinality “ n ” less the matrix Rank “ r ” , or n – r The column space cardinality “ n ” is equal to n = rank ( A ) – nullity ( A ) or n = rank ( A ) – dim ( N ( A ) )
- 192. © Art Traynor2011 Mathematics Solution Space for Non-Homogenous LE Systems ( nHLES ) Solution Space Section 4.6, (Pg. 197) Vector Space For a matrix A composed of m-rows and n-columns Amx n , the non-homogenous system Ax = b with particular solution xp can be expressed in the form x = xp + xh where Ax = 0 represents the associated homogenous system ( HLES ) with solution xh The zero vector 0 = { 0 , 0 , … 0 , 0 } cannot be a solution to an nHLES therefore the set of all solution vectors to the nHLES cannot structure a subspace.
- 193. © Art Traynor2011 Mathematics 1 3 1 0 1 2 – 2 – 5 0 1 0 – 5 Augmented Coefficient Matrix Solution Space Vector Space Particular Solution of a Non-Homogenous LE System ( nHLES ) Example: Find the set of all solution vectors of the LE system Section 4.6, (Pg. 197), Example 9 5 8 – 9 1x1 + 0x2 – 2x3 + 1x4 = 5s 3x1 + 1x2 – 5x3 + 0x4 = 8s 1x1 + 2x2 – 0x3 – 5x4 = – 9s Non-Homogenous System of Linear Equations ( nHLES) A = v1 v2 v3 x1 x2 x3 x4 b Section 6.3, (Pg. 314) The set of all solution vectors to an nHLES of the form Ax = b entails expressing the system as a matrix A composed of m-rows and n- columns Amx n , arrayed in augmented Standard Matrix Form ( aSMF ). ① The aSMF matrix features Row Vectors { v i , v i + 1 ,…vn – 1 , v n } corresponding to the constituent equations of the nHLES. Column Vectors in the aSMF are arrayed to collect each distinct “ unknown ” ( variable ) term { xi , x i + 1 ,…xn – 1 , x n } with each entry populated by their individual coefficients and adjoined by the solution vector of constants { b }.
- 194. © Art Traynor2011 Mathematics 1 3 1 0 1 2 – 2 – 5 0 1 0 – 5 Augmented Coefficient Matrix Solution Space Vector Space Particular Solution of a Non-Homogenous LE System Example: Find the set of all solution vectors of the LE system A = Section 4.6, (Pg. 197), Example 9 Designate each row with an Uppercase Alpha Character…this will allow the Elementary Row Operations (EROs) of the Gauss-Jordan Elimination (GJE) to be applied to be described in a summarized algebraic fashion. 5 8 – 9 1 3 1 0 1 2 – 2 – 5 0 1 0 – 5 5 8 – 9 A1 B1 C1 ② v1 v2 v3 x1 x2 x3 x4 b Gauss-Jordan Elimination (GJE) is an algorithmic scheme applied to a Standard Matrix Form (SMF) representation of a system of Linear Equations resulting in a “row- equivalent” reduced matrix on which the main diagonal entries are all “ones” (pivots in Row Echelon Form - REF) and all entries above and below the “pivots” are populated by “zeros” or Reduced Row Echelon Form (RREF). Section 1.2, (Pg. 19)
- 195. © Art Traynor2011 Mathematics Vector Space Matrix Nullspace, Rank, and Nullity Investigate permutation as a strategy to manipulate the simplest (most easily reduced) rows into the primary and higher positions in the matrix In this case, the first row already features a value of one in the first position along the main diagonal, so all is well to proceed to the next step in the GJE reduction process to arrive at a “Row Equivalent” REF Basis matrix… 1 3 1 0 1 2 – 2 – 5 0 1 0 – 5 5 8 – 9 A1 B1 C1 1 3 1 0 1 2 – 2 – 5 0 1 0 – 5 5 8 – 9 A1 B1 C1 Inspection seems to suggest that scaling Row Three by – 3 and summing with Row Two would yield an appealing reduction of Row Two into the form where the leading entry in the row is rendered into a zero. B2 – 3C1 = B1 Section 4.6, (Pg. 197), Example 9 Solution Space From this established “Pivot” move down to the next row and render the entry there into a “zero” using EROs Example: Find the set of all solution vectors of the LE system ③ ④
- 196. © Art Traynor2011 Mathematics Vector Space Matrix Nullspace, Rank, and Nullity Section 4.6, (Pg. 197), Example 9 Solution Space Example: Find the set of all solution vectors of the LE system Continued… 3 – 3 1 – 6 – 5 0 0 15 8 27 B1 – 3C1 0 – 5 – 5 15 35B1 1 1 0 2 – 2 0 1 – 5 5 – 9 A1 C1 From this established “zero” move down to the next row and render the entry there into a “zero” using EROs 0 – 5 – 5 15 35B1 ④ Inspection suggests that scaling Row Two by – 1 and summing with Row One would yield an appealing reduction of Row Three into the form where the leading entry in the row is rendered into a zero. C2 – 1A1 = C1 ⑤
- 197. © Art Traynor2011 Mathematics Vector Space Matrix Nullspace, Rank, and Nullity Section 4.6, (Pg. 197), Example 9 Solution Space Example: Find the set of all solution vectors of the LE system Continued… C1 – 1A1 0 2 2 – 6 – 14C1 1 0 0 2 – 2 2 1 – 6 5 – 14 A1 C1 With the second column entry in the first row already fixed at a “zero” value, we can proceed down to the next row and render the entry there into a “zero” using EROs 0 – 5 – 5 15 35B1 Inspection suggests that scaling Row Two by a factor of negative one–fifth would reduce Row Two to the desired “Pivot” value of “one” at the next position down along the main diagonal. ⑤ 1 2 0 – 5 – 9 – 1 0 2 – 1 – 5 – B1 = B2 1 5 ⑥
- 198. © Art Traynor2011 Mathematics Vector Space Matrix Nullspace, Rank, and Nullity Section 4.6, (Pg. 197), Example 9 Solution Space Example: Find the set of all solution vectors of the LE system Continued… B1 0B2 1 0 0 2 – 2 2 1 – 6 5 – 14 A1 C1 With the second “pivot” fixed along the main diagonal, we can proceed down to the next row and render the entry there into a “zero” using EROs B2 Now for the coup de grâce – we notice that Row Three is a scalar multiple of Row Two, thus scaling Row Two by a factor of – 2 and summing with Row Three will yield a new Row Three populated entirely of all zero entries 0 1 1 – 3 – 7 ⑥ – B1 1 5 0 – 5 – 5 15 35 1 1 – 3 – 7 0 1 1 – 3 – 7 ⑦ C1 – 2B2 = C2
- 199. © Art Traynor2011 Mathematics Vector Space Matrix Nullspace, Rank, and Nullity Section 4.6, (Pg. 197), Example 9 Solution Space Example: Find the set of all solution vectors of the LE system Continued… C1 0C2 1 0 0 0 – 2 0 1 0 5 0 We now need to examine this reduced matrix to determine an appropriate parameterization of the Solution Vectors of the nHLES The Gauss-Jordan Elimination (GJE) is now complete as we have arrived at a “row-equivalent” matrix to A, now designated B, as the remaining “non- zero” row vectors constitute a Basis set for the system (?) …QED 0 – 2 – 2 6 14 0 0 0 0 0 1 1 – 3 – 7 0 2 2 – 6 – 14 – 2B2 ⑦ B = ⑧ The RREF matrix, thus transformed from the root SMF or TMF (conventionally designated “ A ” with entries “ ci ” ) is now restated as RREF matrix “ B ” with entries “ di ” Section 4.6, (Pg. 193)
- 200. © Art Traynor2011 Mathematics Vector Space Matrix Nullspace, Rank, and Nullity Section 4.6, (Pg. 197), Example 9 Solution Space Example: Find the set of all solution vectors of the LE system X3 and X4 with coefficients ≠ 1 are the best candidates for parameterization. 1 0 0 0 – 2 0 1 0 5 0 0 1 1 – 3 – 7B = RREF ( Basis Set ) for nHLES 1x1 + 0x2 – 2x3 + 1x4 = 5s 3x1 + 1x2 + 1x3 – 3x4 = – 7s x1 + 2x2 + 1x3 + 1x3 = s x1 + 2x2 + 1x3 + 1x4 = t 1x1 + 0x2 – 2s + 1t4 = 5s 1x1 + 0x2 – 2s + 1t4 = 2s – 1t + 5s 3x1 + 1x2 + 1s3 – 3t4 = – 7s 3x1 + 1x2 + 1s3 – 3t4 = – 1ss +3t – 7s 2s – 1s 1s – 1t + 3t + 0t 0s + 1t x = + 5 – 7 + 0 x1 x2 x3 x4 = + 0 ⑨
- 201. © Art Traynor2011 Mathematics Vector Space Matrix Nullspace, Rank, and Nullity Solution Space Example: Find the set of all solution vectors of the LE system Finally we state the solution vectors in terms of their correspondence with the solution for the associated homogenous system 1 0 0 0 – 2 0 1 0 5 0 0 1 1 – 3 – 7B = x = x1 x2 x3 x4 = + 5 – 7 + 0 + 0 – 1t + 3t + 0t + 1t 2s – 1s 1s 0s + + x = x1 x2 x3 x4 = s + 5 – 7 + 0 + 0 – 1t + 3t + 0t + 1t 2s – 1s 1s 0s + t + RREF ( Basis Set ) for nHLES x1 + 2x2 + 1x3 + 1x3 = s x1 + 2x2 + 1x3 + 1x4 = t 1x1 + 0x2 – 2s + 1x1 = 2s – 1t + 5s 3x1 + 1x2 + 1s3 – 3x2 = – 1ss +3t – 7s 10 xi u1 u2 xp x = su1 + tu2 + xp System Solution(s) x = xh + xp xh = su1 + tu2 Ax = 0 ( HLES ) Ax = b ( nHLES ) Section 4.6, (Pg. 197), Example 9 xh thus represents an arbitrary vector in the solution space of Ax = 0 Section 4.6, (Pg. 197)
- 202. © Art Traynor2011 Mathematics 1 3 1 0 1 2 – 2 – 5 0 1 0 – 5 Augmented Coefficient Matrix Solution Space Vector Space Particular Solution of a Non-Homogenous LE System ( nHLES ) Example: Find the set of all solution vectors of the LE system Section 4.6, (Pg. 197), Example 9 5 8 – 9 A = v1 v2 v3 x1 x2 x3 x4 b 1 0 0 0 – 2 0 1 0 5 0 0 1 1 – 3 – 7B = w1 w2 The wi vectors thus form a Basis for the Row Space of A, or the subspace spanned by S = { vi , vi + 1 ,…vn – 1 , vn } Section 4.6, (Pg. 191) Continued… “ Row Equivalent” ( REF Basis Matrix ) Only columns featuring a “leading one” ( “Pivot”) in the RREF matrix B are Linear Independent Section 4.6, (Pg. 192) Linear Independent Linear Dependent x1 x2 x3 x4 b x = x1 x2 x3 x4 = s + 5 – 7 + 0 + 0 – 1t + 3t + 0t + 1t 2s – 1s 1s 0s + t + 10
- 203. © Art Traynor2011 Mathematics Solution Space Consistency Solution Space Section 4.6, (Pg. 198) Vector Space For a matrix A composed of m-rows and n-columns Amx n , of which Ax = b defines a particular solution xp to the nHLES The nHLES is consistent if-and-only-if The solution vector b is among the Column Space of A The solution vector b = [ bi , bi+1 ,…bn – 1 , bn ] T represents a Linear Combination of the columns of A The solution vector b is among those populating the Subspace Rm Spanned by the columns of A There is an additional implication that the respective Ranks of the nHLES coefficient and augmented matrices are equivalent Section 4.6, (Pg. 198) I think what is meant by this is that if b is adjoined to A then reduced by GJE to an RREF matrix B, so long as the column representing b remains, the system is thus demonstrated consistent
- 204. © Art Traynor2011 Mathematics Square Matrix Equivalency Conditions Solution Space Section 4.6, (Pg. 198) Vector Space For a matrix A composed of m-rows and n-columns Amx n , each of the following incidents implies each of the other A is invertible Ax = b has a unique solution for any n x 1 matrix b Ax = 0 has the trivial solution A is “row equivalent” to to In |A | ≠ 0 Rank ( A ) = n The n row vectors of A are Linear Independent The n column vectors of A are Linear Independent “ A ” cannot be empty, or the zero vector
- 205. © Art Traynor2011 Mathematics Coordinate Representation Relative to a Basis Basis Section 4.7, (Pg. 202) Vector Space For an Ordered Basis set B = { vi , vi+1 ,…vn – 1 , vn }, within Vector Space V , there exists a vector x in V that can be expressed as a Linear Combination, or sum of scalar multiples of the constituent vectors of x such that: xn = ci vi + ci+1 vi+1 +…+ cn –1 vn – 1 + cn vn The coordinate matrix, or coordinate vector, of x relative to B is the column matrix in Rn whose components are the coordinates of x the scalars of which ci = [ ci , ci+1 ,…cn – 1 , cn ] are otherwise referred to as the coordinates of x relative to the Basis B c1 c2 cn . . . [ xn ]Bn = Chump Alert: Coordinate matrix representation relative to a Basis ( standard or otherwise) is directly analogous to Normalized ( ? ) Unit Vector representation
- 206. © Art Traynor2011 Mathematics Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis in the vector space to be described as a linear combination of the unit vector set U = { ui , u i + 1 ,…un – 1 , u n } x y O u2 u1 Svn · SBn = SBn ´ v11 v21 vn1 . . . vi 11 02 0n . . . In 01 12 0n . . . 01 02 1n . . . . . . . . . . . . . . . Svn · U = SBn ´ = v1 02 0n . . . 01 v2 0n . . . 01 02 vn . . . . . . . . . . . . . . . SBn ´ We begin by noting that the Standard Basis SBn allows any point v12 v22 vn2 . . . . . . . . . . . . . . . v1n v2n vnn . . . I am switching the text’s designation of B and B´ as it seems more intuitive to think of the “prime” set as the translated, or product set 1a
- 207. © Art Traynor2011 Mathematics Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis This entails that any alternative Basis is necessarily a linear combination of the standard Basis (which is equal to the Identity Matrix) U = In = { ui , u i + 1 ,…un – 1 , u n } x y O u2 u1 Translation of the standard Basis by a set of vector scalars constituting an alternative Basis may thus appear to be trivial Svn · SBn = SBn ´ vi In Svn · U = SBn ´ = SBn ´ 111 112 021 222 xi yj v1 v2 111 012 021 122 111 112 021 222 u1 u2 B´ 1b
- 208. © Art Traynor2011 Mathematics Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis This alternative Basis is akin to a Translation of the standard coordinate origin O → O ´ . x y O O ´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) Svn · SBn = SBn ´ vi Svn · U = SBn ´ = SBn ´ 111 112 021 222 xi yj v1 v2 111 112 021 222 B´ In 111 012 021 122 u1 u2 I’m going to keep the color coding of the B-Prime Matrix as green to emphasize it’s status as a “resultant” (i.e. the transformed, alternative Basis) 1c
- 209. © Art Traynor2011 Mathematics Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis The Translated, or product vector set can be expressed in several equivalent forms. x y O O ´ B´ = { ( 1 , 0 ) , ( 1 , 2 ) } B´ = { v1 , v2 } r11 r12 r21 r22 B´ = v1 v2 xi yj 111 112 021 222 B´ = v1 v2 xi yj Basis constituent vectors expressed in set notation Restated in Transition Matrix Form (TMF) Basis constituent vector components expressed in set notation v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) I’m going to keep the color coding of the B-Prime Matrix as green to emphasize it’s status as a “resultant” (i.e. the transformed, alternative Basis) ②
- 210. © Art Traynor2011 Mathematics Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis Supplying axes through the Translated origin suggests the shifted coordinate system introduced by the alternative Basis x y x´ O O ´ Note that the xi “Unknown” or “Variable” row vector, populated with the multiplicative identity (i.e. all “ones”) shifts the X-Axis of the alternative Basis system of coordinates by a scalar multiple of one 111 012 021 122 B = u1 u2 xi yj u2 u1 111 112 021 222 B ´ = v1 v2 xi yj “Standard” Basis Element 1B 1B v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) 3a
- 211. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis Supplying axes through the Translated origin suggests the shifted coordinate system introduced by the alternative Basis x y x´ O O ´ 111 012 021 122 B = u1 u2 xi yj u2 u1 v2 v1 111 112 021 222 B ´ = v1 v2 xi yj “Standard” Basis Element “Alternative” Basis Element 1B 2B 3B 4B 5B 1B 1B The resultant vector ( v1 + v2 ) indicates the orientation of the “y” coordinate axis and introduces a “skew” in the alternative Basis system by comparison with the standard Basis v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) 1B´ 2B´ 3B´ 4B´ 5B´ 3b
- 212. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis The alternative Basis maps input coordinates from the standard Basis incrementing each “X” value by +1 and each “Y” value by +2 Note that the “two” in the yi “Unknown” or “Variable” row vector also scales or elongates the alternative Basis “y” coordinate x y x´ O O ´ 1B 2B 3B 4B 5B 1B ⇌ ½B´ 1B´ 2B´ 3B´ 2B ⇌ 1B´ 3B ⇌ 1½ B´ 4B ⇌ 2B´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) 111 012 021 122 B = u1 u2 xi yj u2 u1 v2 v1 111 112 021 222 B ´ = v1 v2 xi yj “Standard” Basis Element “Alternative” Basis Element 1B 1B 1B´ 2B ⇌ 1B´ 1B´ 2B´ 3B´ 4B´ 5B´ 3c
- 213. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis Thus the coordinate values of the two systems superimposed will bear the same values for the “X” coordinates, but will feature two values for the “Y” coordinates scaled by factor of two. x y x´ O O ´ 1B 2B 3B 4B 5B 1B ⇌ ½B´ 1B´ 2B´ 3B´ 2B ⇌ 1B´ 3B ⇌ 1½ B´ 4B ⇌ 2B´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) 111 012 021 122 B = u1 u2 xi yj u2 u1 v2 v1 111 112 021 222 B ´ = v1 v2 xi yj “Standard” Basis Element “Alternative” Basis Element 1B 1B 1B´ 2B ⇌ 1B´ 1B´ 2B´ 3B´ 4B´ 5B´ 3d
- 214. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis Any arbitrary vector in the space can be represented as a scalar multiple (i.e. Linear Combination) of the basis vectors xn = ci vi + ci+1 vi+1 +…+ cn –1 vn – 1 + cn vn x y x´ O O ´ 1B 2B 3B 4B 5B 1B ⇌ ½B´ 1B´ 2B´ 3B´ 2B ⇌ 1B´ 3B ⇌ 1½ B´ 4B ⇌ 2B´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) xn = c1 v1 + c2 v2 cn = ( c1 , c2 ) B´ = { ( 1 , 0 ) , ( 1 , 2 ) } 111 112 021 222 B´ = v1 v2 xi yj Basis constituent vector components expressed in Set Notation Form SNF Restated in Transition Matrix Form (TMF) Arbitrary vector in Vector Space V c1 c2 cn . . . cn = ci 311 221 The ordered pair of scalar vector coefficients of x Restated as a Column Vector (scalar coefficients of x) 1B´ 2B´ 3B´ 4B´ 5B´ ④
- 215. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis the ordered pair collection of vector scalar coefficients of the arbitrary vector x is explicitly corresponded with the Basis vectors the products of which form a resultant Linear Combination. x y x´ O O ´ 1B 2B 3B 4B 5B 1B ⇌ ½B´ 1B´ 2B´ 3B´ 2B ⇌ 1B´ 3B ⇌ 1½ B´ 4B ⇌ 2B´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) xn = c1 v1 + c2 v2 cn = ( c1 , c2 ) A scalar coefficient of “one” is implicit c1 c2 cn . . . ci 311 221 The ordered pair of scalar vector coefficients of x Restated as a Column Vector (scalar coefficients of x) B´ = { v1 , v2 } The set of Basis vectors in SNF Juxtaposed as below, it is clear that x is a linear combination of the Basis Vectors B´ = { 1 v1 , 1 v2 } ci ≠ 1 implies that the arbitrary vector is a Linear Combination of the Basis Vector set cn = [ x ]B´ = An illuminating change of notation is here introduced [ x ]B´ whereby 1B´ 2B´ 3B´ 4B´ 5B´ 5a
- 216. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis x y x´ O O ´ 1B 2B 3B 4B 5B 1B ⇌ ½B´ 1B´ 2B´ 3B´ 2B ⇌ 1B´ 3B ⇌ 1½ B´ 4B ⇌ 2B´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) c1 c2 cn . . . ci 311 221 Stated as a Column Vector (scalar coefficients of x) Juxtaposed as below, it is clear that x is a linear combination of the Basis Vectors cn = [ x ]B´ = The novel notation for the set (ordered pair, or column vector form) of vector scalar coefficients ci directs our attention to the specific Basis by which the resulting coordinates are generated. e.g.: x is expressed relative to Basis B “ Relative to ” “ Defined by ” “ Within ” xn = c1 v1 + c2 v2 cn = ( c1 , c2 ) B´ = { v1 , v2 } B´ = { 1 v1 , 1 v2 }1B´ 2B´ 3B´ 4B´ 5B´ 5b
- 217. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2 Example: Find the coordinate matrix of x relative to a non-standard basis ① x y x´ O O ´ 1B 2B 3B 4B 5B 1B ⇌ ½B´ 1B´ 2B´ 3B´ 2B ⇌ 1B´ 3B ⇌ 1½ B´ 4B ⇌ 2B´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) c1 c2 cn . . . ci 311 221 Stated as a Column Vector (scalar coefficients of x) Juxtaposed as below, it is clear that x is a linear combination of the Basis Vectors cn = [ x ]B´ = Vector x therefore can be graphically represented by a Position Vector demarcated in the B-Prime coordinate system [ i.e. three increments of “ x ” from the alternative Basis ( aB ) origin and two increments of “ y ” from the aB origin ]. “ Relative to ” “ Defined by ” “ Within ” 111 012 021 122 B u1 u2 u2 u1 v2 v1 111 112 021 222 B ´ v1 v2 xi yj “Standard” Basis Element “Alternative” Basis Element 1B 1B 1B´ 2B ⇌ 1B´ xB = ( 3 , 2 ) 1B´ 2B´ 3B´ 4B´ 5B´
- 218. © Art Traynor2011 Mathematics y´ Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2Example: Find the coordinate matrix of x relative to a non-standard basis ① x y x´ O O ´ 1B 2B 3B 4B 5B 1B ⇌ ½B´ 1B´ 2B´ 3B´ 2B ⇌ 1B´ 3B ⇌ 1½ B´ 4B ⇌ 2B´ v2 = 2 u2 ( 1 , 2 ) v1 = 1 u1 ( 1 , 0 ) To restate the scalar set of the arbitrary vector coefficients in terms of the standard basis ( sB ), the aB vector components need simply be scaled by the vector equation for x to yield the scalar solution set in terms of the sB xB = ( 3 , 2 ) xn = c1 v1 + c2 v2 B´ = { ( 1 , 0 ) , ( 1 , 2 ) } B´ = { v1 , v2 } xn = c1 ( 1 , 0 ) + c2 ( 1 , 2 ) cn = [ x ]B = ( c1 , c2 ) cn = [ x ]B = ( 3 , 2 ) xn = 3( 1 , 0 ) + 2 ( 1 , 2 ) x = ( 3·1 + 2·1 ) , ( 3·0 + 2·2 ) x = ( 3 + 2 ) , ( 0 + 4 ) x = ( 5 ) , ( 4 ) xn = 3 ( v1 ) + 2 ( v2 ) 1B´ 2B´ 3B´ 4B´ 5B´
- 219. © Art Traynor2011 Mathematics Vector Space Basis Coordinate Matrices and Bases Section 4.7 (Pg. 203), Example 2Example: Find the coordinate matrix of x relative to a non-standard basis ① Note finally that, as B ′ ( B-Prime ) is the Identity Matrix ( IM ), the B matrix is equivalent to the Transition Matrix (from non-standard to standard Basis) conventionally noted as P – 1 Section 4.7 (Pg. 208) [ B′ In ] Adjoin [ In P – 1 ] RREF GJE ( EROs ) P – 1 = ( B′ ) – 1 ′ Change of Basis Non-Standard Standard 111 112 021 222 v1 v2 xi yj P – 1 = ( B′ ) = ′
- 220. © Art Traynor2011 Mathematics Change of Basis ( aka Transformation ) Basis Transformation Vector Space Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis We first restate the given alternative Basis aB = B ´ ( expressed in Set Notation Form – SNF ) into Transformation Matrix Form ( TMF ). ① B´ = { u1 , u2 , u3 } Alternative Basis constituent vectors expressed in Set Notation Form SNF B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) } 1 0 1 0 – 1 2 – 2 – 3 – 5 aB Coefficient Matrix B´ = c1 c2 c3 u1 u2 u3 Given the form of the problem, it is implicit that the arbitrary vector is proffered relative to the standard Basis
- 221. © Art Traynor2011 Mathematics Change of Basis ( aka Transformation ) Basis Transformation Vector Space Section 4.7 (Pg. 203), Example 3 Example: Find the coordinate matrix of x relative to a non-standard basis With this restatement, we next introduce the arbitrary vector coordinates ( to which we are tasked to find the coordinate matrix corresponding to the aB ) rendering it into a column vector form. ① B´ = { u1 , u2 , u3 } B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) } 1 0 1 0 – 1 2 – 2 – 3 – 5 aB Coefficient Matrix B´ = c1 c2 c3 u1 u2 u3 These given arbitrary vector coordinates are presumably introduced relative to the Standard Basis? xn = ( 1 , 2 , – 1 ) c1 c2 cn . . . ci cn = [ x ]B´ = 1 2 c1 c2 c3 xi – 1“ Relative to ” “ Defined by ” “ Within ” Keep in mind that the ‘given’ arbitrary vector is stated relative to the standard Basis and thus represents a solution set by which the set of scalar multiples applicable to the alternative Basis will be derived. Alternative Basis constituent vectors expressed in Set Notation Form SNF
- 222. © Art Traynor2011 Mathematics Change of Basis ( aka Transformation ) Basis Transformation Vector Space Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis We next introduce a column vector of scalars the solution to which will represent the coordinate matrix of the arbitrary vector with respect to the alternative Basis ( aB ) ① B´ = { u1 , u2 , u3 } B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) } 1 0 1 0 – 1 2 – 2 – 3 – 5 aB Coefficient Matrix u1 u2 u3 Alternative Basis constituent vectors expressed in Set Notation Form SNF These given arbitrary vector coordinates are presumably introduced relative to the Standard Basis? 1 2 – 1 c1 c2 c3 xi xn = ( 1 , 2 , – 1 ) Keep in mind that the ‘given’ arbitrary vector is stated relative to the standard Basis and thus represents a solution set by which the set of scalar multiples associated with the alternative Basis will be determined c1 c2 ci – c3 = cn = [ x ]B = ( c1 , c2 , c3 )
- 223. © Art Traynor2011 Mathematics 1 0 1 0 – 1 2 – 2 – 3 – 5 Augmented Coefficient Matrix Vector Space 1 2 – 1 1c1 + 0x2 + 2c3 = 1s 3x1 – 1c2 + 3c3 = 2s 1c1 + 2c2 – 5c3 = – 1s A = c1 c2 c3 u1 u2 u3 b The final “set-up” step entails restating the composed system once again rendering it into a non-Homogenous Linear Equation System ( nHLES) and its corresponding Augmented Coeffficient Matrix (ACM) ① Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis Non-Homogenous Linear Equation System ( nHLES) B´ = { u1 , u2 , u3 } B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) } xn = ( 1 , 2 , – 1 ) cn = ( c1 , c2 , c3 ) The ACM is thus expressed in Standard Matrix Form with the column vectors corresponding to the coefficients of the unknown aB scalars, the solution for which (by means of Gauss Jordan Elimination – GJE) will supply the coordinate matrix relative to the aB that the problem poses for resolution.
- 224. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis 1 0 1 0 – 1 2 – 2 – 3 – 5 Augmented Coefficient Matrix 1 2 – 1 A = c1 c2 c3 u1 u2 u3 b Designate each row with an Uppercase Alpha Character…this will allow the Elementary Row Operations (EROs) of the Gauss-Jordan Elimination (GJE) to be applied to be described in a summarized algebraic fashion. ② 1 0 1 0 – 1 2 – 2 – 3 – 5 1 2 – 1 A1 B1 C1 Gauss-Jordan Elimination (GJE) is an algorithmic scheme applied to a Standard Matrix Form (SMF) representation of a system of Linear Equations resulting in a “row- equivalent” reduced matrix on which the main diagonal entries are all “ones” (pivots in Row Echelon Form - REF) and all entries above and below the “pivots” are populated by “zeros” or Reduced Row Echelon Form (RREF). Section 1.2, (Pg. 19)
- 225. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis 1 0 1 0 – 1 2 – 2 – 3 – 5 1 2 – 1 A1 B1 C1 Investigate permutation as a strategy to manipulate the simplest (most easily reduced) rows into the primary and higher positions in the matrix ③ In this case, the first row already features a value of one in the first position along the main diagonal, as well as a zero directly below in the next row down, so all is well to proceed to the next step in the GJE reduction process to arrive at a “Reduced Row Equivalent” RREF Basis matrix… From this established “Pivot” move down to the next row and render the entry there into a “zero” using EROs ④ 1 0 1 0 – 1 2 – 2 – 3 – 5 1 2 – 1 A1 B1 C1 Inspection suggests that scaling Row One by – 1 and summing with Row Three would yield an appealing reduction of Row Three into the desired form where the leading entry in the row is rendered into a zero. C1 = C2 – 1A1
- 226. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis Continued…④ 1 0 1 0 – 1 2 – 2 – 3 – 5 1 2 – 1 A1 B1 C1 C1 = C2 – 1A1 The resultant should always be stated first (indexed to its succeeding value) in the reduction evolution. The Augend/Minuend term should always be the same term as the resultant (at its prevailing index). The Subtrahend/Summand should thus be the only term subject to any scaling.
- 227. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis We replace the transformed Row then proceed to the next non- one entry (along the main diagonal) or non-zero entry (outside of the main diagonal), working from top-to-bottom, and left-to-right 1 0 0 – 1 – 2 – 3 1 2 A1 B1 C1 With the entry in Row One, Column Two already at zero, we move to reduce the next main diagonal entry to a value of One, which is easily accomplished by scaling Row One by – 1 B1 = – 1B1 Continued…④ 1 2 – 5 – 1C1 – 1A1 – 1 0 – 2 – 1 – 0 2 – 7 – 2C1 – 0 2 – 7 – 2 ⑤
- 228. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis Continued… 1 0 0 – 1 – 2 – 3 1 – 2 A1 B1 C1 – 0 2 – 7 – 2 ⑤ ⑥ Inspection suggests that Row Three be summed with Row Two scaled by a factor of – 2 With the second “pivot” fixed along the main diagonal, we can proceed down to the next row and render the entry there into a “zero” using EROs 1 0 0 – 1 – 2 – 3 1 – 2 A1 B1 C1 – 0 2 – 7 – 2 C2 = C1 – 2B1
- 229. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis With the main diagonal in Column Two fixed at the desired pivot value of “one”, and all other entries in the column reduced to values of “zero”, we proceed next to the top of Column Three to continue our row reduction with further applied EROs 1 0 – 2 1A1 Inspection discloses that the lead entry in Column Three ( j = 3 ) can be reduced to “zero” by adding Row One to Row Three scaled by a factor of Two Continued… – 0 0 – 1 – 2C2 ⑥ C1 – 0 2 – 7 – 2 0 – 2 – 6 – 4– 2B1 ⑦ 0 – 1 – 3 – 2B1 – 0 0 – 1 – 2C2 A1 = A1 + 2C2
- 230. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis Proceeding with EROs down Column Three, we will next address the reduction of the entry in Row Two 1 0 – 0 5A1 Inspection suggests that the entry in Column Three, Row Two ( ij = 23 ) can be rendered into a zero value by the summation of Row Two with Row Three scaled by a factor of – 3. Continued… – 1 0 – 0 – 5A1 0 – 0 – 2 – 4+ 2C2 0 – 1 – 3 – 2B1 – 0 0 – 1 – 2C2 B2 = B1 – 3C2 1 0 – 2 1A1 ⑦ ⑧
- 231. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis Proceeding with EROs down Column Three, we will next address the reduction of the entry in Row Two 1 0 – 0 5A1 Inspection reveals that a simple scaling of Row Three by a factor of – 1 will render the system into its final RREF expression. Continued… – 0 1 – 0 – 8B2 – 3C2 0 – 1 – 0 – 8B2 – 0 0 – 1 – 2C2 ⑧ 0 – 1 – 3 – 2B1 – 0 0 – 3 – 6 ⑨ C3 = – 1C2
- 232. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis Having arrived at the RREF expression of the nHLES, we can explicitly state the values for ci ( the coordinate matrix for the arbitrary vector relative to the alternative Basis ( aB ) ⑨ 1 0 – 0 5A1 0 – 1 – 0 – 8B2 – 0 0 – 1 – 2C3 1 0 – 0 5 0 – 1 – 0 – 8 – 0 0 – 1 – 2 1d1 + 0x2 + 2c3 = 5s 3x1 – 1d2 + 3c3 = – 8s 1c1 + 2c2 – 5d3 = – 2s u1 u2 u3 b d1 d2 d3 No parameterization is necessary as the RREF form gives explicit values for each unknown This differs from the result in example 9, pg 197…it would be interesting to note why this is so? The RREF matrix, thus transformed from the root SMF or TMF (conventionally designated “ A ” with entries “ ci ” ) is now restated as RREF matrix “ B ” with entries “ di ” Section 4.6, (Pg. 193)
- 233. © Art Traynor2011 Mathematics Vector Space Basis Transformation Change of Basis ( aka Transformation ) Section 4.7 (Pg. 203), Example 3Example: Find the coordinate matrix of x relative to a non-standard basis No parameterization is necessary as the RREF form gives explicit values for each unknown 1 0 – 0 5 0 – 1 – 0 – 8 – 0 0 – 1 – 2 1d1 + 0x2 + 2c3 = 5s 3x1 – 1d2 + 3c3 = – 8s 1c1 + 2c2 – 5d3 = – 2s u1 u2 u3 b 10 5 – 2 d1 d2 d3 xi – 1 d1 d2 dn . . . di dn = [ x ]B´ = “ Relative to ” “ Defined by ” “ Within ” 5 – 2 d1 d2 d3 xi – 1 d1 d2 d3 x = d1u1 + d2u2 + d3u3 + xp System Solution(s) x = xh + xp xh = d1u1 + d2u2 + d3u3 Ax = b Ax = 0 ( HLES ) ( nHLES ) RREF ( Basis Set ) for nHLES
- 234. © Art Traynor2011 Mathematics Change of Basis ( aka Transformation ) Basis Transformation Vector Space Section 4.7 (Pg. 203), Example 3 Example: Find the coordinate matrix of x relative to a non-standard basis Continued… A = B´ = 11 1 0 1 0 – 1 2 – 2 – 3 – 5 Augmented Coefficient Matrix 1 2 – 1 c1 c2 c3 u1 u2 u3 b Only columns featuring a “leading one” ( “Pivot”) in the RREF matrix B are Linear Independent Section 4.6, (Pg. 192) Perhaps trivially so, it is worth noting that the wi vectors thus form a Basis for the Row Space of A = B´ , or the subspace spanned by A = B´ = { ui , ui + 1 ,…un – 1 , un } Section 4.6, (Pg. 191) Section 4.5, (Pg. 184) Also recall that the subspace spanned by A = B´ will thus feature precisely three Basis vectors (by the definition of Basis Cardinality ), exactly corresponding to the number of Linear Independent vectors in the space (any greater than which would entail Linear Dependence of one of the vectors). This cardinality will always coincide with that of the “Standard Basis” for the vector space (e.g. R3) 1 0 – 0 5 0 – 1 – 0 – 8 – 0 0 – 1 – 2 u1 u2 u3 b d1 d2 d3 B = “ Row Equivalent” ( RREF Basis Matrix ) w1 w2 Linear Independent = = = w3
- 235. © Art Traynor2011 Mathematics Change of Basis ( aka Transformation ) Basis Transformation Vector Space Section 4.7 (Pg. 203), Example 3 Example: Find the coordinate matrix of x relative to a non-standard basis Continued…12 A = B´ = 1 0 – 0 5 0 – 1 – 0 – 8 – 0 0 – 1 – 2 u1 u2 u3 b d1 d2 d3 B = 1 0 1 0 – 1 2 – 2 – 3 – 5 Augmented Coefficient Matrix 1 2 – 1 c1 c2 c3 u1 u2 u3 b “ Row Equivalent” ( RREF Basis Matrix ) w1 w2 Linear Independent = = = w3 xn = ci · ui + ci+1 · ui+1 +…+ cn –1 vn – 1 + cn vn Arbitrary Vector, generally xn = c1 · u1 + c2 · u2 + c3 · v3 Arbitrary Vector, particular ( 1 , 2 , – 1 ) = c1 ( 1 , 0, 1 ) + c2 ( 0 , – 1, 2 ) + c3 ( 2 , 3, – 5 ) Arbitrary Vector, as nHLES linear combination relative to aB
- 236. © Art Traynor2011 Mathematics Change of Basis ( aka Transformation ) Basis Transformation Vector Space Section 4.7 (Pg. 203), Example 3 Example: Find the coordinate matrix of x relative to a non-standard basis Continued…13 d1 d2 dn . . . di dn = [ x ]B´ = “ Relative to ” “ Defined by ” “ Within ” 5 – 2 d1 d2 d3 xi – 1 xn = ci · ui + ci+1 · ui+1 +…+ cn –1 vn – 1 + cn vn Arbitrary Vector, generally xnn = c1 · u1 + c2 · u2 + c3 · v3 Arbitrary Vector, particular xn = 51 ( 1 , 0, 1 ) + ( – 82 ) ( 0 , – 1, 2 ) + ( – 22 )( 2 , 3, – 5 ) Arbitrary Vector, as nHLES linear combination relative to aB
- 237. © Art Traynor2011 Mathematics Change of Basis ( via Transition Matrix) Basis Transformation Vector Space Section 4.7 (Pg. 204) For an arbitrary vector x in vector space V described by coordinates relative to a Standard Basis B , an ancillary description – in coordinate terms relative to an Alternate Basis B′ ( B-Prime ) – can be determined by operation of a Transition Matrix P ( the entries of which are populated by the components of x ) . The matrix product of P and a column vector [ x ]B′ of scalars ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix of x relative to the alternate basis B′ ( B-Prime ) yield the column vector [ x ]B′ , the “ root ” basis. The “ Change of Basis ” is thus the solution to the unknown column vector [ x ]B′ of scalars ci = [ ci , ci+1 ,…cn – 1 , cn ] .
- 238. © Art Traynor2011 Mathematics Change of Basis ( via Transition Matrix) Basis Transformation Vector Space Section 4.7 (Pg. 204) The matrix product of P and a column vector [ x ]B′ of scalars ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix of x relative to the alternate basis B′ ( B-Prime ) yield the column vector [ x ]B′ . P = B´ = { u1 , u2 , u3 } P = B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) } xn = ci ui + ci+1 ui+1 +…+ cn –1 un – 1 + cn un Note the convention wherein blue font is assigned to those elements of the B-Basis set, whose component vectors are assigned to “ u ” Red font accordingly is assigned to those elements of the B-Prime ( B′ ) Basis set, whose component vectors are assigned to “ v ” P = B´ = Coefficient Matrix u1 u2 u3 d1 d2 d3 xi = [ x ]B′ = p11 Alternative Basis constituent vectors expressed in Set Notation Form SNF Note that Matrix P = B′ is expressed in Transition Matrix Form ( TMF ) wherein the columns are populated by the individual Basis vector components with the rows collecting like “ unknown ” terms. xi = [ x ]B′ c1 c2 c3 Root BasisAlternate Basis p21 p31 p12 p22 p32 p13 p23 p33
- 239. © Art Traynor2011 Mathematics Change of Basis ( via Transition Matrix) Basis Transformation Vector Space Section 4.7 (Pg. 204) The matrix product of P and a column vector [ x ]B′ of scalars ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix of x relative to the alternate basis B′ ( B-Prime ) yield the column vector [ x ]B′ . P = B´ = { u1 , u2 , u3 } P = B´ = { ( 1 , 0, 1 ) , ( 0 , – 1, 2 ) , ( 2 , 3, – 5 ) } xn = ci ui + ci+1 ui+1 +…+ cn –1 un – 1 + cn un P = B´ = 1 0 1 0 – 1 2 – 2 – 3 – 5 Coefficient Matrix u1 u2 u3 d1 d2 d3 xi = [ x ]B′ = xi = [ x ]B′ Root BasisAlternate Basis – 1 – 2 – 1
- 240. © Art Traynor2011 Mathematics Change of Basis ( via Transition Matrix) Basis Transformation Vector Space Section 4.7 (Pg. 204) The matrix product of P and a column vector [ x ]B′ of scalars ci = [ ci , ci+1 ,…cn – 1 , cn ] which forms the coordinate matrix of x relative to the alternate basis B′ ( B-Prime ) yield the column vector [ x ]B′ . P [ x ]B′ = [ x ]B′ P [ x ]B′ = [ x ]B′ P2 P2 P [ x ]B′ = [ x ]B′ P2 P2 P [ x ]B′ = P – 1 [ x ]B′ Whereas the product of the Transition Matrix and the Alternate Basis ( aB ) yields the Root Basis, it is equivalent to state that the product of the Transition Matrix Inverse and the Root Basis will yield the Alternate Basis.Change of Basis B′ B Change of Basis B B′ We recall that to find an inverse matrix we adjoin the Identity Matrix In (on RHS ) to P forming matrix [ A In ] and perform EROs by GJE to arrive at an RREF which will then result in matrix [ In A –1 ] with the inverse occupying the RHS of the resultant matrix
- 241. © Art Traynor2011 Mathematics Change of Basis ( via Transition Matrix) Basis Transformation Vector Space Section 4.7 (Pg. 204) P = B´ = Transition Matrix u1 u2 u3 d1 d2 d3 xi = [ x ]B′ = p11 xi = [ x ]B′ c1 c2 c3 Root BasisAlternate Basis p21 p31 p12 p22 p32 p13 p23 p33 P – 1 = B´ Transition Matrix Inverse d1 d2 d3 xi = [ x ]B′ = p – 1 xi = [ x ]B′ c1 c2 c3 Root BasisAlternate Basis p12 p22 p32 p13 p23 p33 v1 v2 v3 11 p – 1 21 p – 1 31
- 242. © Art Traynor2011 Mathematics Change of Basis ( via Transition Matrix) Basis Transformation Vector Space Section 4.7 (Pg. 206) [ B′ B ] The Transition Matrix ( TM ) from a Root Basis ( RB = B ) to an Alternate Basis ( aB = B′ , B-Prime ) is found, as is the case for similar matrix inverses, by adjoining the Root Basis matrix with the aB matrix ( on the LHS ) and applying EROs via GJE to arrive at a RREF reduced matrix, the result of which will form an adjoined matrix composed of the Identity Matrix ( IM – on the LHS ) and the P – 1 Transition Matrix ( from B to B′ – on the RHS ). Adjoin [ In P – 1 ] RREF GJE ( EROs )
- 243. © Art Traynor2011 Mathematics Transition Matrix Basis Transformation Vector Space Section 4.7 (Pg. 208) When the Root Basis ( RB = B ) is equivalent to the Identity Matrix ( IM ) the process of finding a basis-changing Transition Matrix ( TM ) is simplified to a symmetric operation whereby the Root- Identity Matrix is adjoined ( on the RHS ) to an Alternate Basis Matrix ( ABM = B′ , B-Prime, on the LHS ) and applying EROs via GJE to arrive at a RREF reduced matrix, the result of which will form an adjoined matrix composed of the Root-Identity Matrix ( RIM – on the LHS ) and the P – 1 Transition Matrix ( from B to B′ – on the RHS ). [ B′ In ] Adjoin [ In P – 1 ] RREF GJE ( EROs ) P – 1 = ( B′ ) – 1 ′ Change of Basis Standard Non-Standard