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Lecture 3 section 1-7, 1-8 and 1-9
- 2. Slide 1.7- 2© 2012 Pearson Education, Inc.
LINEAR INDEPENDENCE - Definition
Definition: set of vectors {v1, …, vp} in Rn
is said to
be linearly independent if the vector equation
x1v1 + x1v1+ …+ x1v1 = 0
has only the trivial solution.
Otherwise: linearly dependent
Nontrivial solution exists linear dependence relation
- 3. Slide 1.7- 3© 2012 Pearson Education, Inc.
LINEAR INDEPENDENCE - Example
Example 1: Let , , and .
a. Is the set {v1, v2, v3} linearly independent?
b. If possible, find a linear dependence relation
among v1, v2, and v3.
1
1
v 2
3
=
2
4
v 5
6
=
3
2
v 1
0
=
- 4. Slide 1.7- 4© 2012 Pearson Education, Inc.
LINEAR INDEPENDENCE OF MATRIX COLUMNS
matrix A = [a1a2 … a3]
The matrix equation Ax = 0 can be written as
x1v1 + x1v1+ …+ x1v1 = 0
Each linear dependence relation among the columns of
A corresponds to a nontrivial solution of Ax = 0.
Thus, the columns of matrix A are linearly independent
iff the equation Ax = 0 has only the trivial solution.
- 6. Slide 1.7- 6© 2012 Pearson Education, Inc.
SETS OF ONE VECTOR
A set containing only one vector – say, v – is linearly
independent iff v is not the zero vector.
This is because the vector equation has only
the trivial solution when .
The zero vector is linearly dependent because
has many nontrivial solutions.
1
v 0x =
v 0≠
1
0 0x =
- 7. Slide 1.7- 7© 2012 Pearson Education, Inc.
SETS OF TWO VECTORS
- 8. Slide 1.7- 8© 2012 Pearson Education, Inc.
Characterization of Linearly Dependent Sets
Theorem 7: An indexed set S = {v1, …, vp} of vectors
is linearly dependent iff at least one of the vectors in
S is a linear combination of the others.
In fact, if S is linearly dependent and j > 1, then some
vj (with v1≠ 0) is a linear combination of the
preceding vectors, v1, …, vj-1.
- 9. Slide 1.7- 9© 2012 Pearson Education, Inc.
SETS OF TWO OR MORE VECTORS
Proof: If some vj in S equals a linear combination of
the other vectors, then vj can be subtracted from both
sides of the equation, producing a linear dependence
relation with a nonzero weight on vj.
v1 = c2 v2+ c3 v3
0 = (-1)v1 + c2 v2+ c3 v3+ 0v4 + … 0vp
Thus S is linearly dependent.
Conversely, suppose S is linearly dependent.
If v1 is zero, then it is a (trivial) linear combination of
the other vectors in S.
( 1)−
- 10. Slide 1.7- 10© 2012 Pearson Education, Inc.
SETS OF TWO OR MORE VECTORS
Otherwise, , and there exist weights c1, …, cp, not
all zero, such that
.
Let j be the largest subscript for which .
If , then , which is impossible because
.
1
v 0≠
1 1 2 2
v v ... v 0p p
c c c+ + + =
0j
c ≠
1j = 1 1
v 0c =
1
v 0≠
- 11. Slide 1.7- 11© 2012 Pearson Education, Inc.
SETS OF TWO OR MORE VECTORS
So , and1j >
1 1 1
v ... v 0v 0v ... 0v 0j j j j p
c c +
+ + + + + + =
1 1 1 1
v v ... vj j j j
c c c − −
= − − −
11
1 1
v v ... v .j
j j
j j
cc
c c
−
−
= − + + − ÷ ÷
- 12. Slide 1.7- 12© 2012 Pearson Education, Inc.
SETS OF TWO OR MORE VECTORS
Theorem 7 does not say that every vector in a linearly
dependent set is a linear combination of the preceding
vectors.
A vector in a linearly dependent set may fail to be a
linear combination of the other vectors.
Example 2: Let and . Describe the
set spanned by u and v, and explain why a vector w is
in Span {u, v} if and only if {u, v, w} is linearly
dependent.
3
u 1
0
=
1
v 6
0
=
- 13. Example
Example 2: Let and .
a)Describe the set spanned by u and v
b)explain why a vector w is in Span {u, v} iff
{u, v, w} is linearly dependent.
Slide 1.7- 13© 2012 Pearson Education, Inc.
3
u 1
0
=
1
v 6
0
=
- 14. Slide 1.7- 14© 2012 Pearson Education, Inc.
SETS OF TWO OR MORE VECTORS
So w is in Span {u, v}. See the figures given below.
Example 2 generalizes to any set {u, v, w} in R3
with u
and v linearly independent.
The set {u, v, w} will be linearly dependent iff w is in
the plane spanned by u and v.
- 15. Slide 1.7- 15© 2012 Pearson Education, Inc.
SETS OF TWO OR MORE VECTORS
Theorem 8: If a set contains more vectors than there
are entries in each vector, then the set is linearly
dependent. That is, any set {v1, …, vp} in Rn
is
linearly dependent if .
Proof: Let .
Then A is , and the equation Ax = 0
corresponds to a system of n equations in p
unknowns.
If , more variables than equations, so there
must be a free variable many solutions
p n>
1
v vp
A = L
n p×
p n>
- 16. Slide 1.7- 16© 2012 Pearson Education, Inc.
SETS OF TWO OR MORE VECTORS
Theorem 9: If a set in Rn
contains
the zero vector, then the set is linearly dependent.
Proof: By renumbering the vectors, we may suppose
v1 = 0
1v1 + 0v2 + … + 0vp = 0
Nontrivial solution linearly dependent
1
{v ,...,v }p
S =
- 19. Slide 1.8- 19© 2012 Pearson Education, Inc.
LINEAR TRANSFORMATIONS
A transformation (or function or mapping) T from Rn
to Rm
is a rule that assigns to each vector x in Rn
a
vector T (x) in Rm
Rn
is called domain of T
Rm
is called the codomain of T.
Notation T: Rn
Rm
For x in Rn
, the vector T (x) in Rm
is called the image of
x (under the action of T ).
The set of all images T (x) is called the range of T.
- 20. Slide 1.8- 20© 2012 Pearson Education, Inc.
MATRIX TRANSFORMATIONS
For each x in Rn
, T (x) is computed as Ax, where A is an
mxn matrix.
Notation for a matrix transformation by x Ax
Domain of T is Rn
when A has n columns
Co-domain of T is Rm
when A has m rows
The range of T is the set of all linear combinations of
the columns of A, because each image T (x) is of the
form Ax.
- 21. Slide 1.8- 21© 2012 Pearson Education, Inc.
MATRIX TRANSFORMATIONS - Example
Example 1: Let T: Rn
Rm
by T(x)=Ax
a.Find T (u), the image of u under the transformation T.
b.Find an x in R2
whose image under T is b.
c.Is there more than one x whose image under T is b?
d.Determine if c is in the range of the transformation T.
1 3
3 5
1 7
A
−
=
−
2
u
1
= −
3
c 2
5
=
−
- 22. Slide 1.8- 22© 2012 Pearson Education, Inc.
LINEAR TRANSFORMATIONS
Definition: A transformation (or mapping) T is linear
if:
i. for all u, v in the domain of T:
T(u + v) = T(u) + T(v)
ii. for all scalars c and all u in the domain of T:
T(cu) = cT(u)
Linear transformations preserve the operations of
vector addition and scalar multiplication.
- 23. Linear Transformations
These two properties lead to the following useful
facts.
If T is a linear transformation, then:
T(0) = 0
T(c1v1+ c2v2+…+ cpvp) = c1T(v1)+ c2T(v2)+…+ cpT( vp)
Slide 1.7- 23© 2012 Pearson Education, Inc.
- 24. Slide 1.8- 24© 2012 Pearson Education, Inc.
SHEAR TRANSFORMATION
Example 2: Let . The transformation
defined by T = Ax is called a shear transformation.
1 3
0 1
A
=
- 25. Slide 1.8- 25© 2012 Pearson Education, Inc.
Contraction/Dilation
Given a scalar r, define T : Rn
Rn
by T(x) = rx
contraction when 0 ≤ r < 1
dilation when r > 1
- 26. How to Find Matrix of Transformation
Theorem 10: T: Rn
Rm
is linear
transformation, then a unique matrix A such
that T(x) = Ax for all x in Rn
A is mxn whose jth
column is the vector T(ej) where ej
is the jth
column of In
A = [T(e1) … T(en)]
Examine 1 point in each direction!!
Slide 1.7- 26© 2012 Pearson Education, Inc.
- 27. Existence Question for Transformations – “Onto"
Definition: A mapping T: Rn
Rm
is onto Rm
if
each b in Rm
is the image of at least 1 x in Rn
i.e., range is all of the co-domain
Equivalent: does Ax = b have a solution for all
b
Theorem 1-4
Pivot in every row of A
Theorem 1-12: T is onto iff columns of A span Rm
Slide 1.7- 27© 2012 Pearson Education, Inc.
- 28. Uniqueness Question for Transformations – “one-to-one”
Definition: A mapping T: Rn
Rm
is one-to-
one if each b in Rm
is the image of at most 1 x
in Rn
Equivalent to Ax = b has one or no solution
Pivot every column of A.
Theorem 1-11: T is one-to-one iff T(x) = 0 has
only the trivial solution
Theorem 1-12: T is one-to-one iff columns of A are
linearly independent
Slide 1.7- 28© 2012 Pearson Education, Inc.
- 30. Example
T(x1, x2) = (3x1+x2, 5x1+7x2, x1+3x2)
a)Find standard matrix A of T.
b)Is T one-to-one?
c)Does T map R2
onto R3
?
Slide 1.7- 30© 2012 Pearson Education, Inc.