The document discusses vector spaces and related linear algebra concepts. It defines vector spaces and lists the axioms that must be satisfied. Examples of vector spaces include the set of all pairs of real numbers and the space of 2x2 symmetric matrices. The document also discusses subspaces, linear combinations, span, basis, dimension, row space, column space, null space, rank, nullity, and change of basis. It provides examples and explanations of these fundamental linear algebra topics.

1. VECTOR SPACE
PRESENTED BY :-MECHANICAL
ENGINEERING
DIVISION-B SEM-2
YEAR-2016-17

2. Real Vector Spaces
Sub Spaces
Linear combination
Span Of Set Of Vectors
Basis
Dimension
Row Space, Column Space, Null Space
Rank And Nullity
Coordinate and change of basis
CONTENTS

3. NAME ENROLLMENT NO.
PATEL JAIMIN A 150280119080
PATEL KAUSHAL K 150280119081
PATEL KRUNAL S 150280119082
PATEL MEET B 150280119083
PATEL MILAN V 150280119084
PATEL MILIND B 150280119085
PATEL PRANAV P 150280119086
PATEL RAVI K 150280119087
PATEL RAVI S 150280119088
PATEL UJJWAL G 150280119089
PATIL KRUNAL R 150280119090
PRESENTED BY:-

4. VECTOR SPACE
Let V be an arbitrary non empty set of objects on which two operations are defined,
addition and multiplication by scalar (number). If the following axioms are satisfied by
all objects u, v, w in V and all scalars k and l, then we call V a vector space and we
call the objects in V vectors.
1) If u and v are objects in V, then u + v is in V
2) u + v = v + u
3) u + (v + w) = (u + v) + w
4) There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u
for all u in V
5) For each u in V, there is an object –u in V, called a negative of u, such that
u + (-u) = (-u) + u = 0
6) If k is any scalar and u is any object in V then ku is in V
7) k(u+v) = ku + kv
8) (k+l)(u) = ku + lu
9) k(lu) = (kl)u
10)1u = uz

5. EXAMPLE OF VECTOR SPACE
 Determine whether the set of V of all pairs of real numbers
(x,y) with the operations (𝑥1, 𝑦1) + (𝑥2, 𝑦2) = (x1+x2+1,
y1+y2+1) and k(x,y) = (kx,ky) is a vector space.
Solution let u=(x1,y1), v=(x2,y2) and w=(x3,y3) are objects
in V and k1,k2 are some scalars.
1 . u+v = (x1,y1) + (x2,y2) = (x1+x2+1, y1+y2+1)
since x1+x2+1, y1+y2+1 are also real numbers .
Therefore, u+v is also an object in V.
2. u+v = (x1+x2+1, y1+y2+1)
= (x2+x1+1, y2+y1+1)
= v + u
Therefore , vector addition is commutative.

6. 3. u+(v+w) = (x1,y1)+[(x2,y2) +(x3,y3)]
= (x1,y1)+(x2+x3+1, y2+y3+1)
= [x1+ (x2+x3+1)+1 , y1+(y2+y1+1)+1)
= [(x1+ x2+1)+x3+1 , (y1+y2+1)+y3+1)]
= (x1+x2+1, y1+y2+1)+(x3+y3)
= (u+v)+w
Hence, vector addition is associative.
4. Let (a,b) be in object in V such that (a,b)+u=u
(a,b) +(x1,y1)=(x1,y1)
(a+x1+1,b1+y1+1) = (x1,y1)
a= -1 , b=-1
Hence, (-1,-1) is zero vector in V.
5. Let (a,b) be in object in V such that (a,b)+u=(-1,-1)
(a,b)+(x1,y1)=(-1,-1)
(x1+a+1,y1+b+1)=(1,-1)
a= -x1-2 , b = -y1-2
Hence, (-x1-2,y1-2) is the negative of u in V

7. 6. k1u = k1(x1,y1)
=(k1x1,k1y1)
Since k1x1, k1y1 are real numbers.
Therefore, V is closed under scalar multiplication.
7. K1(u+v)= k1(x1+ x2+1, y1+y2+1)
=(k1x1 + k1x2 + k1, k1y1 + k1y2 + k1)
≠ k1u + k1v
V is not distributive under scalar multiplication.
Hence, V is not a vector space

8. SUBSPACES
If W is a set of one or more vectors in a vector space V,
then W is a sub space of V if and only if the following
condition hold;
a)If u,v are vectors in a W then u+v is in a W.
b)If k is any scalar and u is any vector In a W then ku is in
W.

9. originhethrough tLines(2)
9
Every vector space V has at least two subspaces
(1)Zero vector space {0} is a subspace of V.
(2) V is a subspace of V.
 Ex: Subspace of R2
   00,(1) 00
originhethrough tLines(2)
2
(3) R
• Ex: Subspace of R3
originhethrough tPlanes(3)
3
(4) R
   00,0,(1) 00
If w1,w2,. . .. wr subspaces of vector space V then the
intersection is this subspaces is also subspace of V.

10. Let W be the set of all 2×2 symmetric matrices. Show that W
is a subspace of the vector space M2×2, with the standard
operations of matrix addition and scalar multiplication.
10
sapcesvector:2222  MMW
Sol:
)(Let 221121 AA,AAWA,A TT

)( 21212121 AAAAAAWAW,A TTT

)( kAkAkAWA,Rk TT

22ofsubspaceais  MW
)( 21 WAA 
)( WkA
Ex : (A subspace of M2×2)

11. SPAN
- The set of all the vectors that are the linear
combination of the vectors in the set
S = {v1,v2…..vr} is called span of S
and is denoted by Span {v1,v2…..vr}
- If S = {v1,v2…..vr} is a set of vector in a vector space v
then,
(1)The span is a subspace of v
(2)The span S is the smallest subspace of v that
contains the set S.
- If S1 And S2 are two sets of vectors in v then,
Span S1 = Span S2
if and only if each vector in S1 is a linear combination
of those in S1 And S2 and vice versa

12. METHOD TO CHECK SPAN OF V
1.choose an arbitary vector b in v.
2. express b as alinear combination of v1,v2,…..,vr
b=k1v1+k2v2+…..+krvr
3. if the system of the equation in above equation is
consistent for all choice of b then vectors
v1,v2,…..,vr span v.if it is inconsistent for some
choice of b, vectors do not span v.

14. kkccc uuuv  2211
formin thewrittenbecanifinvectorsthe
ofncombinatiolinearacalledisspacevectorainA vector
21 vuuu
v
V,,,
V
k
Definition
scalars:21 k,c,,cc 
Linear combination
If S = (w1,w2,w3, . . . . , wr) is a nonempty set of vectors in a vector
space V, then:
(a) The set W of all possible linear combinations of the vectors in S
is a subspace of V.
(b) The set W in part (a) is the “smallest” subspace of V that
contains all of the vectors in S in the sense that any other subspace
that contains those vectors contains W.

15.  Example:
Every vector v = (a, b, c) in R3 is expressible as a linear
combination of the standard basis vectors i = (1,0,0), j =
(0,1,0), k=(0,0,1) since
v = (a,b,c) = a(1,0,0) + b(0,1,0) + c(0,0,1)
= ai + bj + ck
 Example: Consider the vectors u=(1,2,-1) and
v=(6,4,2) in R3. Show that w=(9,2,7) is a linear
combination of u and v and that w’=(4,-1,8) is not a
linear combination of u and v.
822
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16. 
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332211 vvvw ccc 
V1=(1,2,3) v2=(0,0,2) v3= (-1,0,1)
Prove W=(1,-2,2) is not a linear combination
of v1,v2,v3
w = c1v1 + c2v2 + c3v3
Finding a linear combination
Sol :

17. Basis
• Definition:
V：a vector space
Generating
Sets
Bases
Linearly
Independent
Sets
 S is called a basis for V
S ={v1, v2, …,
vn}V
• S spans V (i.e., span(S) =
V )
• S is linearly independent
(1) Ø is a basis for {0}
(2) the standard basis for R3:
{i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)

Notes:

18. (3) the standard basis for R
n
:
{e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1)
Ex: R4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}
Ex: matrix
space:
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,
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22
(4) the standard basis for mn matrix space:
{ Eij | 1im , 1jn }
(5) the standard basis for Pn(x):
{1, x, x2, …, xn}
Ex: P3(x) {1, x, x2, x3}

19. Example
Let V = P3 and let S = {1, t, t2, t3}. Show that S is a
basis for V.
Solution
We must show both linear independence and span.
Linear Independence:
Let
c1(1) + c2(t) + c3(t2) + c4(t3) = 0
Then since a polynomial is zero if and only if its
coefficients are all zero we have
c1 = c2 = c3 = c4 = 0
Hence S is a linearly independent set of vectors in V.

20. Span
A general vector in P3 is given by
a + bt + ct2 + dt3
We need to find constants c1, c2, c3, c4 such that
c1(1) + c2(t) + c3(t2) + c4(t3) = a + bt + ct2 + dt3
We just let
c1 = a, c2 = b, c3 = c, c4 = d
Hence S spans V.
We can conclude that S is a basis for V.
In general the basis {1, t, t2, ... , tn} is called the STANDARD
BASIS for Pn.

21. Change of basis problem
You were given the coordinates of a vector relative to one
basis B and were asked to find the coordinates relative to
another basis B'.
Transition matrix from B' to B:
V
BB nn
spacevectorafor
basestwobe}...,,{nda},...,,{etL 2121 uuuuuu 
BB P  ][][hent vv         BBnBB
v,...,,  1 uuu 2
      BnBB
P uuu 2
 ...,,,1
where
is called the transition matrix from B' to B
If [v]B is the coordinate matrix of v relative to B
[v]B‘ is the coordinate matrix of v relative to B'

22. If P is the transition matrix from a basis B' to a basis B in Rn,
then
(1) P is invertible
(2) The transition matrix from B to B' is P
–1
THEOREM 1 The inverse of a transition matrix
THEOREM 2
Let B={v1, v2, … , vn} and B' ={u1, u2, … , un} be two bases
for R
n
. Then the transition matrix P–1 from B to B' can be
found by using Gauss-Jordan elimination on the n×2n
matrix as follows.
Transition matrix from B to B'
 BB   1
PIn 

23. B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two
bases for R
2
(a) Find the transition matrix from B' to B.
(b)
(c) Find the transition matrix from B to B' .
BB ][find,
2
1
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Ex : (Finding a transition matrix)
Sol:
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22
21
22
43
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12
23
10
01
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G.J.E.
B B' I P
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12
23
P (the transition matrix from B' to B)
(a)
Sol:
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22
21
22
43

 G.J.E.
B B'
(a)

24. 
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0
1
2
1
12
23
][][
2
1
][ BBB
P vvv
(b)
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22
43
22
21

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32
21
10
01


G.J.E.
B' B I P
-1
(the transition matrix from B to B')
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32
211
P
 Check: 2
1
10
01
32
21
12
23
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0
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2
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:Check
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vv
vv
B
B
(c)

25. 25
Dimension
 Definition:
The dimension of a finite dimensional vector space V is defined to
be the number of vectors in a basis for V.
V: a vector spaceS: a basis for V
Finite
dimensional
A vector space V is called finite dimensional, if it has a
basis consisting of a finite number of elements
Infinite dimensional
If a vector space V is not finite dimensional,then it is
called infinite dimensional.
• Dimension of vector space V is
denoted by dim(V).

26. 26
Theorems for dimention
THEOREM 1
All bases for a finite-dimensional vector space have
the same number of vectors.
THEOREM 2
Let V be a finite-dimensional vector space, and let be any
basis.
(a) If a set has more than n vectors, then it is linearly
dependent.
(b) If a set has fewer than n vectors, then it does not span
V.

27. 27
Dimensions of Some Familiar
Vector Spaces
(1) Vector space Rn  basis {e1 , e2 ,  , en}
(2) Vector space Mm  basis {Eij | 1im , 1jn}
(3) Vector space Pn(x)  basis {1, x, x2,  , xn}
(4) Vector space P(x)  basis {1, x, x2, }
 dim(Rn) = n
 dim(Mmn)=mn
 dim(Pn(x)) = n+1
 dim(P(x)) = 

28. Coordinates
• Coordinate representation relative to a basis Let B = {v1,
v2, …, vn} be an ordered basis for a vector space V and let x
be a vector in V such that .2211 nnccc vvvx  
The scalars c1, c2, …, cn are called the coordinates of x relative to
the basis B. The coordinate matrix (or coordinate vector) of x
relative to B is the column matrix in Rn whose components are the
coordinates of x.
 
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n
B
c
c
c

2
1
x

29. Find the coordinate matrix of x=(1, 2, –1) in R3 relative to
the (nonstandard) basis
B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)}
Sol:
2100
8010
5001
1521
2310
1201
E.G.J.
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1
2
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310
201
i.e.
152
23
12
3
2
1
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32
31
c
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5
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x
Finding a coordinate matrix relative to a
nonstandard basis

30.  Row space:
The row space of A is the subspace of Rn spanned by the row
vectors of A.
 Column space:
The column space of A is the subspace of Rm spanned by
the column vectors of A.
  },,{ 21
)((2)
2
(1)
1 RAAAACS n
n
n   
}|{)( 0xx  ARANS n
 Null space:
The null space of A is the set of all solutions of Ax=0 and it is a
subspace of Rn.
Let A be an m×n matrix
Row space ,column space and null space
The null space of A is also called the solution space of
the homogeneous system Ax = 0.

31. Find a basis of row space of A =

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2402
1243
1603
0110
3131
Sol:

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2402
1243
1603
0110
3131
A=
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0000
0000
1000
0110
3131
3
2
1
w
w
w
B = .E.G
bbbbaaaa 43214321
Finding a basis for a row space
EXAMPLE
A basis for RS(A) = {the nonzero row vectors of B} (Thm 3)
= {w1, w2, w3} = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)}

32. Finding a basis for the column space
of a matrix

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2402
1243
1603
0110
3131
A
Sol:
3
2
1
..
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11100
65910
23301
21103
42611
04013
23301
w
w
w
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



 BA EGT
EXAMPLE:

33. CS(A)=RS(AT)
(A basis for the column space of A)
A Basis For CS(A) = a basis for RS(AT)
= {the nonzero vectors
of B}
= {w1, w2, w3}

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
1
1
1
0
0
,
6
5
9
1
0
,
2
3
3
0
1

34. Rank and Nullity
If A is an mn matrix, then the row space and the column
space of A have the same dimension.
dim(RS(A)) = dim(CS(A))
THEOREM
The dimension of the row (or column) space of a
matrix A is called the rank of A and is denoted by
rank(A).
rank(A) = dim(RS(A)) = dim(CS(A))
Rank:
Nullity: The dimension of the null space of A is called the nullity of
A.
nullity(A) = dim(NS(A))

35. THEOREM
If A is an mn matrix of rank r, then the dimension of the solution
space of Ax = 0 is n – r. That is
n = rank(A) + nullity(A)
• Notes:
(1) rank(A): The number of leading variables in the
solution of Ax=0.
(The number of nonzero rows in the row-echelon
form of A)
(2) nullity (A): The number of parameters in the general
solution of Ax= 0.

36. 36
Rank and nullity of a matrix
Let the column vectors of the matrix A be denoted by a1, a2, a3, a4,
and a5.




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










120930
31112
31310
01201
A
a1 a2 a3 a4 a5
EXAMPLE
Find the Rank and nullity of the
matrix.
Sol: Let B be the reduced row-echelon
form of A.










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
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
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




00000
11000
40310
10201
120930
31112
31310
01201
BA
a1 a2 a3 a4 a5 b1 b2 b3 b4 b5
G.
E.
rank(A) = 3 (the
number of nonzero
rows in B)
Nullity(A) = n – rank(A
= 5 – 2
= 3

37. THANK YOU
GUIDED BY:-
MR. JAYDEV PATEL