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

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

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:-

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

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.

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

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

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.

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.

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)

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

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.

kkccc uuuv  2211
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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.

 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.
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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 :

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:

(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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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}

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.

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.

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'

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 

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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Sol:
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12
23
P (the transition matrix from B' to B)
(a)
Sol:
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22
21
22
43

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B B'
(a)

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23
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2
1
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P vvv
(b)
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vv
vv
B
B
(c)

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
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
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)) = 

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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c
c
c
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2
1
x

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:
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8010
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2310
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E.G.J.
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x
Finding a coordinate matrix relative to a
nonstandard basis

 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.

Find a basis of row space of A =
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Sol:
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A=
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0000
1000
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3
2
1
w
w
w
B = .E.G
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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)}

Finding a basis for the column space
of a matrix
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A
Sol:
3
2
1
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w
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 BA EGT
EXAMPLE:

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

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))

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
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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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

THANK YOU
GUIDED BY:-
MR. JAYDEV PATEL