The document discusses vector spaces and related concepts. It begins by defining a vector space as a non-empty set V with defined operations of vector addition and scalar multiplication that satisfy certain axioms. Examples of vector spaces include Rn and the set of m×n matrices. A subspace is a subset of a vector space that is also a vector space under the defined operations. Properties of subspaces and examples are provided. Linear combinations, linear independence, spanning sets, and the span of a set of vectors are then defined and explained.
5. CONTENTS
1) Real Vector Spaces
2) Sub Spaces
3) Linear combination
4) Linear independence
5) Span Of Set Of Vectors
6) Basis
7) Dimension
8) Row Space, Column Space, Null Space
9) Rank And Nullity
10)Coordinate and change of basis
5
Dr A K TIWARI
6. Vector Space Axioms
Real Vector Spaces
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 = u
6
Dr A K TIWARI
7. Example: Rn is a vector space
Ku =(ku1,ku2,ku3,. . . . . kun)
Step :1- identify the set V of
objects that will become
vectors.
Let, V=Rn
u+v = (u1,u2,u3,. . . . . un)+(v1,v2,v3,. . . . vn)
= (u1+v1,u2+v2,u3+v3, . . . .un+vn )
Step :2- identify the edition and
scalar multiplication operations
on V.
Step :3- verify axioms 1 & 6, adding two
vectors in a V produces a vector in a V , and
multiplying a vector in a V by scalar
produces a vector in a V.
Step :4- confirm that axioms
2,3,4,5,7,8,9 and 10 hold.
7
Dr A K TIWARI
9. Examples of vector spaces
(1) n-tuple space: Rn
(u1,u2, ,un )(v1,v2, ,vn) (u1v1,u2 v2, ,un vn )
k(u1,u2 , ,un ) (ku1,ku2 , ,kun )
(2) Matrix space: (the set of all m×nmatrices with real values)
V Mmn Ex: :(m = n = 2)
21 22 22
22 21
21 22 21 v u v
v u
u
u v
u11 u12 v11 v12 u11 v11 u12 v12
21 22 21 22
ku
ku12
ku
ku11
u
u12
u11
ku
vector addition
scalar multiplication
vector addition
scalar multiplication
9
Dr A K TIWARI
10. (3) n-th degree polynomial space: (the set of all real polynomials of degree n or less)
n
b )xn
n
b0 ) (a1 b1)x (a
0
p(x) q(x) (a
n
kp(x) ka0 ka1x kan x
(4) Function space:(the set of all real-valued continuous functions defined on the entire
real line.)
V c(,)
( f g)(x) f (x) g(x)
(kf )(x) kf (x)
vector addition
scalar multiplication
vector addition
scalar multiplication
1
0
Dr A K TIWARI
11. PROPERTIES OF VECTORS
Let v be any element of a vector space V, and let c be any
scalar. Then the following properties are true.
(1) 0v 0
(2) c0 0
(3) If cv 0, then c 0 or v 0
(4) (1)v v
11
Dr A K TIWARI
12. Definition:
(V,,) : a vector space
: a non empty subset
W V
(W,,)
W
:a vector space (under the operations of addition and
scalar multiplication defined in V)
Subspaces
W is a subspace of V
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. 12
Dr A K TIWARI
13. 11
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: 2
Subspace of R
(1)
0 0 0, 0
(2) Lines through the origin
R 2
(3)
• Ex: Subspace of R3
(1)
0 0 0, 0, 0
(2) Lines through the origin
(3) Planes through the origin
(4) R3
If w1,w2,. . .. wr subspaces of vector space V then the intersection is this subspaces is also
subspace of V.
Dr A K TIWARI
14. Let W be the set of all 2×2symmetric matrices. Show that W is a subspace of the
vector space M2×2,with the standard operations of matrix addition and scalar
multiplication.
Sol:
W M22 M22 : vector sapces
Let A , A W (AT
A , AT
A )
1 2 1 1 2 2
A W, A W (A A )T
AT
AT
1 2 1 2 1 2
k R, AW (kA)T
kAT
kA
W is a subspace of M22
A1 A2 (A1 A2 W)
(kAW)
Ex : (A subspace of M2×2)
Dr A K TIWARI
15.
0 1
0
W
A B
1
W2 is not a subspace of M22
Let W be the set of singular matrices of order 2 Show that W is not a subspace of M2×
2 with
the standard operations.
Sol:
0 1
0 0
A
1 0
W ,B
0 0
W
Ex : (The set of singular matrices is not a subspace of M2×2)
Dr A K TIWARI
16. 14
v c1u1 c2u2 … ckuk
Definition
A vector v in a vector space V is calleda linear combination of
the vectors u1,u2, ,uk in V if v can be written in the form
c1,c2, ,ck :scalars
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.
Dr A K TIWARI
17. 15
1 0 1 1 1 0 1 1
2 1 0 2
G
u
a
s
s
J
o
r
d
a
n
E
l
i
m
i
n
a
t
i
o
n
0 1 2 4
3 2 1 2 0 0 0 7
this system has no solution ( 0 7)
w c1v1 c2v2 c3v3
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 :
Dr A K TIWARI
18. Definition:
S v1,v2 , , vk : a set of vectors in a vector space V
c1v1 c2v2 ck vk 0
(1) If theequation has only thetrivial solution (c1 c2 ck 0)
then S is calledlinearlyindependent.
(2) If theequation has a nontrivial solution (i.e., not all zeros),
then S is calledlinearlydependent.
Theorem
A set S with two or more vectors is
(a) Linearly dependent if and only if at least one of the vectors in S is expressible as a
linear combination of the other vectors in S.
Linear Independent (L.I.) and Linear Dependent (L.D.):
16
(b) Linearly independent if and only if no vector in S is expressible as a linear
combination of the other vectors in S.
Dr A K TIWARI
19. 1 is linearly independent
(2) 0 S S is linearly dependent.
(3) v 0 vis linearly independent
(4) S1 S2
S1 is linearly dependent S2 is linearly dependent
S2 is linearly independent S1 is linearly independent
Notes
Dr A K TIWARI
20. Determine whether the following set of vectors in R
3
is L.I. or L.D.
S
1, 2, 3,0,1, 2,2, 0,1
v1 v2 v3
c1 2c3 0
2c1 c2 0
c1v1 c2v2 c3v3 0
Sol:
3c1 2c2 c3 0
1 0 2 0 1 0 0 0
2 1 0 0G
aus
s - Jo
rdan
Elim
inati
o
n
0
1 0 0
3 2 1 0 0 0 1 0
c1 c2 c3 0 only thetrivial solution
S is linearly independent
Ex : Testing for linearly independent
Dr A K TIWARI
21. Span of set of vectors
span(S) c1v1 c2v2 ci R
(the set of all linear combinations of vectors in S)
If every vector in a given vector space can be written as a linear combination of vectors in
a given set S, then S is called a spanning set of the vector space.
ck vk
Definition:
If ISf=S{=v{,vv1
,,
v
…
2
,
…
,v,}vk
i}siassaetseotfovfevcteocrtsorisnianvaevcteocrtosrpaspceacVe,
1 2 k
theVn,tthheenspthaensopfanSoifstSheissethteofseatllolfinaellarlicnoemarbinationsof
thecovmecbtionrastiinonSs.
of the vectors in S.
Dr A K TIWARI
22. If S={v1, v2,…, vk} is a set of vectors in a vector space V,
then
(a)span (S) is a subspace of V.
(b)span (S) is the smallest subspace of V that contains S.
(Every other subspace of V that contains S must contain span (S).
Notes:
(1) span() 0
(2) S span(S)
1 2
S1 S2 span(S1) span(S2 )
(3) S ,S V
span (S) V
S spans (generates) V
V is spanned (generated) by S
S is a spanning set of V
Dr A K TIWARI
23. Ex: A spanning set for R3
Show that the set S (1,2,3),(0,1,2), (2,0,1) sapns R3
Sol:
We must determine whether an arbitrary vector u (u1,u2 ,u3 )
in R3
can be as a linear combination of v , v , and v .
1 2 3
u R3
u c v c v c v
1 1 2 2 3 3
3c1 2c2 c3 u3
The problem thus reduces to determining whether this system
is consistent for all values of u1, u2 , and u3.
c1 2c3 u1
2c1 c2 u2
1 0 2
A 2 1 0 0
3 2 1
Ax b has exactlyonesolution for every u.
span(S) R3
Dr A K TIWARI
24. 22
Basis
• Definition:
V:a vector space
S ={v1, v2, …, vn}V
Generating
Sets
Bases
Linearly
Independent
Sets
• S spans V (i.e., span(S) = V )
• S is linearly independent
S is called a basis for V
Notes:
(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)
Dr A K TIWARI
25. (3) the standard basis for Rn
:
{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)}
1
0 0
0 1
0 0
0
0
,
0 1
,
0 0
,
0 0
Ex: matrix space:
1
2 2
(4) the standard basis for mn matrix space:
{ Eij | 1im , 1jn }
(5) the standard basis for Pn(x):
{1, x, x2, …, xn}
Ex: P3(x) {1, x, x2, x3}
Dr A K TIWARI
26. THEOREMS
Uniqueness of basis representation
If S= {v1,v2,…,vn} is a basis for a vector space V, then every vector in V can be
written in one and only one way as a linear combination of vectors in S.
Bases and linear dependence
If S= {v1,v2,…,vn} is a basis for a vector space V, then every set containing more
than n vectors in V is linearly dependent.
Dr A K TIWARI
27. If a vector space V has one basis with n vectors, then every basis for V has n
vectors. (All bases for a finite-dimensional vector space has the same number of
vectors.)
Number of vectors in a basis
An INDEPENDENT set of vectors that SPANS a vector
space V is called a BASIS for V.
Dr A K TIWARI
28. 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 space S: a basis for V
• Dimension of vector space V is denoted by dim(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.
Dr A K TIWARI
29. 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.
Dr A K TIWARI
30. Dimensions of Some Familiar Vector Spaces
(1) Vector space Rn basis {e1 , e2 , , en}
dim(Rn) = n
(2) Vector space Mm basis {Eij | 1im , 1jn}
dim(Mmn)=mn
(3) Vector space Pn(x) basis {1, x, x2, , xn}
dim(Pn(x)) = n+1
(4) Vector space P(x) basis {1, x, x2, }
dim(P(x)) =
Dr A K TIWARI
32. Let A be an m×
n matrix
Row space:
The row space of A is the subspace of Rn spanned by the row vectors ofA.
Column space:
The column space of A is the subspace of Rm spanned by the column vectors ofA.
2
(1)
1
A
CSA{
Null space:
n
(2) (n)
n A 1,2 , R}
A
The null space of A is the set of all solutions of Ax=0 and it is a subspace of Rn.
N S ( A ) { x R n
| A x 0}
The null space of A is also called the solution space of the homogeneous system Ax = 0.
Raw space ,column space and null space
30
Dr A K TIWARI
33.
m
mn
A
A
a
A
⁝
a
⁝
⁝
a
⁝
m2
a a 2
A1
m1
2n
21 22
a1n
a11 a12
a 21 22 2n (2)
Row vectors of A
[a , a ,…, a ] A
11 12 1n (1)
[a , a ,…, a ] A
⁝
[a , a ,…, a ] A
m1 m2 mn (n)
Column vectors of A
Row Vectors:
1 2 n
A ⁝A ⁝ ⁝A
a a
⁝
am2
2n
22
21
⁝
amn
a1n
⁝
am1
a11 a12
a
A
mn
a a a
⁝ ⁝ ⁝
m1 m2
a a a
21 22 2n
a11 a12 a1n
Column Vectors:
|| ||
A
(1)
A
(2)
||
A
(n)
Dr A K TIWARI
34. Elementary row operations do not change the null space of a matrix.
THEOREM 2
The row space of a matrix is not changed by elementary row operations.
RS(r(A)) = RS(A) r: elementary row operations
THEOREM 3
If a matrix R is in row echelon form, then the row vectors with the leading 1′s (the nonzero
row vectors) form a basis for the row space of R, and the column vectors with the leading 1′s
of the row vectors form a basis for the column space of R.
32
THEOREM 1
Dr A K TIWARI
35. 33
Find a basis of row space of A =
0
Sol:
1 3 1 3
0 1 1 0
3 0 6 1
3 4 2 1
2 0 4 2
a1 a2 a3 a4
A=
0
0
0
1
0
0 w 3
2
w1
w
G
.E.
B =
1 3 1 3
0 1 1
3 0 6 1
3 4 2 1
2 0 4 2
1 3 1 3
1 1
0 0
0 0
0 0 0 0
b1 b2 b3 b4
Finding a basis for a row space
EXAMPLE
Abasis 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)}
Dr A K TIWARI
36. THEOREM 4
If A and B are row equivalent matrices, then:
(a)A given set of column vectors of A is linearly independent if and only if the corresponding
column vectors of B are linearly independent.
(b)A given set of column vectors of A forms a basis for the column space of A if and only if the
corresponding column vectors of B form a basis for the column space of B.
THEOREM 5
If a matrix A is row equivalent to a matrix B in row-echelon form, then the nonzero
row vectors of B form a basis for the row space of A.
Dr A K TIWARI
37. Finding a basis for the column space of a matrix
0
1 3 1 3
0 1 1
A 3 0 6 1
3 4 2 1
2 0 4 2
Sol:
3
2
1
2 w
1 w
0
6 w
2
0 3 3 2 1 0 3 3
1 0 4 0 0 1 9 5
B
1 6 2 4 0 0 1 1
0 1 1
0 0 0 0
G
.E.
3
1
3
1
AT
EXAMPLE
Dr A K TIWARI
38. (a basis for the column space of A)
1
2 6 1
3
5
3, 9 , 1
CS(A)=RS(AT)
ABasis For CS(A) = a basis for RS(AT)
= {the nonzero vectors of B}
= {w1, w2, w3}
1 0 0
0 1 0
Dr A K TIWARI
39. Finding the solution space of a homogeneous system
Ex:
Find the null space of the matrix A.
Sol: The null space of A is the solution space of Ax = 0.
3
1
4
1 2 2 1
6 5
2 0
A 3
0
0
1
3
1
1 2 2 1 1 2 0 3
6 5 0 1
2 0 0 0
4 G
.J
.E
0
A 3 x1 = –2s – 3t, x2 = s, x3 = –t, x4 = t
2
1
sv tv
t
s
s
t
t
x
x
x
1
1
3
0
0
1 0
2
2s 3t
x4
3
2
x1
NS(A) {sv1 tv2 | s,t R}
Dr A K TIWARI
40. Rank and Nullity
If A is an mn 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))
Dr A K TIWARI
41. THEOREM
If A is an mn 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 free variables in the solution of Ax = 0.
Dr A K TIWARI
42. Rank and nullity of a matrix
EXAMPLE
Let the column vectors of the matrix A be denoted by a1, a2, a3, a4, and a5.
0
A
1 0 2 1
0 1 3 1 3
2 1 1 1 3
0 3 9 0 12
a1 a2 a3 a4 a5
Find the Rank and nullity of the matrix.
Sol: Let B be the reduced row-echelon form of A.
0
1
0
B 0
A
1 0 2 1 0
0 1 3 1 3
2 1 1 1 3
0 3 9 0 12
a1 a2 a3 a4 a5
1 0 2 0 1
1 3 0 4
0 0 1
0 0 0
b1 b2 b3 b4 b5
G.E.
rank(A) = 3 (the number
0 of nonzero rows in B)
nuillity(A) n rank(A) 53 2
nuillity (A) n rank(A) 5 3
40
2
Dr A K TIWARI
43. 41
Coordinates and change of basis
B
c
n
c
⁝
2
• 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
x c1v1 c2v2 cn vn.
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.
c1
x
Dr A K TIWARI
44. 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:
x c u c u c u (1, 2, 1) c (1, 0,1) c (0, 1, 2) c (2, 3, 5)
1 1 2 2 3 3 1 2 3
1
0
1
3
2
0
1
5c 1
3c 2
2c 1
3
2
1
c 2c 5c 1
0 2 1 1 0 0 5
1 3 2 G.J
.E.
0 1 0 8
2 5 1 0 0 1 2
c 2c 1 1 0
1 3
c 3c 2 i.e. 1
2 3
1 2
2
5
B
[ x ] 8
Finding a coordinate matrix relative to a nonstandard basis
Dr A K TIWARI
45. Change of basis problem
B B B B
1
u
u ,..., v
u , 2 n
then [v]B P[v]B
where
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:
Let B {u1,u2 ,...,un} and B {u1
,u2...,un} be two bases
for a vector space V
If [v]B is the coordinate matrix of v relative to B
[v]B‘ is the coordinate matrix of v relative to B'
n
B B
P u , u ,..., u
1 B 2
is called the transition matrix from B' to B
Dr A K TIWARI
46. If P is the transition matrix from a basis B' to a basis B in Rn, then
(1) P is invertible
–1
(2) The transition matrix from B to B' is P
THEOREM 1 The inverse of a transition matrix
THEOREM 2
n
Let B={v1, v2, … , vn} and B' ={u1, u2, … , un} be two bases for R . 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'
B⁝B 1
n
I ⁝P
Dr A K TIWARI
47. 2
B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two bases for R
(a) Find the transition matrix from B' to B.
B
B'
(b) Let [v] find [v]
2
1
,
Ex : (Finding a transition matrix)
0
1 0 ⁝ 3 2
I
1 ⁝ 2 1
P
(c) Find the transition matrix from B to B' .
Sol:
2
2
2
2
3
3 4
4 ⁝
⁝
1
1 22
2
2
2
2 ⁝
⁝
B
1
3 2
P 2
(the transition matrix from B' to B)
G.J.E.
B'
(a)
Dr A K TIWARI
48.
2 12 0
3 21
1
2 B B
B
[v]
1
[v] P[v]
(b)
2
1 2 ⁝ 3 4
0 1 ⁝ 2 3
1 0 ⁝ 1 2
G.J.E.
B'
2 ⁝ 2 2
B I P
-1
3
2 (the transition matrix from B to B')
2
P1
1
Check: 2
1
3 0 I
1 2
3 2 1 2 1 0
PP 2
1
0
2
Check:
[v]
1
v (1)(3,2) (0)(4,2) (3,2)
[v]
1
v (1)(1,2) (2)(2,2) (3,2)
B
B
(c)
Dr A K TIWARI