Vector Spaces

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Vectors in Rn
Vector Spaces
Subspaces of Vector Spaces
Spanning Sets and Linear Independence
Basis and Dimension
Rank of a Matrix and Systems of Linear Equations
Coordinates and Change of Basis
Vectors in R

n
An ordered n-tuple:
a sequence of n real number ( x1 , x2 ,  , xn )
 n-space: R
n
the set of all ordered n-tuple
Dr. Abhishek Das, SAS, VIT Vellore

Ex:
1
n=1
R = 1-space
= set of all real number
n=2
R = 2-space
= set of all ordered pair of real numbers ( x1 , x2 )
n=3
2
3
R = 3-space
= set of all ordered triple of real numbers ( x1 , x 2 , x3 )
n=4
4
R = 4-space
= set of all ordered quadruple of real numbers ( x1 , x2 , x3 , x4 )
Dr. Abhishek Das, SAS, VIT Vellore

Notes:
n
(1) An n-tuple ( x1 , x2 ,  , xn ) can be viewed as a point in R
with the xi’s as its coordinates.
(2) An n-tuple ( x1 , x2 ,  , xn ) can be viewed as a vector
x  ( x1 , x2 ,  , xn ) in Rn with the xi’s as its components.

Ex:
x1 , x2 
x1 , x2 
a point
0,0
a vector
Dr. Abhishek Das, SAS, VIT Vellore
u  u1 , u 2 ,  , u n , v  v1 , v2 ,  , vn 

Equal:
u  v if and only if
(two vectors in Rn)
u1  v1 , u 2  v2 ,  , u n  vn
 Vector
addition (the sum of u and v):
u  v  u1  v1 , u 2  v2 ,  , u n  vn 

Scalar multiplication (the scalar multiple of u by c):
c u  cu 1 , cu 2 ,  , cu n 

Notes:
The sum of two vectors and the scalar multiple of a vector
n
in R are called the standard operations in Rn.
Dr. Abhishek Das, SAS, VIT Vellore

Negative:
 u  (u1 ,u2 ,u3 ,...,un )

Difference:
u  v  (u1  v1 , u 2  v2 , u3  v3 ,..., u n  vn )

Zero vector:
0  (0, 0, ..., 0)

Notes:
(1) The zero vector 0 in Rn is called the additive identity in Rn.
(2) The vector –v is called the additive inverse of v.
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Properties of vector addition and scalar multiplication)
n
Let u, v, and w be vectors in R , and let c and d be scalars.
(1) u+v is a vector in Rn
(2) u+v = v+u
(3) (u+v)+w = u+(v+w)
(4) u+0 = u
(5) u+(–u) = 0
(6) cu is a vector in Rn
(7) c(u+v) = cu+cv
(8) (c+d)u = cu+du
(9) c(du) = (cd)u
(10) 1(u) = u
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (Vector operations in R4)
Let u=(2, – 1, 5, 0), v=(4, 3, 1, – 1), and w=(– 6, 2, 0, 3) be
4
vectors in R . Solve x for x in each of the following.
(a) x = 2u – (v + 3w)
(b) 3(x+w) = 2u – v+x
Sol: (a) x  2u  ( v  3w )
 2u  v  3w
 (4,  2, 10, 0)  (4, 3, 1,  1)  ( 18, 6, 0, 9)
 (4  4  18,  2  3  6, 10  1  0, 0  1  9)
 (18,  11, 9,  8).
Dr. Abhishek Das, SAS, VIT Vellore
(b) 3(x  w )  2u  v  x
3x  3w  2u  v  x
3x  x  2u  v  3w
2x  2u  v  3w
x  u  12 v  32 w
 2,1,5,0   2, 23 , 21 , 12   9,3,0, 29 
 9, 211 , 92 ,4 
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Properties of additive identity and additive inverse)
n
Let v be a vector in R and c be a scalar. Then the following is true.
(1) The additive identity is unique. That is, if u+v=v, then u = 0
(2) The additive inverse of v is unique. That is, if v+u=0, then u = –v
(3) 0v=0
(4) c0=0
(5) If cv=0, then c=0 or v=0
(6) –(– v) = v
Dr. Abhishek Das, SAS, VIT Vellore

Linear combination:
The vector x is called a linear combination of v 1 , v 2 ,..., v n ,
if it can be expressed in the form
x  c1 v 1  c2 v 2    cn v n
c1 , c2 ,  , cn : scalar
 Ex 6:
Given x = (– 1, – 2, – 2), u = (0,1,4), v = (– 1,1,2), and
3
w = (3,1,2) in R , find a, b, and c such that x = au+bv+cw.
Sol:
 b  3c 
a 
4a 
b 
1
c  2
2b  2c   2
 a  1, b  2, c  1
Thus x  u  2 v  w
Dr. Abhishek Das, SAS, VIT Vellore

Notes:
A vector u  (u1 , u 2 ,  , u n ) in R n can be viewed as:
a 1×n row matrix (row vector): u  [u1 , u 2 ,  , u n ]
or
u1 
u 
a n×1 column matrix (column vector): u   2 
 
 
un 
(The matrix operations of addition and scalar multiplication
give the same results as the corresponding vector operations)
Dr. Abhishek Das, SAS, VIT Vellore
Vector addition
Scalar multiplication
u  v  (u1 , u 2 ,  , un )  (v1 , v2 ,  , vn )
cu  c(u1 , u 2 ,  , u n )
 (u1  v1 , u 2  v2 ,  , un  vn )
u  v  [u1 , u2 ,  , u n ]  [v1 , v2 ,  , vn ]
 [u1  v1 , u 2  v2 ,  , un  vn ]
u1  v1  u1  v1 
u  v  u  v 
u v   2   2   2 2
      
    

un  vn  un  vn 
 (cu1 , cu2 ,  , cun )
cu  c[u1 , u2 ,  , u n ]
 [cu1 , cu2 ,  , cun ]
u1  cu1 
u  cu 
cu  c  2    2 
   
   
un  cun 
Dr. Abhishek Das, SAS, VIT Vellore
Vector Spaces

Vector spaces:
Let V be a set on which two operations (vector addition and
scalar multiplication) are defined. If the following axioms are
satisfied for every u, v, and w in V and every scalar (real number)
c and d, then V is called a vector space.
Addition:
(1) u+v is in V
(2) u+v=v+u
(3) u+(v+w)=(u+v)+w
(4) V has a zero vector 0 such that for every u in V, u+0=u
(5) For every u in V, there is a vector in V denoted by –u
such that u+(–u)=0
Dr. Abhishek Das, SAS, VIT Vellore
Scalar multiplication:
(6) cu is in V.
(7) c(u  v )  cu  cv
(8) (c  d )u  cu  du
(9) c( du)  (cd )u
(10) 1(u)  u
Dr. Abhishek Das, SAS, VIT Vellore

Notes:
(1) A vector space consists of four entities:
a set of vectors, a set of scalars, and two operations
V：nonempty set
c：scalar
 (u, v)  u  v： vector addition
 (c, u)  cu： scalar multiplication
V ,
, 
is called a vector space
(2) V  0： zero vector space
Dr. Abhishek Das, SAS, VIT Vellore

Examples of vector spaces:
(1) n-tuple space: Rn
(u1, u2 ,, un )  (v1, v2 ,, vn )  (u1  v1, u2  v2 ,, un  vn ) vector addition
scalar multiplication
k (u1 , u2 , , un )  (ku1 , ku2 ,, kun )
(2) Matrix space: V  M mn (the set of all m×n matrices with real values)
Ex: ：(m = n = 2)
u11 u12  v11 v12  u11  v11 u12  v12 
u u   v v   u  v u  v 
 21 22   21 22   21 21 22 22 
u11 u12  ku11 ku12 
k



u
u
ku
ku
22 
 21 22   21
vector addition
scalar multiplication
Dr. Abhishek Das, SAS, VIT Vellore
(3) n-th degree polynomial space: V  Pn (x)
(the set of all real polynomials of degree n or less)
p ( x)  q ( x)  (a0  b0 )  (a1  b1 ) x    (an  bn ) x n
kp( x)  ka0  ka1 x    kan x n
(4) Function space: V  c(, ) (the set of all real-valued
continuous functions defined on the entire real line.)
( f  g )( x)  f ( x)  g ( x)
(kf )( x)  kf ( x)
Dr. Abhishek Das, SAS, VIT Vellore

Notes: To show that a set is not a vector space, you need
only find one axiom that is not satisfied.
 Ex: The set of all integer is not a vector space.
Pf:
1  V , 12  R
( 12 )(1)  12  V (it is not closed under scalar multiplication)
  noninteger

scalar
integer
 Ex: The set of all second-degree polynomials is not a vector space.
Pf:
Let p ( x)  x 2 and q ( x)   x 2  x  1
 p ( x)  q ( x)  x  1  V
(it is not closed under vector addition)
Dr. Abhishek Das, SAS, VIT Vellore

Ex:
V=R2=the set of all ordered pairs of real numbers
vector addition: (u1 , u 2 )  (v1 , v2 )  (u1  v1 , u 2  v2 )
scalar multiplication: c(u1 , u2 )  (cu1 ,0)
Verify V is not a vector space.
Sol:
1(1, 1)  (1, 0)  (1, 1)
 the set (together with the two given operations) is
not a vector space
Dr. Abhishek Das, SAS, VIT Vellore
Subspaces of Vector Spaces

Subspace:
(V ,,) : a vector space
W  
 : a nonempty subset
W V
(W ,,) ：a vector space (under the operations of addition and
scalar multiplication defined in V)
 W is a subspace of V
 Trivial subspace:
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.
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Test for a subspace)
If W is a nonempty subset of a vector space V, then W is
a subspace of V if and only if the following conditions hold.
(1) If u and v are in W, then u+v is in W.
(2) If u is in W and c is any scalar, then cu is in W.
Dr. Abhishek Das, SAS, VIT Vellore

Ex: Subspace of R2
(1) 0
0  0, 0
(2) Lines through the origin
(3) R 2

Ex: Subspace of R3
(1) 0
0  0, 0, 0 
(2) Lines through the origin
(3) Planes through the origin
(4) R 3
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (A subspace of M2×2)
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.
Sol:
W  M 22
M 22 : vector sapces
Let A1, A2  W ( A1T  A1, A2T  A2 )
A1  W, A2  W  ( A1  A2 )T  A1T  A2T  A1  A2 ( A1  A2  W )
k  R , A  W  (kA)T  kAT  kA
(kA W )
W is a subspace of M 22
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (The set of singular matrices is not a subspace of M2×2)
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:
1 0 
0 0
A
W , B  
W


0 0 
0 1 
1 0
A B  
W

0 1 
W2 is not a subspace of M 22
Dr. Abhishek Das, SAS, VIT Vellore

2
Ex: (The set of first-quadrant vectors is not a subspace of R )
Show that W  {( x1 , x2 ) : x1  0 and x2  0} , with the standard
operations, is not a subspace of R2.
Sol:
Let u  (1, 1)  W
  1u   11, 1   1,  1  W
(not closed under scalar
multiplication)
W is not a subspace of R 2
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (Determining subspaces of R2)
Which of the following two subsets is a subspace of R2?
(a) The set of points on the line given by x+2y=0.
(b) The set of points on the line given by x+2y=1.
Sol:(a)
Let
We have to show that,
(closed under addition)
Dr. Abhishek Das, SAS, VIT Vellore
W is a subspace of R 2
Dr. Abhishek Das, SAS, VIT Vellore
(b) W   x, y  x  2 y  1
(Note: the zero vector is not on the line)
Let v  (1,0)  W
  1v   1,0  W
W is not a subspace of R 2
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (Determining subspaces of R3)
Which of the following subsets is a subspace of R 3?
(a) W  ( x1 , x2 ,1) x1 , x2  R
(b) W  ( x1 , x1  x3 , x3 ) x1 , x3  R
Sol:
(a) Let v  (0,0,1)  W
 (1) v  (0,0,1)  W
W is not a subspace of R 3
(b) Let v  ( v1 , v1  v 3 , v 3 )  W , u  (u1 , u1  u 3 , u 3 )  W
 v  u  v1  u1 , v1  u1   v 3  u 3 , v 3  u 3   W
kv  kv1 , kv1   kv 3 , kv 3   W
W is a subspace of R 3
Dr. Abhishek Das, SAS, VIT Vellore
 Thm:
(The intersection of two subspaces is a subspace)
If V and W are both subspaces of a vector space U ,
then the intersection of V and W (denoted by V  U )
is also a subspace of U .
Dr. Abhishek Das, SAS, VIT Vellore
Spanning Sets and Linear Independence

Linear combination:
A vector v in a vector space V is called a linear combination of
the vectors u1,u 2 , ,u k in V if v can be written in the form
v  c1u1  c2u 2    ck u k
c1,c2 , ,ck : scalars
Dr. Abhishek Das, SAS, VIT Vellore

Ex 2-3: (Finding a linear combination)
v1  (1,2,3) v 2  (0,1,2) v 3  ( 1,0,1)
Prove (a) w  (1,1,1) is a linear combination of v1 , v 2 , v 3
(b) w  (1,2,2) is not a linear combination of v1 , v 2 , v 3
Sol:
(a) w  c1 v1  c2 v 2  c3 v 3
1,1,1  c1 1,2,3  c2 0,1,2  c3  1,0,1
 (c1  c3 , 2c1  c2 , 3c1  2c2  c3 )
c1
 c3
 2c1  c2
3c1  2c2  c3
1
1
1
Dr. Abhishek Das, SAS, VIT Vellore
1 0  1 1
Guass  Jordan Elimination



 2 1 0 1      
3 2 1 1
1 0  1 1 
0 1 2  1


0 0 0 0 
 c1  1  t , c2  1  2t , c3  t
(this system has infinitely many solutions)
t 1
 w  2 v1  3 v 2  v 3
Dr. Abhishek Das, SAS, VIT Vellore
(b)
w  c1 v1  c2 v 2  c3 v 3
1 0  1 1 
GuassJordan Elimination



 2 1 0  2 
3 2 1
2 
1 0  1 1 
0 1 2  4 


0 0 0
7 
 this system has no solution ( 0  7)
 w  c1 v1  c2 v 2  c3 v 3
Dr. Abhishek Das, SAS, VIT Vellore

the span of a set: span (S)
If S={v1, v2,…, vk} is a set of vectors in a vector space V,
then the span of S is the set of all linear combinations of
the vectors in S,
span(S ) c1 v1  c2 v 2    ck v k
ci  R
(the set of all linear combinations of vectors in S )

a spanning set of a vector space:
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.
Dr. Abhishek Das, SAS, VIT Vellore
 Notes:
span ( S )  V
 S spans (generates) V
V is spanned (generated) by S
S is a spanning set of V
 Notes:
(1) span( )  0
(2) S  span( S )
(3) S1 , S 2  V
S1  S 2  span( S1 )  span( S 2 )
Dr. Abhishek Das, SAS, VIT Vellore
 Ex: (A spanning set for R3)
Sol:
We must determine whether an arbitrary vector u  (u1 , u2 , u3 )
in R 3 can be as a linear combination of v1 , v 2 , and v 3 .
u  R 3  u  c1 v1  c2 v 2  c3 v 3
 c1
2c1  c2
 2c3  u1
 u2
3c1  2c2  c3  u3
The problem thus reduces to determining whether this system
is consistent for all values of u1 , u2 , and u3 .
Dr. Abhishek Das, SAS, VIT Vellore
1 0 2
A2 1
3 2
0 0
1
 Ax  b has exactly one solution for every u.
 span( S )  R 3
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Span(S) is a subspace of V)
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).)
Dr. Abhishek Das, SAS, VIT Vellore

Linear Independent (L.I.) and Linear Dependent (L.D.):
S  v1 , v 2 ,  , v k  : a set of vectors in a vector space V
c1 v1  c2 v 2    ck v k  0
(1) If the equation has only the trivial solution (c1  c2    ck  0)
then S is called linearly independent.
(2) If the equation has a nontrivial solution (i.e., not all zeros),
then S is called linearly dependent.
Dr. Abhishek Das, SAS, VIT Vellore

Notes:
(1)  is linearly independent
(2) 0  S  S is linearly dependent.
(3) v  0  v is linearly independent
(4) S1  S 2
S1 is linearly dependent  S 2 is linearly dependent
S 2 is linearly independent  S1 is linearly independent
Dr. Abhishek Das, SAS, VIT Vellore
 Ex: (Testing for linearly independent)
3
Determine whether the following set of vectors in R is L.I. or L.D.
S  1, 2, 3, 0, 1, 2,  2, 0, 1
v1
v2
v3
c1
 2c3  0
Sol:
0
c1 v1  c2 v 2  c3 v 3  0  2c1  c2 
3c1  2c2  c3  0
1 0  2 0 
1 0 0 0 
- Jordan Elimination
 2 1 0 0 Gauss


 0 1 0 0
3 2 1 0
0 0 1 0
 c1  c2  c3  0 only the trivial solution 
 S is linearly independent
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (Testing for linearly independent)
Determine whether the following set of vectors in P2 is L.I. or L.D.
Sol:
S = {1+x – 2x2 , 2+5x – x2 , x+x2}
v1
v2
v3
c1v1+c2v2+c3v3 = 0
i.e. c1(1+x – 2x2) + c2(2+5x – x2) + c3(x+x2) = 0+0x+0x2
2 0 0
1
c1+2c2
=0
G.J.
 c1+5c2+c3 = 0   1 5 1 0 
–2c1 – c2+c3 = 0
 2  1 1 0
1 2 0 0 


1
1
1
0


3
0 0 0 0 


 This system has infinitely many solutions.
(i.e., This system has nontrivial solutions.)
 S is linearly dependent.
(Ex: c1=2 , c2= – 1 , c3=3)
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (Testing for linearly independent)
Determine whether the following set of vectors in 2×2
matrix space is L.I. or L.D.
Sol:
2 1 3 0 1 0 
S  
,
,



0
1
2
1
2
0
 
 


v1
v2
v3
c1v1+c2v2+c3v3 = 0
2 1
 3 0
1 0  0 0 
c1 
 c2 
 c3 





0
1
2
1
2
0
0
0





 

Dr. Abhishek Das, SAS, VIT Vellore

2c1+3c2+ c3 = 0
c1
=0
2c2+2c3 = 0
c1 + c2
=0
 2
1

0

1
3 1 0
1
0 0 0 Gauss- Jordan Elimination 0

 
0
2 2 0


1 0 0
0
0 0 0
1 0 0
0 1 0

0 0 0
 c1 = c2 = c3= 0 (This system has only the trivial solution.)
 S is linearly independent.
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (A property of linearly dependent sets)
A set S = {v1,v2,…,vk}, k2, is linearly dependent if and
only if at least one of the vectors vj in S can be written as
a linear combination of the other vectors in S.
Pf:
() c1v1+c2v2+…+ckvk = 0
 S is linearly dependent
 ci  0 for some i
c1
ci 1
ci 1
ck
 v i  v1   
v i 1 
v i 1    v k
ci
ci
ci
ci
Dr. Abhishek Das, SAS, VIT Vellore
()
Let
vi = d1v1+…+di-1vi-1+di+1vi+1+…+dkvk
 d1v1+…+di-1vi-1-vi+di+1vi+1+…+dkvk = 0
 c1=d1, …,ci-1=di-1, ci=-1,ci+1=di+1,…, ck=dk (nontrivial solution)
 S is linearly dependent

Corollary to Theorem 4.8:
Two vectors u and v in a vector space V are linearly dependent
if and only if one is a scalar multiple of the other.
Dr. Abhishek Das, SAS, VIT Vellore
Basis and Dimension

Basis:
V：a vector space
S ={v1, v2, …, vn}V
Generating
Sets
Bases
Linearly
Independent
Sets
(a ) S spans V (i.e., span(S) = V )

(b) 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. Abhishek Das, SAS, VIT Vellore
n
(3) the standard basis for R :
{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)}
(4) the standard basis for mn matrix space:
{ Eij | 1im , 1jn }
Ex: 2 2 matrix space:
 1 0   0 1   0 0   0 0  
,
,
,





0
0
0
0
1
0
0
1
 
 
 


(5) the standard basis for Pn(x):
{1, x, x2, …, xn}
Ex: P3(x)
{1, x, x2, x3}
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Uniqueness of basis representation)
If S  v1 , v 2 ,  , v n  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.
Pf:
 1. span(S) = V
 S is a basis  
 2. S is linearly independent
 span(S) = V Let v = c1v1+c2v2+…+cnvn
v = b1v1+b2v2+…+bnvn
 0 = (c1–b1)v1+(c2 – b2)v2+…+(cn – bn)vn
 S is linearly independent
 c1= b1 , c2= b2 ,…, cn= bn
(i.e., uniqueness)
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Bases and linear dependence)
If S  v1 , v 2 ,  , v n is a basis for a vector space V, then every
set containing more than n vectors in V is linearly dependent.
Pf:
Let S1 = {u1, u2, …, um} , m > n
 span ( S )  V
u1  c11 v1  c21 v 2    cn1 v n
uiV

u 2  c12 v1  c22 v 2    cn 2 v n

u m  c1m v1  c2 m v 2    cnm v n
Dr. Abhishek Das, SAS, VIT Vellore
Let k1u1+k2u2+…+kmum= 0
 d1v1+d2v2+…+dnvn= 0
(where di = ci1k1+ci2k2+…+cimkm)
 S is L.I.
 di=0
i
i.e.
c11k1  c12 k 2    c1m k m  0
c21k1  c22 k 2    c2 m k m  0

cn1k1  cn 2 k 2    cnm k m  0
 If the homogeneous system has fewer equations than
variables, then it must have infinitely many solution.
m > n  k1u1+k2u2+…+kmum = 0 has nontrivial solution
 S1 is linearly dependent
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Number of vectors in a basis)
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.)
Dr. Abhishek Das, SAS, VIT Vellore


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:
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
 dim(V) = #(S)
S: a basis for V
(the number of vectors in S)
Dr. Abhishek Das, SAS, VIT Vellore
 Note:
dim({0}) = 0 = #(Ø)
Dr. Abhishek Das, SAS, VIT Vellore

Ex:
(1) Vector space Rn
 basis {e1 , e2 ,  , en}
 dim(Rn) = n
(2) Vector space Mmn  basis {Eij | 1im , 1jn}
 dim(Mmn)=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. Abhishek Das, SAS, VIT Vellore

Ex 9: (Finding the dimension of a subspace)
(a) W={(d, c–d, c): c and d are real numbers}
(b) W={(2b, b, 0): b is a real number}
Sol: (Note: Find a set of L.I. vectors that spans the subspace)
(a) (d, c– d, c) = c(0, 1, 1) + d(1, – 1, 0)
 S = {(0, 1, 1) , (1, – 1, 0)} (S is L.I. and S spans W)
 S is a basis for W
 dim(W) = #(S) = 2
(b)  2b, b,0   b2,1,0 
 S = {(2, 1, 0)} spans W and S is L.I.
 S is a basis for W
 dim(W) = #(S) = 1
Dr. Abhishek Das, SAS, VIT Vellore

Ex: (Finding the dimension of a subspace)
Let W be the subspace of all symmetric matrices in M22.
What is the dimension of W?
Sol:
a b 

W  
a , b, c  R 

b c 

a b
1 0  0 1   0 0 

 a
 b
 c




b
c
0
0
1
0
0
1



 
 

 1 0   0 1   0 0  
 S  
,
,
 spans W and S is L.I.



  0 0  1 0   0 1  
 S is a basis for W
 dim(W) = #(S) = 3
Dr. Abhishek Das, SAS, VIT Vellore

Thm: (Basis tests in an n-dimensional space)
Let V be a vector space of dimension n.
(1) If S  v1 , v 2 ,  , v n  is a linearly independent set of
vectors in V, then S is a basis for V.
(2) If S  v1 , v 2 ,  , v n  spans V, then S is a basis for V.
Dr. Abhishek Das, SAS, VIT Vellore
Thm: (Extension theorem and Deletion theorem)
Let V be a finite-dimensional vector space.
(i) Any linearly independent set in V can be extended to a basis by
adding more vectors if necessary
(ii) Any set of vectors that spans V can be reduced to a basis by
discarding vectors if necessary.
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore
Dr. Abhishek Das, SAS, VIT Vellore