Module 4: Vector Spaces
n
n
A real coordinate space of dimension n (we write it as R or
, where n is a natural
number) is a coordinate space over the real numbers. This means that it is the set of the ntuples of real numbers (sequences of n real numbers).
For example,


R1 is the real line with points as a real numbers on it (R1: set of all real number).
R2 is a plane with points as ordered pair of real numbers  x1 , x2  (R2: set of all ordered
pair of real numbers).

R3 is of three-dimensional space with points as ordered triple of real numbers  x1 , x2 , x3 

(R3: set of all ordered triple of real numbers).
R4 is a four-dimensional space with points as ordered quadruple of real numbers
 x1 , x2 , x3 , x4  (R4: set of all ordered quadruple of real numbers).

Rn is an n-dimensional space with points as ordered n-tuple of real numbers
 x1 , x2 , x3 ...., xn  (Rn: set of all ordered n-tuple of real numbers).
Standard operations in Rn
Dr. PARVEZ ALAM
The sum of two vectors and the scalar multiple of a vector in Rn are called the standard
operations in Rn.
For example:
Let u=(2, 3, 1, 5) and v=(-2, -3, -1, -5); here v is additive inverse of u.
If we add zero vector(additive identity) 0=(0, 0, 0, 0) in u it will be same, i.e. 0+u=u+0=u
Dr. PARVEZ ALAM
(0+2, 0+3, 0+1, 0+5)= (2+0, 3+0, 1+0, 5+0)= (2, 3, 1, 5)=u.
Note:
Dr. PARVEZ ALAM
The matrix operations of addition and scalar multiplication give the same results as the
corresponding vector operations.
Vector Space
A non-empty set V is said to be a vector space over
(field) if there exist two maps
 : V×V  V defined by  v1 , v2  v1  v2 called addition, and  : ×V  V defined by
 , v 
  v called scalar multiplication, satisfying following properties
(w.r.t. addition  : V×V  V )
u  v V,
V1 Closure:
for all elements u , v of V
V2 Commutativity:
uvvu,
for all elements u , v of V
V3 Associativity:
 u  v   w  u   v  w ,
for all elements u , v, w of V
V4 Existence of
additive identity:
There exists an 0 ∈ V, such that 0 v  v ,
V5 Existence of
additive inverse:
For every vV, there exists an u ∈ V, such that v  u  0  u  v .
This u is denoted by v .
for all elements v of V
*It means V is Abelian w.r.t. ‘+’ addition.
(w.r.t. scalar multiplication  : ×V  V )
V6 Closure w.r.t.
for all elements v of V and scalar  
  v V,
scalar multiplication
V7 Associative w.r.t.
scalar multiplication
    v        v , for all elements v of V and scalar , 
V8 Distributive
   u  v     u    v , for all elements u, v of V and scalar  
V9 Distributive
     v    v    v ,
V10 Action of 1
1  v  v , for all elements v of V
for all elements v of V and scalar , 
Dr. PARVEZ ALAM
Note:
V ,
, ,   ,  and  are two defined

Usually we are writing a vector space as



operations on which set V satisfy all above 10 axioms. These two operations may be
different for other vectors spaces.
The elements (u,v,w,……. or x,y,z……., etc.) of a vector space V are called vectors.
If any set that not satisfying any one condition then the set will not be a vector space.
V  0; is a vector space and we say it zero vector space. (Check all conditions of VS

Exercise)
V  is a vector space over itself
. (Check all conditions of VS Exercise)
Dr. PARVEZ ALAM
Question 1: Show that V 
2
  x   x1, x2  ; x1, x2 
 is a vector space. V.S. 
2
,
, , 
Dr. PARVEZ ALAM
Question 2: Show that V 
2
Similar to the
n
 x   x1, x2 , x1......xn ; x1, x2 , x1......xn 
, we can show that
n
 is a vector space.
is also a vector space.
Hint: We take 3 vectors x   x1 , x2 , x1......xn  , y   y1, y2 , y1...... yn  & z   z1, z2 , z1......zn  and 2
scalars , 

Dr. PARVEZ ALAM
V.S. 
n
,
.
, , 
Question 3: Show that set of all 2x2 square real matrices
Dr. PARVEZ ALAM
u
u 
V  M 22    11 12  ; u11 , u12 , u21 , u22 
 u21 u22 

 is a vector space. V.S.  M 22 , , ,  

Dr. PARVEZ ALAM
Question 4: Show that set of all real matrices of order m x n
  u11 u12 ... u1n 


  u21 u21 ... u2 n 
V  M mn  
; u11 , u12 , u21 , u22 ....umn 


...
...
...
...

 um1 um1 ... umn 
mn

Similar to the M 22 , we can show that M mn is also a vector space.



 is a vector space.



Hint: We take 3 matrices
 a11 a12
a
a21
A   21
 ... ...

 am1 am1
,  .
Dr. PARVEZ ALAM
V.S.  M mn ,
... a1n 
 b11 b12
b
... a2 n 
b21
 , B   21
 ... ...
... ... 


... amn 
bm1 bm1
, ,  
... b1n 
 c11 c12
c
... b2 n 
c21
 & C   21
 ... ...
... ... 


... bmn 
cm1 cm1
... c1n 
... c2 n 

... ... 

... cmn 
and
2
scalars
Question 5: Show that n-th degree polynomial space: (It consist all degree polynomials upto
degree  n ) V.S.  Pn ( x), , ,  
Dr. PARVEZ ALAM
V  Pn ( x)  an x n  an1x n1  ....  a2 x 2  a1x  a0 ; an , an1 ,....., a2 , a1, a0 
 is a vector space.
Dr. PARVEZ ALAM
Note: To show that a set is not a vector space, you need only
find one axiom that is not satisfied.
Ex1.The set of all integers (V  ) is not a vector space.
Proof: It is not closed under scalar multiplication (Failing V6 condition)
Ex2: The set of all second-degree polynomials is not a vector
space.
Proof: It is not closed under vector addition (Failing V1 condition)
Ex3:
with defined operations
Dr. PARVEZ ALAM
Proof: (Failing V10 condition; 1.v=v)
Vector Subspace
Definition: Let V be a vector space over
, then W is said to be a subspace of V if and only if
W is a subset of V (i.e, W  V) and W is itself a vector space over
of V.
with same operations
Note: Every vector space V has at least two subspaces known as trivial subspace of V.
(1) Zero vector space W={0} is a subspace of V.
(2) V is a subspace of V (itself).
Theorem: (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 for
every elements w1 , w2 W and scalars   ; following two conditions should satisfy
1. (V1 Closure property w.r.t. vector addition): w1  w2  W
2. (V6 Closure property w.r.t. scalar multiplication): w1  W
Alternate: Some time we use combined form of V1 and V6 to show subspaces as; w1 , w2 W
and scalars ,  , w1  w2  W.
Dr. PARVEZ ALAM
Note: We can any of one to check subspaces, either test by showing V1 and V6 or together
(Alternate). But some time Alternate one is being complex see Ex. 1,2,6, however it’s time
saving. But go for normal test by showing V1 and V6 see Ex. 7,8,910.
Dr. PARVEZ ALAM
Dr. PARVEZ ALAM
Dr. PARVEZ ALAM
Homework Questions:
Intersection of two vector subspaces
Theorem1: The intersection W1  W2 of two vector subspaces W1 and W2 a vector space V
is also vector subspace of V.
Union of two vector subspaces
Theorem2: The union W2  W1 of two vector subspaces W1 and W2 a vector space V is also
Dr. PARVEZ ALAM
vector subspace of V if and only if, either W1  W2 or W2  W1 .
What is ‘Linear combination’?
Let
u1 , u2 , u3 , u4 , ...., un are n vectors in vector space V and c1 , c2 , c3 , c4 , ...., cn are n
scalars in
, then the combination c1u1  c2 u2  c3 u3  c4 u4 ...., cn un will give a new
vector x (say) and this x vector is called linear combination of vectors u1 , u2 , u3 , u4 , ...., un .
i.e., c1u1  c2 u2  c3 u3  c4 u4 ...., cn un  x V.
Dr. PARVEZ ALAM
and this x also belongs to the vector space V.
Dr. PARVEZ ALAM
Linearly Span
If we consider a set S  s1 ,s 2 ,s3 ,....s n   V , then the set of all linear combinations (for different
set of scalars we will have different linear combinations) of these vectors si's called linear span
of set S, we write it by symbol L(S) and given by
n

L(S)   isi 1s1   2s 2 + 3s3 +...+ ns n ;  i  
 i1

Note: 1. This L(S) will also form a vector subspace of V.
2. L(S) is smallest vector subspace of V containing S.
Dr. PARVEZ ALAM
A spanning set of a vector space: If every vector in a given vector space V 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 and we can write as
L(S)=V
It means V is spanned (generated) by S.
Linearly Independent (L.I.)
&
Linearly Dependent (L.D.)
Let a set S  s1 ,s 2 ,s3 ,....s n   V , then vectors s1,s2 ,s3 ,....s n are said to be linearly
independent, if the vector equation
1s1  2s2 + 3s3 +...+nsn  0
………..(1)
has only the trivial solution 1  2 =3 =....=n  0 .
If the equation (1) has non-trivial solution, i.e., all scalars 1, 2 , 3 ,....n are not zero, then
we say that vectors s1,s2 ,s3 ,....s n are linearly dependent (L.D.).
Remarks:
Dr. PARVEZ ALAM
1.  is L.I.
2. If the set S is L.I. then every subset of S is L.I.
3. If the set L.D. then every superset of S is L.D.
4. Every singleton set of non-zero vector is L.I.
5. The singleton set S  {v} is L.D. if and only if v is zero vector.
6. If a non-zero vector v is multiple of a non-zero vector u, then u, v are L.D. i.e., set {u, v}
is L.D.
For example; {u=(1, 2, 3), v=(2, 4, 6)} is L.D., because v is multiple of u, v=2u.
Ex.2 {u=(1, 2, 3), v=(1, 2, 3)} is also L.D. same element, v=1.u.
Dr. PARVEZ ALAM
Dr. PARVEZ ALAM
Dr. PARVEZ ALAM
Dr. PARVEZ ALAM
Basis and dimension of a vector space
Basis: Let V be a vector space. A subset S of V is called basis if
1. S spans V, i.e., L(S)=V
2. Subset S is linearly independent
Dimension: The number of elements (cardinality) in a basis of V is called the
dimension of V, and is denoted by dim V.
Notes:
1.  is basis for V.S. V={0}, so dim 0  0.
 
2
So, the dimension of
3
2
; B  e1  (1, 0, 0), e2  (0, 1, 0), e3  (0, 0, 1) .
is 3, i.e., dim
4
4. The standard basis for
   2.
is 2, i.e., dim
3
3. The standard basis for
So, the dimension of
; B  e1  (1, 0), e2  (0, 1) .
2
2. The standard basis for
   3.
3
;
B  e1  (1, 0, 0, 0), e2  (0, 1, 0, 0), e3  (0, 0, 1, 0), e4  (0, 0, 0, 1) .
4
So, the dimension of
5. The standard basis for
is 4, i.e., dim
n
   4.
4
;
B  e1  (1, 0, 0, 0, ...,0), e2  (0, 1, 0, 0,...,0), e3  (0, 0, 1, 0,...,0), .......... en  (0, 0, 0, 0,...,1) .
So, the dimension of
3
is 2, i.e., dim
   n.
n
6. The standard basis for 2 x 2 matrix space M 22 :
Dr. PARVEZ ALAM

1 0 
0 1 
0 0 
0 0 
B  e1  
, e2  
, e3  
, e4  




0 0 
0 0 
1 0 
0 1  

So, the dimension of M 22 is 4, i.e., dim  M 22   2  2  4.
7. The standard basis for 3 x 2 matrix space M 32 :

1 0 
0 1 
0 0
0 0
0 0
0 0 


B  e1  0 0 , e2  0 0 , e3  1 0 , e4  0 1  , e5  0 0 , e6  0 0 













0 0
0 0
0 0
0 0
1 0
0 1  

So, the dimension of M 32 is 2, i.e., dim  M 32   3  2  6.
8. So, the dimension of M mn is given by dim  M mn   m  n.

 . So, dim  P ( x)   2  1  3.
10. The standard basis for P ( x) ; B  e  1, e  x, e  x , e  x  . dim  P ( x)   3  1  4 .
11. The standard basis for P ( x) ; B  e  1, e  x, e  x ,..., e  x  . dim  P ( x)   n  1.
12. The standard basis for P ( x ) ; B  e  1, e  x, e  x ,......... . So, dim  P ( x)    .
9. The standard basis for P2 ( x) ; B  e1  1, e2  x, e3  x
2
2
2
3
1
2
3
n
1
2
3
3
3
4
2
n
n
n
2
1
2
3
n
12. Generally we are representing standard basis by B  e1 , e2 , e3 , .......... en  .
Theorem: (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 have the same number of vectors.)
Finite dimensional: A vector space V is called finite dimensional, if it has a basis consisting of
a finite number of elements.
For example:  Pn ( x) set of polynomials of all degrees less or equals to n is dim  Pn ( x)   n .
Dr. PARVEZ ALAM
Infinite dimensional: If a vector space V is not finite dimensional, then it is called infinite
dimensional.
For example:
Question 1: Find the basis and dimension of subspaces U and W;
Dr. PARVEZ ALAM
(a). W   d , c  d , c  ; c, d 
,
(b). U   2 x, x, 0 ; a, c 

Que 2: Let W be the subspace of all symmetric matrices in M 22 . What is the dimension of W?
OR
a b 

such
that
b

c
;
a
,
b
,
c
,
d

.

c
d



Find the basis and dimension of subspace W   
OR
a b 

;
a
,
b
,
c
,
d

.

b
d




Dr. PARVEZ ALAM
Find the basis and dimension of subspace W   
Note: If in some problems conditions are given in equation form, we are solving the
condition(s) to get values of coordinates to put in the set.
Ques 3: Find the basis and dimension of the plane x + 3y + z = 0 in
OR

Find the basis and dimension of set W   x, y, z  
3
3
.
; x  3 y  z  0 .
OR
Find the basis and dimension of subspace W
Solution: First we solve the equation(s) x  3 y  z  0 given as condition to obtain the
value of x, y, z, then we put in the coordinate of the set W.
x  3 y  z  0; let y=t and z=s, then we can obtain x  3t  s . Putting in set W, we get
W   3t  s, t , s  
3
; x  3 y  z  0 .
Now this problem changes to last problem (1)
 3t  s, t , s    3t , t , 0    s, 0, s 
 t  3, 1, 0   s  1, 0, 1
Therefore;
S   3, 1, 0    1, 0, 1  L( S )  W

Now, we check its L.I.;

Hence, S is the basis of W, and Dim W=2.
c1  3, 1, 0   c2  1, 0, 1   0, 0, 0 
On simplifying and solving we are getting c1  c2  0
Dr. PARVEZ ALAM
Note: If only one condition given we directly put any one variable in set and proceed.
See the above examples solution;
Dr. PARVEZ ALAM
Que 4: (HW) Find the basis and dimension of the plane x + 2z = 0 in
OR

Find the basis and dimension of set W   x, y, z  
3
3
.
; x  2 z  0 .
Hint: Let y  s and z  t then by x  2 z  0, we get x  2s

Then we get the set W as W   2t , s, t  ; s, t 
OR
.
Or directly we can replace x  2 z by x  2 z  0, and putting y=y and z=z so we get W as
W   2 z, y, z  ; y, z 
.
Remark: If more than one equations are in condition then we have to solve the system
of equations. See the next example
3
Que 5: Find the basis and dimension of subspace W of
W   x, y, z  
3
; x  y  z  0, x  2y  2 z  0, 2 x  y  z  0 .
Solution: First we solve the equations
x  y  z  0,
x  2y  2 z  0,
2x  y  z  0
given in set, then we replace in set.
1 1 1 0 
1 1 1 0 




1 2 2 0  by Gauss Ele. method (Row Echelon form), we get 0 1 1 0 
 2 1 1 0 
0 0 0 0 
 Here, n    3  2  1 so, assuming one variable z  t , then we get x  0, y  t .

Now the subspace W can be written as;
W   0,  t , t  
3
; t
.
Now, we find set of vectors that spans W;
 0,  t, t   t  0,  1, 1
S   0,  1, 1  L( S )  W
Dr. PARVEZ ALAM


Now, we check its L.I.; S is singleton set of a non-zero element, so it is L.I.
Hence, S is the basis of W, and Dim W=1.
Corollary 1. Let S be a set of n vectors in an n-dimensional vector space V. If S is
independent, then S is a basis for V.
Corollary 2. Let S be a set of n vectors in an n-dimensional vector space V. If S spans V, then
S is a basis for V.

Que 5: Is the set of vectors W  1, 1, 2  , 1, 2, 5  ,  5, 3, 4 
Dr. PARVEZ ALAM
for
3
 in V ( )  i.e.,  form a basis
3
3
? (Use of Corollary 1)
Important Note: If vectors after putting matrix are giving square matrix, the by finding
determinant only we can tell about L.I and L.D.
1. If Determinant is zero the vectors are L.D.
2. If Determinant is non-zero the vectors are L.I.
Dr. PARVEZ ALAM