Chapter

Vector Spaces

honoisllqsltesaoigles

o far we have studied groups and rings, Now we shall study another important
Salgebraicstructureknown as vector space. Before givingthe definitionof avector
space we shall make a distinction between internal and external compositions.

Let A be any * be A a,be A,and a* b is unique then *is said to be an

set. If a

internal composition in the set A.Here a and b are both elements of the setA.

Let V and F be any two sets. If aoeVfor all a F and for all a e V and aoa is
unique, then o is said to be an external composition in V overF. Hereais an element
of the set F and a is an element of theset V and the resulting element a oa is an
element of the set V.eNwoNN
11 Vector Space

Definition : Let (F, +,)be a field. The elements of F will be called scalars. Let V be a
non-empty set whose elements will be called vectors. Then V a vector spaceover thefield F, if
is

1. There is defined an internal composition in V called addition of vectors and denoted

by '+'. Also for this composition V is an abelian group.
 L4
2. and denoted
There is an external composition in Vover Fcalled scalar multiplication

mtipicatively i.e., a ae V for all aeF and for all ae

V. In other Words V is closed

with respect to scalar multiplication.

3. scalar multinlication and addition of vectors satisfy the
1he tvo compositions i.e.,

following postulates

(1) a(oa = aa+aß taeFand ta,ße V.

+ B)

(i) (a +b) a =aa +ba a,be and ¥ ae F V.

(iii) (ab)a =a (ba)V a, be FandVae V.

(iv) la =aVaeV and I is the unity element of the feld F.

When V isa vector space over the fieldE,we shall say that V (F)is a vector space. If
 13 Vector Subspaces ve bae Porgoe
Definition: Let Vbe a vector spaceoverthe feldFand WV. Then Wis called a subspace
of Vif W itself is a vector space over F with respect to the operations of vector addition and scalar

multiplication in V.

Theorem l: The necessary and sufficient condition fora non-empty subset W of a vector W
Space V (F) to be a subspace of V is that W
is closed under vector addition and scalar

multiplication in V. ads hoat

Proof:o If W itself is a vector space over F with respect to vector addition and scalar
multiplication in V,then W must be closed with respect to these two compositions.
Hence the condition is necessary.
The condition is sufficient. Now suppose that Wis a non-empty subset of V and
Wis closed under vector addition and scalar multiplication in V.
then- le F. Now W

:
Let ae W.If I is the unity element of F, is closed under scalar

multiplication. Therefore

-le F, ae W >(-)ae W-(lo)-0e W. s e W

We V and la =ain V.]
a

Thus the additive inverse of each element of W is also in W.

Now W is closed under vector addition.

Therefore aeW, -aeW a +(-a)e W 0e W,hodhott

where 0 is the zero vector of V.
Hence the zero vector of V is also the zero vector of W.Since the elements of W are
also the elements of V, therefore vector addition will be commutative as well as

associative in W. Hence Wis an abelian group with respectto vector addition. Also it

is given that W is closed under scalar multiplication. The remaining postulates ofa
vector space will hold in W since they hold in Vof which Wis a subset.

.
Hence
Wis
W itself is

asubspace
a vector space
of V.
for the two compositions.
 15 Linear Combination of Vectors and Linear Span of aSet
. Linear combination. Definition. Let V (F)be a vector space. If ay, a 0.,e V,
then any vector

is calleda linear combination of the vectors j,O2,..,

Linear span. Definition: Let V (F)
be a vector space andS be any non-empty subset of

V. Then the linear span of S is the setof alllinear combinations of finitesets of elements of Sand
is denoted by L (S). Thus we have

is any arbitrary finite subset of S and a, ,...4

a)

is any arbitrary finite subset of F}.

Theorem l: The linear span L(S)ofany subset Sofa vector space V (F)isa subspace of V
generated by S i.e., L(S) = (S}.

Proof: Let a ,ßbe any two elementsof L (S). Ne noolo ve sd (a.Ai

Then
and
where the a'sand b's are elements of F and the a's and B's are elements of S

If a ,bbe any two elements of F, then

aa + bß =a (a a t ay ay t...+ amO.m) + b (bj Bi + B2t..+b,
b, B)
=a (a a )+ a(ay az) +..ta (a,a)+ b (bi Bi) + b (b B2)t+...
...+b (b,B)
= (aa) a +(aay) ay +...+ (aam) +(bb)B +(bb,) B2 t...
.,

.+(bb) B

Thus aa +bß has been expressed as a linear combination of a finite set

aq .az ..„.a,m .B.B2 ...p,of the clements of S.Consequently aa bß eL (S). +

Thus +
a,be F and a ,ße L (S) aa bße L(S). Hence L (S)is a subspace of

V (F).
Also each element of S belongs to L (S)because if a., e S,then a,, =la, and this

implies that .,e L (S). Thus L (S)is a subspace of V and S is contained in L(S).
Now W
if is any subspace of V containing S,then each element of L (S)must bein W
because W is to be closed under vector addition and scalar multiplication. Therefore

L (S)
will be contained in W.
Hence L (S)=(S)i.e., L (S)is the smallest subspace of V containing S.
 Evample 23: In the vector space R'express the vector -2,5)as a linear combination
(J, of the

vectors (1, l1,), (1,2,3) and (2,- 1, I).

Solution: Let a =(l,- 2,5), =(1,1,), az =(1,2,3), ky =(2,- 1,1).

We wish to expressa as a linear combination of o,a2 3

Then (,-2,5) = a1 (,1,1) + ay (1,2,3)

=(a1, aj, a)+(az ,2ay,3ag)+(2ag- ag,a3)

+dg (2,-1, 1)8
=(a +a +2a3, aj +2a) -ag,aj +3ay t a3)desot
a +ag +2ag = 1
b-(1)
boign ohier
a +2a2
a +3ay taz =5 .(3)
Method 1. The augmented matrix of the system of equations (1),(2)and (3)is
2:
[A:B] =|12 -1 : Sa(0,
13
We shall reduce the matrix [A:B]to Echelon form by applying the row
transformations only.
1 1 2:
We have [A:B]o 1 3-3
0 2-1: 4

1 1 2:
0 1-3:-3
00 5: 10
Thus we have rank A= rank [A:B] i.e., the system of equations (1), (2). (3) is

consistent and so has a solution ie.,acan be expressed as a linear combination of

, a) ,g.The equations (1), (2) and (3) are equivalent to equations

aj t ay t+ 2a3 = 1, ay -3a3 =-3, 5a, =10.

Solving for unknowns, we get az 2,a, =3, = -6.ofeeet
= a
h

1) + 3(1,2,3)+ 2 (2,-1,).bos
s
Hence (1, -2,5)=-6 (1, 1,

Method 2. Eliminating a, between (1) and (2), we get

34 + 5a, =-3 ..(4)

Eliminating az between (2) and (3), we get

+5a, =3 .5)

Putting

Hence
4
2a
Solving (4) and (5), we get

=-6,
(,-2,5)=-6 (4 1
, a =-6, a) =3.
=3 in (1), we get ag =2.ioiaos
) +3(1,2,3) +2(2,-1).
sotac
 00ot
(m. 3.1)is a linear combination oj the vectors
pe24: For what value of m.the vector
(3.2,1) and (2,1,0) ?

Solution: Let aj=(3,2,l) and a, =

=(2,1,0),1
(2,1,0).

,)= aa +ba, where a. heRosistdioo3E 9 O S

Then (m, 3,1) =aa (3, 2, 1) + b (2,1,0) = (3a,2a, a) +(2b,b,0)

=(3a +2b, 2a +b,a)
3a =m, 2a +b=3, a =1.
+ 2b
These give a =1,b=l,m=5.
Hence the required value of m is 5.

Example 25: In the vector space R determine whether or not the vector (3,9,-4,-2) is a
linear combination of the vectors (1,-2,0,3).(2,3,0, -1) and (2,- ,2,1).
Solution: Let a =(3,9, - 4,- 2), =(1,-2,0,3),
ay =(2,3, 0, - =(2,-1,2,1). 1), ag
Let
.=aa +ba, +ca3 where b,ce R. a,

(3,9,- -2)=a (1,- 2,0,3) +b (2,3,0,- +c (2,-

Then
or
4,
) 1) 1,2,
(3,9, -4,- 2)=(a 2b + 2c,- +3b - 2c,3a -b+c)
+ 2a c,
a+2b =3 + 2c
...(1)
- 2a + 3b-c=9
2c =-4 avo .(2)
..(3)
3a -b+c=-2
.(4)
Solving (1), (2)and (3), we get a = 1, b=3,c = -2.These values satisfy (4). Thus the
system of equations (1),(2), (3)and (4)has a solution i.e.,a can be written as a linear
combination of a ,a2 and ag.
Also we have a = +3y -2a3.
 17 Linear Dependence and Linear Independence of
Vectors

Linear dependence. Definition. Let V (F)be a vector space. A finite set

(a.ay ..., anl of vectors of V is said to be dependent

linearly if there exist scalars

41, 42 ,..., a,EFnot all of them O (some of them mavbe zero) such thatt dattar

Linear
a +ay Oy
independence. Definition. Let
t az 3 t...+ a, a., =0.
V (F) be a vector space. A finite set

(a,ay .c,)of vectórs of V is said to be linearly independentifevery relation of theform

aj +ay l) +... + 4,, a.,n =0, a; E F, lsiSn

=0foreach ls is n.
a,

Any infinite set of of V said

vectors is to be linearly independentif its every finite subset is

linearly independent, otherwise it is linearly dependent.

Note: The empty set Ø is defined as a linearly independe

 Evample 35: Show thatthe system of threevectors (1,3,2),(1,-7,–8),(2,1,- )ofVs (B)is
linearly dependent.

Solution: Let a,b,c be scalars i.e., real numbers such that 9vsd oW ()
ooa (1,3,2) + b (1,-7,- 8) + c (2,1,– 1) =(0,0,0)
ic.,ato +b+2c, 3a
(a 7b +c,2a 8c= 0.0.0) utcl irs r sa

Then a+b+2c =0,

3a-7b +c=0, .(2)
and 2a- 8b -c=0. ...(3)

Ifthe equations (1),(2), (3) possess a non-zero solution, then the given vectors are

linearly dependent.

Adding (2)and (3), we get

5a-15b =0 or a-3b = 0. ...(4

Multiplying (3)by2 and adding to (1), we get

5a-15b =0 or a-3b=0.
The equations (4) and (5)arethe same and give a =3b.
Putting a = 3bin (),we get 2c +4b =0 or c=-2b.
If = get
we take b 1, we a =3,c=-2.
Thus a =3,
b=l,c= -2 is a non-zero solution ofthe equations (1),(2)and (3). He

the given set of vectors is linearly dependent.

 Erample 36: In V (R),where R is thefieldofreal numbers, examineeach ofthe followingsets

of vectors forlinear dependence:

() {(2, 1, 2), (8,4, 8)}

(ü){(1,2, 0), (0, 3, 1),(-1,0, 1)}Nsbsodastson ti sitoai
(ii) {(-1,2, I). (3,0, - I),(-5,4, 3))

(iv) (2,3, 5),(4, 9,25)})hasltot4 sh snd3.l

() 15),(3, 4,7)).
{1,2,1). (3, -
Solution: (i) We have 4 (2,1,2)+(-1)(8,4,8)=(8,4, 8)+(-8,-4,- 8)
=(0,0,0) i.e.,the zero vector.
Since in this relation the scalar coefficients4, -lare not both zero,thereforethe given
set is linearly dependent.
(ii) Let a, b,c be scalars i.e., real numbers such that
+b (0,3, + c(-1,0,1) =(0,0,0)
i.e.,

i.e.,

These equations
a (14,2,0)
(a-c,2a + 3b, b +c) =(0,0, 0)
I)

a+0b-c=0,2a+3b + 0c = 0,0a +b+c=0.

will have a non-zero solution i.e., a solution in which a,b,c are not all
p sl

zero if the rank of thecoefficient matrix is less than three i.e., the number of unknowns
a,b,e. If the rank is 3, then the zero solution a =0, b=0, c=0 will be the only
solution.

I0
10 -1
Coefficient matrix A=2 3

We have |A|=1(3-0)-2 (0+I) =l+0.

: Rank
ystem
A=3. Hence a=0, b =0,c =0 is
is linearly independent.
the only solution. Therefore the given

iii),Let 4,b, c be scalars such that

 (0,0,0)
oa(-12,)++3,0,- )+c(-5,4,3)
(-4+3h-5e, 2a +0b + 4c, a -b +3¢) (0,0,0)
-a+3b-Sc =0,2a +0b + 4e=0,a - b+ 3e=0.
The coefficient matrix of these equations is

[-1 3 -5
A= 2 0 4
1 -1 3

.
We have| A|=-1(0
Rank
equations
A is<3
will
i.e.,
+4) – 2 (9-5)+
the number of unknowns
poSsess a non-zero
1 (12 -0)=0.
a,b, c.

solution. For example

Therefore the givern system of
a =-2, b =1,c= 1 is a

non-zero solution. Hence the given system of vectors is linearly dependent.

(iv) Let a,b be scalars i.e., real numbers such that

a (2,3, 5)+b (4,9,25) =(0,0,0)
i.e., (2a + 4b, 3a +9b,5a+25b) =(0,0,0)
i.e., 2a + 4b=0,3a +9b =0,5a + 25b=0.
[2 47
The coefficient matrix of these equations is A =3. 9
|5 25|
Obviously rank A =2 i.e., equal to the number of unknowns a and b.
Therefore these equations have the only solution a =0,b =0.
Hence the given set of vectors is linearly independent.
(v) Let a, b, c be scalarsi.e, real numbers such that

a (1,2, l) + b (3,1,5)+c (3,-4,7) =(0,0,0)

i.e, (a+3b +3c, 2a + b-4c, a +5b +7c)=(0,0,0)
ie, a+ 3b +3c =0, ...(1)

2a +b - 4c=0, ..(2)

a+5b +7c =0. ...(3)

Multiplying (1) by 2, we get

2a + 6b +6c =0. ..(4)

Subtracting (4)from (2), we get

-5b l0c =0
--

Or b+2c=0. a.(5)
Again subtracting (3) from (1), we get
2b -4c - or b+2c =0.=0 ..(6)

The equations (5) and (6) are the same and give b =-2c.
Putting b=-2cin (1), we get a =3c. If we take c =l,we get b= -2 and a = 3. Thus

a=3, b=-2,c =lis a non-zero solution of theequations (1),(2) and (3). Hence the
given set of vectors is linearly dependent.
 Evample 42: Showthat the three vectors (1,1,-),(2,-3,5) and(-2.1.4) ofR° are linearly

independent.
Solution: Let a,b,c be scalars i. e.,real numbers such that
a (1, 1,-1) +b (2,-3,5) + c (-2,1,4) =(0,0,0)
ie., (a+2b-2 c,a-3b + c,-a+5b +4c) =(0,0,0)
ie., a+ 2b - 2c =0 ...(1)
a-3b +c =0 ...(2)
-a+5b + 4c=0 .3)
Now we shall solve the simultaneous equations (1),(2) and (3).
Multiplying (2) by 2 and adding to (1l), we get

Again multiplying
3a –4b =0.

Saa+ 95=0.
(1)by 2 and adding to (3). we get cdsotodaal.(4)

oodrntn t(5)
gvol aW

Multiplying (5) by 3 and subtracting from (4),we get

-31b =0 or b=0.

Putting b=0in (5), we get a =0.

Now putting a =0,b =0 in (1), we get c =0.
Thus a = 0, b =0,c= 0is the onlysolution of the equations (1),(2)and (3).

a (1, 1,-1) +b (2,-3, 5)+c (-2,1,4) =(0,0,0)

a=0,b=0,c =0.
Hence the vectors 1-),(2,-3,5), (-2,1,4)of
(1, R are linearly independent.
 ooViono
19 Basis of A Vector Space
Definition: A subset'S of a vector space V (F ) said to be a basis
is of V (F), if
() Sconsists of linearly independent vectors,
(ii) S generates V (F )
i.e., L =
(S) Vi.e, each vector inVis alinear combination of afinite

number of elements of S.
 The number of
Dimension of a finitely generated vector space. Definition.
elements in any basis of a finite dimensional vector space V (F)is called the dimnension of the

vector space V (F) and will be denoted by dim V.

The vector space V, (F) is of dimension n.The vector space Va (F)is of dimension 3.

If a field F is regarded as a vector space over F,then F will be of dimension I and the

set

element of F will form a basis of F.tosut st

S =(1} consisting of unityelement of F alone isa basis ofF. In fact every non-zero
 Example 49: Show that the vectors (1,2, ),(2,1,0), (1,- 1,2) forma basis of R°.

(Meerut 2002)

Solution: We know that the set {(l,0,0), (0,1,0).(0,0, forms a basis forR'.
1)}

Therefore dim R=3. If show that the set S =((1,2,1), (2,1,0),(1,-1,2)) is linearly
we

independent, then this set willalso form a basis for R.

(See theorem 4 of 1.11]|

We have 4 +a2 (2,

(1,2, 1) 1, 0) +az -l2) =(0,0,0)0
(1,

(a +2a) +a3,2a +t ay -ag ,a + 2a3)=(0,0,0).

 a +2a9 +dg =0
2a, +ay -a =0 .2)
...3)
+ 2a3 =0.
Now we shallsolve these equations to get the values of ay, ay a,. Multiplying the

equation (2) by 2, we get

4a +2a, -2a3 = 0.
...(4)

Subtracting (4) from (1), we get

-3a +3ag = 0
-a + az =0.
Adding (3) and (5),we get 3az =0 or az =0. Putting az=0in (3),we get aj =0.Now
putting ag =0 and a =0in (1),we get a =0.
Thus solving the equations (1),(2) and (3),we get a =0,a, =0,az = 0.Therefore the
set S is lineatly independent. Hence it forms a basis for R.

Example 50: Determine whether or not the following vectors form a basis of R:
(1,1,2).(1,2,5),(5,3,4).

Solution: We know that dimnR =3.If thegiven set of vectors is linearly independent,

it willform a basis of R otherwise not. We have

4 (1,1,2)+ a (1,2, 5) +az (5,3,4) =(0,0,0)
(a +ay 5ag ,a +2a, +3a3 ,24 +5ay +4a3) =(0,0,0).
t+

a +ay +5az = 01 .)
a + 2a) +3a3 =0 .2)
2a +5a +4az =0 ...3)

Now we shall solve these equations to get the values of aq, ay,az. Subtracting (2)from
(1), we get

-4y +2az =0. ...(4)

Multiplying (1) by 2, we get

2aj + 2ay +10a3 =0. ...(5)

Subtracting (5)from (3), we get

3ay -6d =0
Or ay- 2a =0. ..(6)

We see that the equations(4) and (6) are the same and give ay =2ag Puting
ay =Zaz in (1), we get aj =-7a3. If we put ag = 1, we get ay =2 and
a =-7. Thus
44 =-7,ay = 2, a3 =lis a non-zero solution of theequations (1),(2) ánd (3). Hence

the given set is linearly dependent so it does not form a basis of R'
 4.2 Inner Product Spaces

(F)bea vector space where F is eitherthe feld real numbersorthe

Definition: Let V of

field of
complex numbers. An inner product on V is afunction from V x V into F which assigns ta

each ordered pair of vectors a., B in Vascalar (a, B)in such away that
=
(1) (a,B) (B.«) [Her(B,«) denotes the conjugate complex of the number (B. «)].
(2) (ac +bB, y) =a (a,Y) +b (B, Y)

(3) (a, a) >0and (a, a)=0 a=0ISNP

for any a,B, ye V and a, b e F.
Also the vector space Vis then said to be an inner product space wvith respect to thespecified
innerproduct defincd on it.
 43 Norm orLength of a Vectorin an Inner Product Space
Consider the vector space V3 (R) with standard inner product defined
a=(a,ay ,az) E V (R), we have
2
(a,a) =a+a, + ag2.)2
Now we know that in the three dimensional Euclidean space (4 + ay +ag)is the
length of the vector a =(a,a ,aq ). Taking motivation from this fact, we
make the

following definition.

Definition: Let Vbean inner product space.Ifa e V, thenthe norm or the lemgth of the vector

a,written as || all,is defined as the positive square root

||a|| =(a,a).
of (a,a) i.e.,

Unit vector. Definition: Let Vbe an inner product space. Ifae V is such that a||=1,
thenaiscalled a unit vector or is said to be normalized. Thus in an innerproduct space a vector
called aunit vector if itslength is 1.
| is
 Example 13: Apply the Gram-Schmidt process to the vectors B =(.0,),By =(40,-h
B =(0,3,4), toobtain anorthonormal Vg (R)with the standard inner
basis for produg.

Solution: We have||B|| =(B, .B)=1.1+0.0+1.l= (1) + (0) + -2. (1)

Let
V2
Now let
Y2 -B -(B2 ,) a

We have (B2 ,a4) =l.

V2 +0.0+(-1).N2

9 -(4.0,-I)-o0=(.0,-
l2
).S
Now ll Yel =(.Y2) =(0 +0f +(-I =2.ooo to
 L273
Y2
Let -0,-b-(0.
N2

Now let Y3 =B3 , )

-(B3 ,a)a B3 az)

We
we have@sm)=(0.3.)( 0-0+30+42i2.
N2 V2

N2 +3-0-4=22.

Y3 -(0,3,4) -2v2.02120-)
=(0,3,4)-(2,0,2)+(2,0,-2) (0,3,0).=
Nowot |Y3 |= =
(Y3,Y3) (o +(3) +(0=9.
Put
Y313(0,3,0) = (0,1,0).

Now

is the required orthonormal basis for V (R).