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MATHEMATICS IA

VECTOR SPACES
FChapter at a Glance

pefinition: (V,O) be a commutative group and (F, t, ) be a ficld. An external

Let
jon iknown as "scalar multiplication" is defined as a mapping (FXV) > V denoted by
composition

o Then V is
defined to be a vector space over the field F if the scalar multiplication
conditions:
satisfying the following
1)xOa E V Vx E F; Va E V
x O r 0 a ) = (x.y)Oa Vx, y E F; Va E V

x0(a®P) (X©a)e(EOB)
=
Vx E F; Va,ß EV
9(+y)Oa= (zO«)®VOa) Vx,y EF; Va EV
= a Va EV
) 1a
[l is the identity element in F.]
The elements of V and
Note: A vector spaceV over the field F is written as V(F). are vectors

scalars.
those of Fare

A vector space the field of Real Numbers is defined as a Real

Real Vector Space: over

Vector Space.
An Important Vector Space:
ThesetR= { , x 2 . . ,x,)/x, E R} is a real vector space with respect to operations
(vector addition) and O (scalar multiplication) in R" defined as follows:

CX,); Vc E R
cO2 X) (cX1,CX2
nd(1,2 ) E R".

Some Elementary Properties:

IfV(F) is a vector space, then
()0.a 0, Va EV
i) c.0 = 6, Vc EF

am)C.a 6 eitherc 0 ,or a= 6 =

i-(c.a) c.(-a) (-). a; Vc E F, Va EV

= =

Sut
pace: Let V (F) be a vector space and W be a non-empty subset of V. If W forms a vector
ae over the same field F and with respect to the same compositions (Addition and

multip
pcation) as in V, then W is said to be the subspace of the vector space V(F).

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Theorem: A non-empty subset W of a vector space v(F) is a subspace of V if and only if
i ) a E W,BE W *a+ß ew
i) cE F, a E W = ca E W
Equivalently, the two conditions can also be written as ca + dß E W, Va,ß ¬

Vc, dEF W,
Theorem: The intersection of two subspaces of a vector space V (F) is a subspace of .
In other words, if S and T two
are
subspaces of a vector space V (F), then S N T is a
subsna
of V. ospace
Note: The union of two subspaces of a vector
space V (F) is not, in general, a subspace of V,
Linear Independence and
Dependence
of Vectors
In a
V(F), a finite subset of n vectors {1,V2
vector space
Vn is said to be
Linea y
Independent(L. ) if c^v tc2v2 t* + U , =6 (c ¬ F)
G =0, (i =1,2,.., n)
In a vector space V(F), a finite subset of
{1,V2 ,
n vectors
V,ns is said to be Lineaty
Dependent (L. D) if there exists scalars C (i=1,2,.,n) not all 0 (zero) such th
C1 tC22 t**+ c,v, = 6 ( ; ¬ F)
An infinite subset S of a
V(F) is said to be Linearly Independent (L. I) if even
vector space
finite subset of S is linearly independent and S is said to be
Linearly Dependent (L. D) if some
finite subset of S is linearly dependent.
A few important results:
If a finite set S contains null vector 6, then S is linearly dependent. (In other words, an LI
set cannot contain the null vector.)
IfS =
{v}, v #0;i.e., S is a singleton set of non-null vector, then S is L.I
Subset of an L.I set is L.I and Superset of an L.D set is L.D.
Theorem In the vector space R" (R)the VEcto

(11,a21 n1); 2 (12a22**,n2) =

(1n a2n *,an )are linearly independent iff|a,,| *#0.

Linear Combination of Vector:
A vector v is said to be linear combination of S =
{vi, V2, ....., v,}, if there exists st
alar

EF (i= 1, 2,.. n) such that CqV, tc,v2 t * t c,v,

The set of vectors{(C1tC2V2 t****+ Gqv,)/c; ¬ F} is known as linear
spau
o f a

et S and denoted by L(5).

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D i m e n s i o n
of a Vector Space:
vector space (F) is said to be of dimension n if there exists finite subset
he
The
Sspans V; and
having n linearly independent and also the set
S {2 Va vectors
=

written as dinm(V) n.

Vector Space:
Basis of a
Tet V(F) be a vector space. If S be a subset of V such that

)V=2(S))
and (i) S is linearly
independent set
defined as a Basis of .
Then Sis
Note:
basis containing n elements, then dim (V) =
n.
1.fVbe a vector space having a
fthe vector space{6}, (where 6is the null vector) then dim(V)
=
0.

Linear Transformation

Definition:
Let Vand W be vector space over a field F. A mapping T:V > W is said to be linear mapping

or linear transformation if it satisfies the following conditions:

0T(a+B) =T(«)+ T(B), Va,f EV
Va ¬EV
ciyT(ca) cT (a), VcEF;
=

In particular, if V
=
W, then 7 is said to be a linear operator on V.

Equivalent condition:T(ca + dß) = dT(a) + dT(B), V ,d E F; Va,ß EV

Kernel of a linear mapping and Nullity:

Let V and W be vector space over a fieldF. Let T: V> W be a linear mapping. The set of all
vector a E V such that T(a) = p, p being the null vector in W, is said to be the Kernel of T
and is denoted by Ker (T) = {a ¬ V\T(a) = p}.

Ker(Tis also called Null Space ofTand dimension of Ker (T) is known as Nullity(T).

Image of a linear mapping and Rank:

Let V and W be vector space over a fieldF. Let T: V > W be a linear mapping. The images of
the elements of V under the mapping T forms a subset of W. This subset is said to be the image

of7 and denoted by Im(T) = {T(«)\a ¬ V}

m(T) is also called Range ofTand dimension of Im(T) is known as Rank(T).

neorem (Sylvester's Law of Nullity): Let T: V> W be a linear mapping, where V is finite
imensional. Then Rank (T) + Nullity(T) = dim V.

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Theorem: Let V and W be fieldF and T': V > W be
vecto space over a a linear
T is injective if and only if Ker (T) =
{p}.
pping T
Theorem: Let V andW be vector space over fieldF and T: V W be
a a linear
that (T) = {4}. Then the of images linearly independent set of vectors in
mapping s
mann
V are 1.
independent in W. linea
Theorem: Let T: V W be a linear mapping and a1,2.*, nbe a basis of V
{T(«4),T(a,),.T(a,) Jgenerates Im (T)
Theorem: If V and W be both finite dimensional vector spaces of same
to one
dimension, then Tic.
implies that 7 is onto and vice-versa.
Inverse of Linear
Transformation:
Let Vand W be vector space over a fieldF and a linear
T(x) y) is invertible if and only if T is one to one
= mapping T: V W (defined
and onto mapping, and
denoted as T:W> V and inverse of 7:
defined as =
T(y) x.

Multiple Choice Type Questions

1.1. The set of
a) linearly dependent
vectors{(2, 1,1). (1, 2,2). (1, 1,1)} in Ris WBUT 200
c) basis of R b) linearly independent
Answer: (a) d) none of these

1.2. The value of

kfor which the
dependent is vectors(1,2,1), (k,1,1) and (1,1,2) are linear
a) 2 WBUT 2006
b) c)-1
Answer: (6)

1.3. The value of K

for which the
vectors
linearly dependent (1,2,1). (K, 1,1) and (0,1,1) a=
a) 1
b) 2 WBUT 2007
Answer: (c) c) 0
d) 3
1.4. The four vectors
(1, 1, 0, 0 ), ( 1, 0, 0, 1 ). (1, 0, a, 0
independent if ), (0, 1,
a) a, b) are linea
a0,b+2 b) a+2,b+ 0 c) WBUT 2008]
Answer: (c) a0,b*-2 d} a-2,b#
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10 0
lo o 0
1 0 a O =0 a 0-a
a b 0 a b
0 1a
=-a-(ab +a) = -2a-ab = -a(2+b)
The vectors are L.I if a(2+b)0 i.e. a #0,b+-2

following
i K. IfS and T are two subspaces of
1.5.
a vector space V, then which of the
subspace of V also?
WBUT 2009, 20181
isa
a) SUT b) SnT c)S-T d) T-S

Answer: (b)
the following sets is linearly independent? WBUT 2010]
1.6. Which of
a){(1,2),(2, 4)] b) {(1,2,3), (2,4,6), (1,1,1)}
c){(2,0,0), (0,3,0),(0,0,4)} d) none of these

Answer: ()

17.7:RRis defined by T(%,a)= (2x,-A, t s ) . Thenkernelof Tis

WBUT 2008]
a{1.2)) b) {(1,-1)} o{(0,0)} d)(1.2)).(1,-1).
Answer: (C)
T)=(2,-g, +2).7,)eR?
Le, (,x,)e Ker(T)
Then, T,)=(0,0)
2- =0 x + =0

=0,x =0
1.8. If a linear transformation T: R R be defined by T(x1,x2) = (x1,x2), then

Ker(T) is WBUT 2006]

((-1,-1).(1,1)} b2)(1)}
(1,0).(0,1)} d) f(o,0))
Answer: (d))

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1.9. If a

then ker(T) is
linear transformation )=%+3
T: R2 - R2 be defined by 7T(7, *2)=
WBUT 2007
{(1,-1)} b)(1,0)o{0,0)} d 10). (0,1
Answer: (a)

LShort Answer Type Questions

EXamine if the set of vectors ((1,2,2), (2,1,2), (2,2,1)) is linearly independent in

R3 WBUT 2004
Answer:
I 2 2
212 2-1-3)-2(-2)+2(2)=5*0
2
Hence, the given vectors are linearly independent.

2.2. Does S = {(x,y,z) E R:2x -y+ 3z = 0) form a subspace of H*? Justify

[WBUT 2006
Answer:
Let, a, peS. Then we may write

If a,be R°,then we have

aa+bß=al-y +z. , Z)+b-y +Z2, 2, 2)
(a- +z)+b-y, +Zz) ay +by2, az, +bz,)
=--by,)+(az, +bz,)) ay +by2, azj +bz,) =(-1+m,1, m)e S
Since =ay, +by, e R and m =az, +bz, e R°
Therefore, a,beR' and a, ße S a a +bßeS
Therefore, Sis a vector subspace of R°

2.3. Showthat (3, 1,-2).(2,1,4).(1,-1,2) form a basis of R

WBUT 2007, 2008, 2010
Answer:
1
Since, |2 1 4=3(2+
2
4)-1(4-4)+ (-2)(-2-1) =24 (#0)
1 -1
So, the given set of vectors S is L.I set and number of vectors in S = 3 dim R*.
Therefore, S is a basis of R°(R).

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24. Show that the vectors (1, -2, 3). (2, 3, 1), (-1, 3, 2) form a basis of R.

[WBUT 2018]
Answer:

Let S={1, -2, 3). (2, 3, 1). (-1, 3, 2)}

|I 2-
Since,-23 3=1(6-3)-2.(-4-9)+(-1)(-2-9)-3+ 26+11= 40 (#0)
Since, -2 3
3 1
So, the given vectors S is L.I set
and number ofvectors in S=3=dim(R').
Therefore, S is a basis of R'.

2.5. In the vector space R°(R), show that the vectors E, = (1,0), E, = (0,1) are
linearly independent. [MODEL QUESTION]
Answer:
LetcE+c,E2 =6, c C2 E R
G1,0)+c2(0,1) = (0,0)

( ) =(0,0)
0,c=0
Thus, both c C2 0 and hence, the given set of vectors {E, E,} is L.I set.

2.6. In the
space R vector (R), showthat the vectors
E-(1,0,,0),E = (0,1,,0), E(0,0
(0, ,0,,1) are

linearly independent. [MODEL QUESTION]

Answer:
LetcEt CEz + + E, =
0, cyz ., E RR
q ( 1 , 0 , , 0 ) +c2(0,1,,0) + + ,(0,0,,1) = (0,0,,0)

(c, c ,n)= (0,0,,0)

q =0,c2 0,,=
=0
Hence, the set ofvectors {E, E E,} is L.I set.

2.7. In the vector space R(R), verify whether the vectors v = (1,1),vz = (2,2)

are linearly independent/ dependent. [MODEL QUESTION]

Answer:
Let cv + c2v, = 0 , c,c2 E R
G(1,1)+ c(2,2) = (0,0)

G+22 C + 2c,) =
(0,0)
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C+2c2 = 0, C+2c, = 0
C t2c2 =0 c + 2c, = 0 a = = k ¬ R (Say)

=-2k, c2 =k
Hence, the given set of vectors {v1, v2} is L.D set.

the set of
2.8. In the vector space R (R), verify whether
vectors
S{(1,1,0), (1,0,1), (0,1, 1)} is linearly independent/ dependent.
[MODEL QUESTIONI
Answer:
1
Since,1
Since, o 11(0-1) - (1-0) +0 = -2 (+0)
lo 1 1
So, S is an L.I set.

2.9. In the vector space R°(R), verify whether the set of vectors
S= {(1,2,-3), (2,-3,1),(-3, 1,2)} is linearly independentI dependent.
[MODEL QUESTION
Answer:
Since,
1 2
-3 = 1(-6-1)-2(4+3) 3(29)= -7-14 +21 =0
-3
So, S is an L.D set.

2.10. Examine whether the set of vectors S= {(2,-3,1),.(3,-1,5).(1,-4,3)} is

linearly independent / dependent. [MODEL QUESTION
Answer:
3
Since, -4= 2(-3+20)-3(-9+4) +1(-15+1) = 35(#0)
5
So, S is an L.I set.

2.11. Find k so that the vectors (1,-1,2). (0,k, 3) and (-1,2,3) are lineary
dependent. [MODEL QUESTION]
Answer:
10
For linearly dependence ofthe given vectors, |-1 k = 0
2 3
or, 1(3k - 6) - 0 +(-1)(-3-2k) =0
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sk-3 0 k=
or,

212. Ifa , , y .are

42.f a,,Y linearly independent vectors, show that a +B,8 + YYta are
linearly independent. [MODEL QUESTION
Answer:

A(a+) + %(B +v) + oa(r ta) =

t + (c tc2)ß+ (c2 + cg)y =0

Since, a,B, are linearly independent vectors, we get
t 0 , t a 0,C2tG=0
Solving three equations, cg 0 .
we get
the c C2
Lence, the vectors a + B,B +7y t a are linearly independent.

213.Show that S={(x,y,z) ¬ R°:2x 4+y-z = 0}is a subspace of R.

[MODEL QUESTION
Answer:
tis obvious that, (0,0, 0) E S, so, S is non-empty.
Leta = , Z , ) E S; B=22z2) E S, where xY,z, ER

2x1t 1-Z10)
0}(1)
2x t y-2 =
Therefore,
tz2 ¬ K
Now, a +B (x1tz» Y1 ty2»z1 t+z2)and x t+ Y t
=
ya71
2(4+ 2)+ ty)-(4+z,)
(2x +y-z)+ (2x2 +y2-z) =0+0
= = 0 using (1
Thus,a +B ES,V a,ß ES
) Again, let CE H Now, ca = (Cx1,y, Cz1)

Sine, cx,, cz, E R and 2(x) + ey-(cz,)= c(2x,+ - z ) = c .0 = 0

Tby(1)
Thus, ca ES, Vc E R; a ES
Hence, S is a subspace of R".

214. In the vector space R"(R), show that the

the vectors
( 1 , 0 , . . , 0 ) , E , = ( 0 , 1 , , 0 ) , , E , = (0,0, * * , 1) forms a basis

K and hence find the dimension of R". [MODEL QUESTION]

Answer:
) Let
S={E, E ,En
ER", then v =
(xX2,"" ,x,): where x, ER.
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***z,,
E L($)
+x,Ez +
Therefore,
OTe, v =(x,xz,x,)
(X1 = x,E,
=

Thus, R" EL(S)

S R"
Also, as S C R " > L(s)
R* E R
Therefore, L(S)
=

=
0, c1,C2C
Let cE + c2E,
+ +cE, 1) (0,0,..,0) =

(i) ++C, (0,0,",

c(1,0,,0) + c(0,1,,0)
(C2,ca)
=
(0,0,, 0)
O

C2 0,**,n is L.I set.

=

0,
E , , En
So, the set of vectors {E,
,E}is a basis of R".
Hence, S {E, E, vectors in basis n.
= =

Number of =

contains n vectors; so, dim(R")

Since, S
,E,J is known
as standard
basis
of R.
Note:, E2
is a basis of R(R), and :find
S {(0,1,-1),(1,1,0), (1,0,2)}
2.15. Show that [MODEL QUESTION
relative to this
basis.
the co-ordinates of (1,0,-1)
Answer:
0 1 1
0=0(2-0)-1(2-0)+ 1(0 =--1(#0)
+1)=
It
is observed that 1 1
-1 0 2 3 dim R3
in S = =

is L.I set and number of

vectors
So, the given set of vectors S

Therefore, S is a basis of R$ (R).

Now, let (1,0,-1) =

(0,1,-1) + cz(1,1, 0) +3(1,0,2)

Therefore, c2 + 1, CtC2 0, -C1 t 2c3

Solving, we get c=-3, c2= 3,cg-2
(-3,3,-2)are the coordinates.

and find te
2.16. Show that S {(1,0,0), (1,1,0),(1,1,1)} is a basis of C(C),
basis. [MODEL QUESTION
co-ordinates of (3 + 4i,6i,3 +7i) with respect to this

Answer:
11 1 1
It is observed that 0 1 1=1 (#0)
lo 1
dim
So, the given set ofvectors S is L.I set and number of vectors in S= 3 =

Therefore, S is a basis of c(C).

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Now, let (3 + 41,61,3 + 71)= c,(1,0,0)+ ez(1,1,0) + ca(1,1,1)

No
(tCa t C C2 t g, c)
Therefore, C t C2 tCa3+4i, c2 +c 6i, c = 3+ 71
Solving, we get C 3 2i, c2 =-3- i,c=3+7t
(3-2i,-3 arethe coordinates.

217. Show that the transformation T:R3» R2 defined by 7(x,y,z) = (z.y),

where (z,y, 7) ER° is a linear mapping from R to R? [MODEL QUESTION]
Answer:

Let a (Y1,Z1) E R° and ß (*2,2, 22) E R3

= =

.Tla)= T(zY1,z2)= (z1Y1) and T(B) = T(z2y2»z2) = (7¥2)

Again, a += (7itz2% th»z1tz2) ER'

andT(a+ B) = T(x, t x 2 1 ty2,Z1tz2)

= *1)+(TV2)= T(a)+T(B) condition ()]

vcEF,T(ca)= T(cx1,¬Y,cz) = ( c x 1 ) =c ( * , ) = cT(«)

[condition (i)]
Thus, both the conditions are satisfied and hence, Tis a linear mapping.

Long Answer Type Questions

3.1. Prove that the set of all second order real square matrices is a vector space
with respect to matrix addition and multiplication by a real number. WBUT 2006]
Answer:
a b
Let be the set of all 2x2 real matrices. Then A |
=

e d
and
B-
where a, b, c, d, x, y, z, w are real numbers

Then 4+B= atx b+y].

Dcdz wc+z d+w]
t can be easily shown A+B = B+A for all 4, BeV
For every n V there exist 0 o0 00 in V such that

A+0=
leallo oe
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--a
a -b
For every
every A=| in V, there exist -A-c in Vsuch that A+(-A)=0
A= d
Let e be any real number then
a b ae be
e A=e
ce de
1A=1 " where 1is the real number
1ACd
Now for any real number k and k2
e

a . k,a k=k(k)
(,) 4=kl dkc kd
For any real number k and 4, B in V

a+x b+y ka +kx kb+ ky ka

k(A+ B)=c+z d+
(d+ kc +kz kd +kwkc kd|kz kw
x Y=kA +kB

Again for any real numbers x, x and A in V

+6Ja (4 +4,)]
(6+h)4-(+k +k)e (6+k,)d|
4+)4=(6+k
ka+ ka kb+k^bka kbk kb
ke+ke kd+k^d] ke kd] ke kd

Hence all the conditions are satisfied.

i s a vector space.

3.2. Examine whetherthe set {(1, 2, 1), (2, 1, 0), (1, -1, 2)) is a basis of R.
WBUT 2005]
Answer:
Let a (1,2,1), a, = (2,1,0),a, = (1,-1,2).
Let us consider the relation
Ca +Ca, +Ca, =6
where Ci, C2, C3 are real.
Then C(1,2,1) + C(2,1,0) + C(1,-1,2) (0,0,0)
=

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This gives
C+2C% +C =0
2C +C2-C =0
and C +2C =0
This is a homogeneous system of equations in Ci, C2, C3.
2
Here the
coefticient determinant2 1 -1=-9 #0
O 2
Cramer's rule, there exists a unique solution and the solution is
By
C0, C2 =0, C3 = 0.
This proves the set is linearly independent.
Also, number vectors in S= 3 dim R3. =

of
Therefore, S is a basis of R3

a set. of n-vectors? Show

that the
3.3. What is meant by linear independence of
2008, 2009]
vectors (1, 2, 1), (-1, 1, 0) and (5, -1, 2) are linearly independent.[WBUT
Answer:

Let be a vector space over a field F. A finite subset {a, a2,...a, of vectors of V is
said to be linearly independent, if every relation of the form
iaiSn implies
a,a +a,, t...+a,a, =0, a, ¬F,
4 =0 for each 1sisn

Let, a, b. c be scalars such that

af,2,1)+b(-1, 1, 0)+c(5, -1, 2) =(0, 0, 0)
ie.(a-b+Sc,2a +b-c, a+2c) =(0,0, 0)
which gives
a-b+5c=0
2a+b-c=0
a+2c= 0
(1 -1 5)
Theco-efficient matrix is A =|2 I - |
io 2
1-1 5
Now,det A=2 1-1=2#0
0 2
Therefore rank of A =3 =number of variables and hence the system of equations have

uniquesolution. Hence a=b=c= 0 is the unique solution,

Thus the given set of vectors is linearly independent.
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3.4. Show subspace of H°.

that. =
{(x,y, z) E R:2x +y -

z =
0} is a
its
Find its
dimension. WBUT 2005
Answer:
S={(x,y,z) ER' |2x+y-z =0}
Let a=
(x},Z) and f=(7%,2,z2)¬S
2 + - z =0and 2x +2 -zz =0
These
imply that 2(% +*a)+( +%)-(+2)=0
Thus a +B= (x +*2»y+y2.z +Z) ES.
Let C be a real number and aeS.
Then Ca=C(hMa)= (Cxj, C,Cz)
Now 2Cx +Cy-Cz =C(2x +-z)=C.0=00
Thus CaeS.
Therefore S is a subspace of R'.
Let = (a,b,c) ES.
Then a,b,c ER and 2a+b-c=0
5 = (a,b,2a + b) =a(1,0,2) + b(0,1,1)
Let y = (,0,2) and S = (0,1,1).

Then=ay +bÑ EL{y,}

Therefore ScL{y,8}
Again yeS and SeS. This implies that L{y,6}ES.
Consequently S= L{y.ö}.
To examine linear independence of the set {y.8}, let us consider the relation
Ca7+C8=0 where Cj,C ER
C1,0,2)+C,(0,1,1) = (0,0,0).
We have C = 0, C =0 and this proves thatthe set {y.8} is linearly independent. Hence
{7.8} is a basis of S and since the basis contains two elements, therefore the dimension
of S is 2.

3.5. Showthat W= {(x,y,z) ¬ R°:2x - y + 3z = 0} is a subspace of R°. Find a

basis of w. What its dimension? WBUT 2007, 2008]

Answer:
1s Part: Refer to Question No. 2.2 of ShortAnswer Type Questions.

2nd Part:
Let a = (a, b, c) be an arbitrary element of W.

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

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=a(1,2,0) + c(0,3, 1) = av t+ cvz (say)

vector a E W can be expressed linear combination of set of vectors
Thus, any as

Now, C tG2=
c(1,2,0) +ca(0,3,1) = (0,0,0)
(C 2e, +3ca, c2) = (0,0,0)
CC2=0
Therefore, {vi,"2} is linearly independent.
Hence, (,,a}= {(1,2,0), (0,3, 1)]is a basis of W.
Therefore, dim W = No. of vectors in basis =

3.6. If W {(zy.z) ¬ R:x +y +z = 0}, show that W is a subspace of *°,

and find a basis of WV. WBUT 2008]
Answer:
1' Part:
It is obvious that, (0,0,0) E S; so, S is non-empty.
Let a=(1z2) E S; B= (x22,z,) E 5, where z,z, E R

Therefore, 1tt20 ....(1)

() Now, +ß =(*1tX2 Y1 ty2»Z1 t22)and xj t*2 Y1+y71 tz2 E K

(z1t x2) t+ C1 ty2)+ (7tz)

( 1 t 1 +z2)+ (x2+y2 t z) =
0+0 =0
[using (1)]
Thus, a + ß ES,V a,ßES
i) Again, let cE R. Now, ca = (cx,Y , Cz)
Since, cx cy cz, E R and cx +cy t cz c(x + + z ) = c.0 = 0
by
())
Thus, ca E S, VcE R; a ES
Hence, S is a subspace of R°.

2nd Part:
Let a = (a, b, c) be an arbitrary element of W.

Therefore, atb+c = 0c= -a - b

a = (a,b,c) = (a,b, - a - b)
a(1,0,-1) + b(0,1,-1) = av, t bv, (say)

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 AIONS
trickstarvivek.com
combination or
set
of vector
Tnus, any vector linear
can be expressed
as
a E W

Now, C tc2v2 = 6
c(1,0,-1) + c,(0,1,-1) = (0,0,0)
CC2 =0
2-c-C2)= (0,0,0) independent.
is linearly
nerefore, {vi,v} {(1,0,-1),(o,1,-1))
=

Hence, (v, v2} is a basis of W. 2 basis=

Therefore, dim W =
No. of vectors in
for all (x,y) E R*; Show that T
R> Ris given by T(x,y) (x, 0),
=
3.7. If T:
dim Im(7,)=2.
dim Ker (T) +
is linear transformation and verify that
a
[WBUT 2004
Answer:
(xzYz) E R
Let a=
( )E R and =

T(a)= T(zn)= (70) and T(8) =T(z22)= (xz,0)

ER
Again, a +B (x1 +z2 y1 +y2)
=

and T(a + B) = T(x1 txz¥1 t )

= (z +x,0)
(x1,0)+ (x2,0) = T(a) +T(B) condition ()]
Let cE R,T(ca) T(cx, cy)
=
=
(cx,0) =
c(z,0) =
cf(a) condition (i)]
Tis linear mapping.
Thus, both the conditions are satisfied and hence,
a

Tofind Ker(T):
Ker(T)-.v)e R3 :T(x,y)= (0,0)}
Let us consider T(x,y) =(0,0)
(x,0)= (0,0)
=0
Therefore, (0,y) = y(0, 1) E Ker(T) for any y ER

Thus, Ker(T) = {(0,1)}

Therefore, dimKer(T) =
1.

Tofind Im(T):
Letv= (a, b) ¬ R?
T(v) = (a,0)
Im(T) = {T(v):v ¬R} = {(a,0)} = {a(1,0):a ¬ R}

Thus, Im(T) = {(1,0)}

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Hence, dimim(T) = 1,

Hence, dimKer (7) + dim/m(T) = 1+1 2

3.8. If T: R° R< be a linear mapping such that T(1,0) = (3,2),

r(1,1)= (2,3). Find T(x,y).

Answer:

Let a ( 1 , 0) E R*and ß = (1,1) ¬ R?

First ofal, we have to show that fa, }is a basis of R.

Since. 1 (#0)
a , i s linearly independent set containing two vectors and dim R' =2.
{a, Pjis a basis of R2
So, any vector of R* can be expressed as the linear combination of {a,B}.
Let (7,) E R° be an arbitrary vector and
let(x, y) = ca + dß

( z y ) = e(1,0) + d(1,1)

%y) (c + d, d)=

=c+d= x,d=y
c=X-y,d=y
Therefore, (-v)a
(,y) =

+yß
Hence, T(z,y) = (x-y) T(«) + y T(B)
T(zy)= (x-y)(3,2) +y(2,3) = (3x - ,2x +y)
This gives the transformation T.

3.9. Determine the Linear transformation T:R' R* which maps the basis
vectors (1,0,0), (0,1,0), (0,0,1) of R to the vectors (0,1,0), (0,0,1), (1,0,0)
respectively. Find Ker(T), Im(T). WBUT 2004]
Answer:
Let a = (x,y,z) E R* be an arbitrary vector.

Then a =
(x, y,z)x(1,0,0) + y(0,1,0) +z(0,0, 1)
=

Since T is a linear
mapping
T(a) = x T(1,0, 0) + y T(0,1,0) + z T(0,0,1)
=x (0,1,0)+y (0,0, 1) +z (1,0, 0) =
(zxy)
herefore, the linear transformation is T(x,y, z) = (z,x,y), for all (x, y, z) ¬ R'

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Tofind Ker(T):
Let us consider T(x,Y, z) = (0, 0, 0)

(z,x, y) = (0,0,0)
* =y =z = 0

Therefore, Ker(T) =
{(0,0,0))
Tofind Im(T):
{r(v): v ¬ R'} =
{r(1,0, 0),r(0,1,0), T(0,0,1)
Im(T) =

((0,1,0), (0,0, 1),(1,0, 0)}

is defined by
T:R3R2
A linear transformation
3.10. WBUT 2006
T(7,y,2) =(x+y,x-z).
Find its rank and nullity.
Answer:

Let (x, y, z)e KerT.

Then r+y, x-z)= {0, 0)
and x- z
= 0
X+y= 0
y=-*, z=x
E R
x, y, =)= (x, -x, x)= x{1, -1, 1)wherex
KerT=Lla|/wherea = (1, -1,1)
vector a
KerT is linear span of the
Dimension of KerT is 1. Thus,
the Nullity(T)
=
1
1,0)and T(0,0,1).
ImTis the linear span of the vectors T(1,0,0), T(0,
T(1,0,0)=(1,1),
r(o,1,0)=(1,0),
T(0,0,1)=(0,-1)
and (0,-1)
But (1, 1) is linearcombination of vectors (1,0)
are linearly independent.
Again vectors (1, 0) and (0,-1
and (0, -1), that is Im(T) ((1,0),(0,-1
=

Im T islinearspan ofthe vectors (1,0)

that is Rank (T) =2.
Thus, dimension of Im T is 2,

T: R > R defined by T(x,y)

=
(2x-Y,7)
3.11. Examine whether the mapping WBUT 2008]
is linear.
Answer:
Iet a=(x,M} &ß=(xz,V) be two vectors in R

a)=T%n)=(2x-)
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MATHEMATICS IA

T(-T()=(2 -J,i)
Let a,b, ER

then aa+bB ER?

Tlaa+b)=T{al4-M)+b{*.))=T(ay +bx2,ay +bys)

2i ay +bs--ba +bx=al2-ya)+b{2,-,a)
ar ()+bTr() = o7(a) + br()
Tis lineartransformation.
3.12. a) Define linearly dependence and independence of vectors. Prove that a set
of vectors containing null vector is linearly dependent. WBUT 2018]
Answer:
In a vector space V(F), a finite subset of n vectors {v,v2, .. . , V} is said to be Linearly

Independent (L ) ifca"i tev2 t****+ caVn=0 (cEP)

0 , =1,2,.,n)
In a vector space V{F), a finite subset of n vectors {V,V2,..,V,} is said to be Linearly
Dependent (L. D) if here exists scalars ci.( i =1,2,...,n) not all 0 (zero) such that

(cE F)
Let,V2..,V,,6} be a set of (n +1) vectors having a null vector 0 in a vector space
V(F).
Since, in vector space V(F)
) 0.a = 0, Va E V
(i) c.6 = 04, Vc EF

So, O.y +0.v, +....+0.v, + C.0 =0+0+.+6+0=6

From the above relation, it is clear that all the scalar are 0, except C(which may be
non-zero).
Hence, from the definition of L.D the set of vectors containing the null vector is
linearly dependent.
b) Find a basis of real vector space R'containing the vectors (1,1, 2), (3,5,2).
WBUT 2018]
Answer:
We know that {0,0,0),(0,1, 0), (0, 0,1)} is a basis of R.

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 SLICATIONS
Now, we have to select any one from the above basis such that trickstarvivek.com
it will forms
s a basis
alo
with the given
set of vectors (1,1,2). (3,5,2). whether the
Now, let and then verily vec.
etors
us consider the first vector (1,0,0) a
d,0,0),(1,1, 2), (3, 5, 2)} are linearly independent.

Now, |1 1 2=1.(2-10)=-8 (*0)

3 5 2
is linearly independent.
Hence, the set S ={(1,0,0), (1, 1,2), (3,5, 2)}
Again, dim(R') =3 = No.of Vectorsin S.
basis of R' and hence the required basis
containing (1,1, 2), (3, 5,2) i
So, S forms a

S={0.0,0),(1,1,2), (3,5,2)}
c) Let T:V->W be a linear transformation such
that T(%)=B, 7()=P+
and T(a,)=A+B+B where (a,,a,,a,) and (B.B,B) are basis of V and w
basis. Hence find whether 7
matrix of T relative to these
respectively. Find the
exist. Find the matrix which will represent 7 WBUT 2018
Answer:
The linear transformation can be expressed as
T(4)=1 +0.4, +0,4
T(a)=1,A +1,, +0,,
T(a)=1,6, +1, +14,
to the given basis is represented
. The matrix of linear transformation T relative
(1 1
as
[r]=0 1
1
0 0 1
1
Now, det([7T]) =|0 1 =1(#0)
0
1- 0
IT] exists, that is, [T]" :W>V existsand given by [7]' =|0 -
0 0 1

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If T be a
3.13. linear operator on R defined by
f+.Y,2)= (x,y-z,x tytz), (x, y, z) ¬R3, Show that Tis invertible and find
(»yz)={
[MODEL QUESTION
Answer:

To verify One
ro verify One-to-One mapping, let (x, y,z) ¬Ker (T)
Therefore, x,y,z) = (0,0,0)
( , y - z , x + y t z ) = (0,0,0)
X = 0,y-z = 0,x+ytz =0

x = 0,=0, z = 0

Therefore, Ker(T) = {{0,0,0)) and hence, T is One-to-One.

Therefore, dimKer(T) = 0, that is Nullity(T) = 0

Again, from Sylvester's law of Nullity, we know that

Rank(T)+ Nullity(T) = dim R3

3
Rank (T) +0=
Rank (T) =3
Therefore, the range/ image of T is Im(T) = R and hence, 7 is Onto mapping.
So,T is One-to-One and Onto, and hence invertible.
Let (a,b, c) ¬ R' (co-domain set) be an arbitrary element and (t,y,z)E R' (domain
set) be the pre-image of (a, b, c) such that
T(%y,z)= (a,b,c)
(,y-z,x +y +z) (a, b,c)
a, y -2= b,x +y+z=c
C-atb C-a-b
x a, y 2 2 2
Therefore, T":R3> R" is defined by
T(a,b,c) = (z,y,2)
- atb
Tbc)=(

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