Regulation: IFETCE R-2019
UNIT 3
Academic Year: 2022-2023
LINEAR TRANFORMATION
INTRODUCTION TO LINEAR TRANSFORMATION
Transformation between two vector spaces is a rule that assigns a vector in one space to a vector in the
other space. Linear transformations are transformations that satisfy a particular property around addition and
scalar multiplication. In this lesson, we will look at the basic notation of transformations, what is meant by
“image” and “range”, as well as what makes a linear transformation different from other transformations.
3.1 LINEAR TRANSFORMATION
Let V and W be vector spaces over F. we call a function T:
V→W a linear transformation from V to W if, for all u,v 𝜖 V and c 𝜖 𝐹, we have
(i) T(u+v) = T(u) + T(v)
(ii)T(cu) = cT(u).
(or)
It is enough to show that T(au+bv) = aT(u) + bT(v)
ROTATION
For any angle  , define Tθ: R2 → R2 by the rule: Tθ(a1, a2) is the vector obtained by rotating (a1, a2)
counterclockwise by θ if (a1, a2) ≠ (0, 0), and Tθ (0, 0) = (0,0).
Then Tθ: R2 → R2 is a linear transformation that is called the rotation by θ.
REFLECTION
If we define T: R2 → R2 by T(a1, a2) = (a1, - a2). Then T is called the reflection about the x-axis.
If we define T: R2 → R2 by T(a1, a2) = (-a1, a2). Then T is called the reflection about the y-axis.
PROJECTION
Let T: R2 → R2 be defined by T(a1 , a2 )= (a1 , 0). Then T is called the projection on theX-axis.
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Let T: R2 → R2 be defined by T(a1 , a2 )= (0, a2 ). Then T is called the projection on the Y- axis.
Worked Examples 3.1(A)
Example 1
Let S and T be two linear transformation from 𝐑𝟑 𝐭𝐨 𝐑𝟑 such that S(1,0,0) = T(0,1,0,) S(1,1,0)= T(1,1,0).
Does this imply S=T on 𝐑𝟑 ? Justify your answer.
Solution:
Given, S(1,0,0) = T(0,1,0,)
 a1 b1 c1  1   x1 y1 z1  0 
 a 2 b 2 c2  0    x 2 y 2 z 2  1 
 a 3 b3 c3  0   x 3 y3 z3  0 
 a1  y1
a 2  y 2 ..............(1)
a 3  y3
Given, S(1,1,0) = T(1,1,0,)
 a1 b1 c1  1   x1
 a 2 b 2 c2  1   x 2
 a 3 b3 c3   0  x 3
y1
y2
y3
z1  1 
z 2  1
z3   0
 a1  b1  x1  y1
a 2  b 2  x 2  y 2 ............(2)
a 3  b3  x 3  y3
sub (1) in (2) we get
b1  x1
b2  x 2
b3  x 3
 y1 x1 c1 
S   y2 x 2 c2   T


 y3 x 3 c3 
 S  T in R 3
Example 2
Write down the matrix form of the linear transformation that is described by rotation about an angle 𝛉
in the counter clockwise in xy-plane.
Solution:
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Let 𝛼 be the angle that (a1 ,a2 ) makes with the positive x axis ,
and let r  a12  a 2 2
Then a1 = r cos𝛼and a2 = r sin𝛼. also 𝑇𝜃 (a1 ,a2 ) has length r and makes an angle 𝛼 + 𝜃 with
the positive X axis. Then the matrix form of linear transformation will be,
cos   sin   a1 
 T  
 
 sin  cos   a 2 
Example 3
𝟐
Does there exist linear transformation T: 𝐑 → 𝐑𝟐 for which T(𝟏, 𝟑)= (𝟕, 𝟗).
Solution:
Given, T(1,3)= (7,9).
 a1 b1  1 7 
 a b   3   9 
 
2 
 2
 a1  3b1  7
a 2  3b 2  9
Where a1,b1,a2,b2 are taking any real values.
It is a linear transformation.
Example 4
Let T: 𝐑 → 𝐑 be define d by T(𝐚𝟏 , 𝐚𝟐 )= (𝟐𝐚𝟏 + 𝐚𝟐 , 𝐚𝟏 ). Prove that T is linear.
Solution:
Given T: R3 → R2 be defined by T(a1 , a2 )= (2a1 + a2 , a1 ).
To prove T is linear
let 𝑢 = (a1 , a2 ) ∈ R2 and 𝑣 = (a1 ′, a2 ′)ℇ R2 ,For c ∈ R
𝟐
𝟐
T(u  v)  T (a1,a 2 )  (a1 ',a 2 ') 
 T(a1  a1 ',a 2  a 2 ')
  2(a1  a1 ')  (a 2  a 2 '),a1  a1 ' 
  (2a1  a 2 )  (2a1 ' a 2 '),a1  a1 ' 
 (2a1  a 2 ,a1 )  (2a1 ' a 2 ',a1 ')
T(u  v)  T(u)  T(v)
T(c u)  T c(a1,a 2 ) 
 T (c a1 ,c a 2 ) 
 (2 c a1  c a 2 ,c a1 ) 
 c(2a1  a 2 ,a1 ) 
 cT(u)
This implies T is linear
Example 5
If T is linear then prove that T(0)=0
Solution:
Given T is linear transform., Then by the definition of linear transform
we have T(0) = T(0.x) = 0T(x) = 0,
where x is an arbitrary element in V,
Therefore, T(0)=0.
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Example 6
If T is linear if and only if T(cx+y) = cT(x)+T(y), for all x,y 𝛜 𝐕 and c 𝛜 𝐅.
Solution:
If T is linear, then T(cx+y) = T(cx)+T(y)
=cT(x) + T(y)
Conversely, If T(cx + y) = cT(x) + T(y),
then we may take c = 1 or y = 0 and conclude that T is linear.
Hence proved.
Example 7
If T is linear then T(x-y) = T(x)-T(y) , for all x,y𝛜 𝐕.
Solution:
The definition of subtraction is given by x-y=x+(-y)
Also x  V, 1.x   x
1  (1)  x  0
1.x  (1)x  0
Since the additive inverse in unique, we conclude that (-1).x=-x
Assume T is linear T(X-Y) =T(x+-1.y)
=T(x) + (-1) T(y)
=T(x) + (-T(y))
=T(x)-T(y) for all x,y V
Example 8
Let T: 𝐑𝟐 → 𝐑𝟐 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 − 𝐚𝟐 ). Prove that T is linear.
Solution:
Given,T(a1 ,a 2 )  (a1  a 2 )
let 𝑢 = (a1 , a2 , )ℇ R2 and 𝑣 = (a1 ′, a2 ′)ℇ R2 , For c ℇ R
T(u  v)  T  (a1 ,a 2 )  (a1 ',a 2 ')
 T(a1  a1 ',a 2  a 2 ')
  (a1  a1 ')  (a 2  a 2 ') 
 (a1  a 2 )  (a1 ' a 2 ' )
T(u  v)  T(u)  T(v)
T(c u)  T  c(a1 , a 2 ) 
 T  (c a1 , c a 2 ) 
  (c a1  c a 2 ) 
  c(a1  a 2 )   cT(u)
Hence it’s a linear transformation .
Example 9
Suppose that T: 𝐑𝟐 → 𝐑𝟐 is linear, T(1,0) = (1,4), and T(l, 1) = (2,5).What is T(2,3)?
Is T one-to-one?
Solution:
Given that T is linear
To find: T (2,3)
Let (a, b) be any vector in R2, then
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(a, b)  x(1,0)  y(1,1)
x y a
yb
Solving above two equations for x and y, we get x  a  b, y  b
Therefore (a, b)  (a  b)(1,0)  (b)(1,1)
Taking T on both sides, we get
T (a, b)  T (a  b)(1,0)  (b)(1,1)
Since T is linear
T (a, b)  (a  b) T (1,0)  (b) T (1,1) , Given that T ( l , 0) = (1,4)
and T(1,1) = (2,5).
T (a, b)  (a  b) (1,4)  (b) (2,5)
 (a  b,4a  4b)  (2b,5b)
T (a, b)  (a  b,4a  b) , Given a  2, b  3
T (2,3)=(5,11)
Example 10
Prove that there exists a linear transformation T: 𝐑𝟐 → 𝐑𝟑such that T ( l , l ) = (1,0,2) and
T(2,3) = (1,-1,4). What is T(8,11)?
Solution:
(i)To prove that T linear
Consider the set S  1,1, (2,3),now we will prove S is linearly
independent
Let a(1,1)  b(2,3)  (0,0)
 (a, a)  (2b,3b)  (0,0)
 (a  2b, a  3b)  (0,0)
 (a  2b)  0 and (a  3b)  0
 a  0, b  0
Whenever a(1,1)  b(2,3)  (0,0) , the result will be a=0 and b=0
Therefore S  1,1, (2,3)is linearly independent.
Furthermore, the dimensions of R2=2 represents the number of vectors in R
 S spans R2
So every vector in the domain R2 can be written as a linear combination of these vector
Therefore, it proves the transformation T is lineally independent
(ii)To find the value of T(8,11)
Let (a, b) be any vector in R2, then
(a, b)  x(1,1)  y(2,3)
x  2y  a
x  3y  b
Solving above two equations for x and y, we get
x  3a  2b, y  b  a
Therefore (a, b)  (3a  2b)(1,1)  (b  a)(2,3)
Taking T on both sides, we get
T (a, b)  T (3a  2b)(1,1)  (b  a)(2,3)
Since T is linear
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T (a, b)  (3a  2b) T (1,1)  (b  a ) T (2,3) ,
Given that T ( l , l ) = (1,0,2) and T(2,3) = (1,-1,4).
T (a, b)  (3a  2b) (1,0,2)  (b  a) (1,1,4)
 (3a  2b,0,6a  4b)  (b  a, a  b,4b  4a)
T (a, b)  (2a  b, a  b,2a) , Given a  8, b  11
T(8,11)=(5,-3,16)
Example 11
𝟑
𝟐
Is there a linear transformation T: 𝐑 → 𝐑 such that T(l, 0,3) = (1,1) and T ( - 2 , 0 , - 6 ) = (2,1)?
Solution:
We know that, If T is linear, then
(a) T (x + y) = T(x) + T(y)
(b) T(cx) = c T(x)
Given: T(l, 0,3) = (1,1)
…(1)
T ( - 2 , 0 , - 6 ) = (2,1)
…(2)
T ( - 2 , 0 , - 6 ) = T ( -2 (1, 0, 3))
= - 2 T (1, 0, 3) = -2 (1, 1) by (1)
= (-2, -2) ≠ (2, 1)
So, T is not linear.
Example 12
𝟐 𝟐
𝟐
𝟐
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 , 𝐚𝟐 ) ,check whether it is linear or not.
Solution:
let 𝑢 = (a1 , a2 , ) ∈ R2 and 𝑣 = (a1 ′, a2 ′) ∈ R2 , For c ℇ R
T(u  v)  T (a1,a 2 )  (a 1 ',a 2 ' ) 
 T(a1  a1 ',a 2  a 2 ')
2
2
  a 1  a 1 '  ,  a 2  a 2 '  



since T(a1 , a 2 )  T(a12 , a 2 2 )


  a12  2a1a1 ' a1 '2 , a 2 2  2a 2a 2 ' a 2 '2 
T(u)  T(v)  T(a1,a 2 )  T(a1 ',a 2 ')
 (a12 , a 2 2 )  (a1 '2 , a 2 '2 )
Therefore T(u  v)  T(u)  T(v)
This implies T is not linear.
Example 13
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝟏, 𝐚𝟐 ) ,check whether it is linear or not.
Solution:
Let T: R2 → R2 be defined by T(a1 , a2 )= (1, a2 )
let 𝑢 = (a1 , a2 , )ℇ ∈and 𝑣 = (a1 ′, a2 ′) ∈ R2 , For c ℇ R
T(u  v)  T (a1 ,a 2 )  (a1 ',a 2 ') 
𝟐
𝟐
 T(a1  a1 ',a 2  a 2 ')
 1,(a 2  a 2 ') 
T(u  v)  (1,a 2 )  (1,a 2 ') ...............(1)
T(u)  T(v)  T(a1,a 2 )  T(a1 ',a 2 ')
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 (1,a 2 )  (1,a 2 ').............(2)
T(u  v)  T(u)  T(v)
Therefore T is not linear.
Example 14
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝐬𝐢𝐧𝐚𝟏 , 𝟎) ,check whether it is linear or not.
Solution:
Let T: R2 → R2 be defined by T(a1 , a2 )= (sina1 , 0)
let 𝑢 = (a1 , a2 , ) ∈ R2 and 𝑣 = (a1 ′, a2 ′) ∈ R2 , For c ℇ R
𝟐
𝟐
T(u  v)  T (a1,a 2 )  (a 1 ',a 2 ') 
 T(a1  a1 ',a 2  a 2 ')
T(u  v)   sin(a1  a1 '),0  ...............(1)
T(u)  T(v)  T(a1,a 2 )  T(a 1 ',a 2 ')
 (sin a1 ,0)  (sin a1 ',0).............(2)
T(u  v)  T(u)  T(v)
Therefore T is not linear.
Example 15
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 , 𝐚𝟏 𝟐 ) ,check whether it is linear or not.
Solution:
let 𝑢 = (a1 , a2 , ) ∈ R2 and 𝑣 = (a1 ′, a2 ′) ∈ R2 , For c ℇ R
T(u  v)  T (a1,a 2 )  (a 1 ',a 2 ' ) 
𝟐
𝟐
 T(a1  a1 ',a 2  a 2 ')
2
  a1  a1 '  ,  a1  a1 '   since T(a1 , a 2 )  T(a1 , a12 )


  a1  a1 '  ,  a12  2a1a1 ' a1 ' 2  
T(u)  T(v)  T(a1,a 2 )  T(a1 ',a 2 ')
 (a1 , a12 )  (a1 ', a 1 ' 2 )
Therefore T(u  v)  T(u)  T(v)
This implies T is not linear
Example 16
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (⌊𝐚𝟏 ⌋, 𝐚𝟐 ) ,check whether it is linear or not.
Solution:
let 𝑢 = (a1 , a2 , ) ∈ R2 and 𝑣 = (a1 ′, a2 ′) ∈ R2 , For c ℇ R
𝟐
𝟐
T(u  v)  T  a1 ,a 2   (a1 ',a 2 ' ) 
 T(a1  a1 ',a 2  a 2 ')
  a1  a1 ' ,  a 2  a 2 '  since T(a1 , a 2 )  T( a1 , a 2 )
=  a1 ,a 2    a1 ' ,a 2 ' 
  a1  a1 '  ,  a 2 ,a 2 ' 
 T(u  v)  T(u)  T(v) . This implies T is not linear
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T(u)  T(v)  T  a 1, a 2   T(a 1 ', a 2 ' )
Regulation: IFETCE R-2019
Academic Year: 2022-2023
Example 17
Verify that T: 𝐑𝟑 → 𝐑 and T(𝐮)=‖𝒖‖ is a linear or not.
Solution:
let 𝑢, 𝑣 ∈ R and For c ∈ R
T(u  v) || u  v ||
sin ce T(u) || u ||
|| u ||  || v || [By Triangleinequality]
 T(u)  T(v)
T(u  v)  T(u)  T(v)
This implies T is not linear.
3.2 NULL SPACES AND RANGES
NULL SPACE
Let V and W be vector space, and let T: VW is linear.
We define the null space (or kernal) N(T) of T to be the set of all vectors x in V such that T(x)=0:
That is , N(T)={x ∈ V ;T(x)=0}
RANGE
Solution:
Let V and W be vector space, and let T: VW is linear.
We define the range (or image) of T to be subset of W consisting of all images (under T) of vectors in V;
That is, R (T) = {T(x): x ∈ V}.
RANK AND NULLITY
RANK OF T
The dimension of range space is called the Rank of T.
It is denoted by R(T).
NULLITY OF T
The dimension of null space is called the Nullity of T.
It is denoted by N(T).
DIMENSION THEOREM.(OR) RANK-NULLITY THEOREM
Statement:
Let V and W be vector spaces and T:𝐕 → 𝐖 be linear. If V is finite dimensional, then,
Nullity(T) + rank(T) = dim(V)
Proof:
Suppose that dim(V )  n, dim( N (T ))  k and {v1 , v2 ,......vk } is a basis for N (T ) .
By the corollary to Theorem , if W is a subspace of a finite- dimensional vector space V , then any basis for W
can be extended to a basis for V .
From this we have extend {v1 , v2 ,......vk } to a basis   {v1 , v2 ,......vn } for V .We claim that
S  {T (vk 1 ), T (vk 2 ),....T (vn )} is a basis for R(T ) . First we prove that S generates R(T ) .
Using Theorem , [Let V and W be vector spaces, and let
T : V  W be linear. If   {v1 , v2 ,......vn } is a basis for V , then T (vi )  0 for 1  i  k ]
we have R(T )  span(T ())  span({T (v1 ), T (v2 )...T (vn )})  span(S ) .
Now we prove that S is linearly independent.
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n
Suppose that
 bT (v )  0
i
i  k 1
i
for bk 1 , bk  2 ....bn  F
n
Using the fact that T is linear, we have T (  bi vi )  0
i  k 1
n
so
 b v  N (T )
i  k 1
i i
Hence there exist c1 , c2 .......ck  F such that
n
k
i  k 1
i 1
 bi vi   ci vi (or )
k
 (ci )vi 
i 1
n
 bv
i  k 1
i i
0
Since  is a basis for V , we have bi  0 for all i
Hence S is linearly independent. This shows that T (vk 1 ), T (vk 2 ),......T (vn ) are distinct.
 rank (T )  n  k
 rank (T )  dim(V )  dim( N (T ))
 dim( N (T ))  rank (T )  dim(V )
Worked Examples 3.2(A)
Example 1
Give an example of a linear transformation T: 𝐑𝟐 → 𝐑𝟐 such that N(T) = R(T).
Solution:
By the rank nullity theorem and by the hypothesis N(T)=R(T)
we get dim[𝑁(𝑇)]=dim[𝑅(𝑇)]=1,
and as you mentioned we have T2=0 which means that T is nilpotent .
Let N(T)=span(v) and complete it on a basis (v,w) so T(w)=αv
0  
and then the matrix of T relative to this basis is , 

0 0 
Example 2
Give an example of distinct linear transformations T and U such That N(T)=N(U) and R(T)=R(U).
Solution:
Let T : R2  R2 and T ( x, y)  (y, x) .
Let U is the identity map from R2  R2 .
Then we have N (T )  N (U )  0 and R(T )  R(U )  R2
Example 3
State the dimension theorem.
Solution:
Let V and W be vector spaces, and let T : V  W be linear.
If Vis finite- dimensional then, the dimension theorem will be,
nullity(T )  rank (T )  dim(V )
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Regulation: IFETCE R-2019
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Worked Examples 3.2(B)
Theorem 1
Let V and W be vector spaces and T:𝐕 → 𝐖, If B={𝐯𝟏 , 𝐯𝟐 ,……𝐯𝐧 } is a basis for V, Then show that
range(T)= Span({𝐓(𝐯𝟏 ), 𝐓(𝐯𝟐 ),……𝐓(𝐯𝐧 )})
proof:
Clearly T (vi )  R(T ) for each i.
Because R(T ) is a subspace,
R(T ) contains span({T (v1 ), T (v2 )...T (vn )})  span(T ( B))
Now suppose that w  R(T ) . Then w  T (v) for some v V . Because B is a basis for V ,
n
we have v   ai vi for some a1 , a2 ,....an  F . Since T is linear, it follows that
i 1
n
w  T (v)   aiT (vi )  span(T ( B))
i 1
So R(T ) is contained in span(T ( B)) .
Theorem 2
Let V and W be vector spaces over the field F and T:𝐕 → 𝐖 be a linear transformation. Show that N(T)
and R(T) are subspaces of V and W respectively.
proof:
Let us consider the symbols 0 v and 0 w to denote the zero vectors of V and W , respectively.
Since T (0v )  0 w , We have 0v  N (T )
Let x, y  N (T ) and c  F.
Then T ( x  y)  T ( x)  T ( y)  0w  0w  0w and T (cx)  cT ( x)  c0w  0w .
Hence x  y  N (T ) and cx  N (T ) ,
so that N (T ) is a subspace of V .
Because T (0v )  0 w , we have 0w  R(T )
Now let x, y  R(T ) and c  F .
Then there exist v and w in V such that T (v)  x and T (w)  y .
So T (v  w)  T (v)  T (w)  x  y
T (cv)  cT (v)  cx .
Thus x  y  R(T ) and cx  R(T ) ,
so R(T ) is a subspace of W .
Theorem 3
Let V and W be vector spaces. AndT:𝐕 → 𝐖 be a linear transformation. If T is one-to-one if and only if
N(T)={0}.
proof:
Suppose that T is one-to-one and x ϵ N(T). Then T(x) = 0 = T(0).
Since T is one-to-one, we have x = 0. Hence N(T) = {0}.
Now assume that N(T) = {0} and suppose that T(x) = T(y).
Then 0 = T(x) – T(y) = T(x – y) by property.
Therefore x – y ϵ N(T) = {0}.
So,x – y = 0 or x = y.
This means that T is one-to-one.
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Theorem 4
Let V and W be vector spaces of equal(finite) dimension: 𝐕 → 𝐖 be a linear. Then the following are
linear. (i)T is one-to-one (ii) T is onto (iii) Rank(T)=dim(V)
proof:
(i)  (iii)
From the dimension theorem, we have
nullity(T) + rank(T) = dim(V).
Let V and W be vector spaces and T : V  W be a linear transformation.
If T is one-to-one if and only if N (T )  {0} .
Now, with the use of the above result,
we have that T is ono-to-one if and only if rank (T) = dim(V)
(iii)  (ii)
rank (T) = dim(V), if and only if rank (T) = dim (W),
and if and only if dim (R(T)) = dim(W).
By Theorem W be a subspace of a finite dimensional vector space V. Then W is finite dimensional and dim(W)
 dim(V). Moreover, if dim(W) = dim(V), then V=W.
This equality is equivalent to R(T) = W,
T is onto.
Theorem 5
Let V and W be vector spaces over F, and suppose that{𝐯𝟏 , 𝐯𝟐 ,……𝐯𝐧 } is a basis for V. For
𝐰𝟏 , 𝐰𝟐 ,……𝐰𝐧 in W, there exists exactlyone linear transformation T: 𝐕 → 𝐖 such that T(𝐯𝐢 )=𝐰𝐢 ,
for i= 1,2,3,,n.
proof:
Let x ϵ V. Then,
n
x   ai vi ,
i 1
Where a1, a2, …, an are unique scalars.
Define T: V →W by
n
T ( x)   ai wi
i 1
(a) T is linear: Suppose that u, v ϵ V and d ϵ F. Then we may write
n
n
u   bi vi and v   ci vi for some scalars b1, b2, …, bn, c1, c2, …, cn.
i 1
i 1
n
Thus, du  v   (dbi  ci )vi .
i 1
n
n
n
i 1
i 1
i 1
So, T (du  v)   (dbi  ci ) wi d  bi wi   ci wi  dT (u )  T (v) .
(b) Clearly
T (vi )  wi for i  1,2,....n.
(c) T is unique: Suppose that U: V→W is linear and U(vi) = wi for i = 1, 2, …, n. Then for x ϵ V with
n
x   ai vi ,
i 1
n
n
i 1
i 1
We have,U ( x)   aiU (vi )  ai wi  T ( x). Hence U = T.
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Theorem 6
Let V and W be vector spaces over a field F, and let T,U: 𝐕 → 𝐖 be linear (i) For all a𝛜 F, aT+Uis linear.
(ii)Using the operations of addition and scalar multiplication in the preceding, definition the collection
of all linear transformations from V to W is a vector space over F.
proof:
(a) For x,y ∈ V and c ∈ F
 aT  U  cx  y   aT(cx  y)  U  cx  y 
 a  T(cx  y)   cU(x)  U(y)
 a  cT(x)  T(y)   cU(x)  U(y)
 acT(x)  cU(x)  aT(y)  U(y)
 c(aT  U)(x)  (aT  U)(y)
Therefore, aT+U is linear.
(b) Noting that T0, the zero transformation, plays the role of the zero vector. It is easy to verify that the axioms
of a vector space are satisfied, and hence that the collection of all linear transformations from V into W is a
vector space over F.
Hence proved.
Theorem 7
Let 𝐰𝟏 , 𝐰,……, 𝐰𝐤 be subspaces of a finite dimensional vector space V such that ∑𝐤𝐢=𝟏 𝐖𝐢 = 𝐕 prove that
V is the direct sum of 𝐰𝟏 , 𝐰,……, 𝐰𝐤 if and only if dim(V)=∑𝐤𝐢=𝟏 𝐝𝐢𝐦(𝐖𝐢 )
proof:
First we can check that,
w1  w2  w3 =(w1  w2 )  w3
If V is direct sum of w1 , w,……, wk , the dimension of V would be the sum of of the dimension of each W, by
using the induction method.
Conversely we first prove that if we have ∑ki=1 Wi = V
Then we must have, dim(V)=∑ki=1 dim(Wi )
We induct on k. for k=2, we may use the known formula,
dim(w1 + w2 ) =dim(w1 )+dim(w2 )-dim(w1 ∩ w1 )
Suppose it holds for k=m. we know that,
m
∑m−1
i=1 Wi + Wm = ∑i=1 Wi
m−1
m
∑m
i=1 Wi ≤ ∑i=1 Wi + Wm ≤ ∑i=1 Wi
By the induction hypothesis and case for k=2.
To prove the original statement, suppose,
by contradiction, that W1∩ ∑ 𝑖=2𝑘 Wi has nonzero element.
Then by known formula
dim(w1 + w2 ) =dim(w1 )+dim(w2 )-dim(w1 ∩ w1 ),
We have, dim(∑ 𝑖=2𝑘 Wi ) > dim(V)-dimW1=∑ki=2 Wi
This is impossible. So we get the desired result.
3.3 MATRIX REPRESENTATION OF LINEAR TRANSFORM
Each linear transformation L : R n → R m can be written as a matrix multiple of the input:
L(x) = Ax,
th
where the i column of A, namely the vector ai = L(ei),
where {e1, e2, . . . , en} is the standard basis in R n .
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That is, to find the columns of A one must find L(e i) for each 1 ≤ i ≤ n.
Worked Examples 3.3(A)
Example 1
Let T: 𝐑𝟑 → 𝐑𝟐 be defined by T(𝐚𝟏 , 𝐚𝟐,𝐚𝟑 )= (𝐚𝟏 − 𝐚𝟐, 𝟐𝐚𝟑 ). Compute the matrix of transformation with
respect to the standard bases of 𝐑𝟑 𝐚𝐧𝐝 𝐑𝟐 .
Solution:
Given, T (a1,a 2 ,a 3 )  (a1  a 2 ,2a 3 )
  {1, x, x 2} be the basis of 𝐑𝟐 .
Let,
T (1,0,0)  (1,0)
T (0,1,0)  (1,0)
T (0,0,1)  (0,1)
1 1 0 

0 0 1 
T B  
r
Example 2
Let T: 𝐏𝟐(R) → 𝐏𝟐(R) defined as T ( f ( x))  f(x)  xf ( x), Compute the matrix of the transformation
Compute the matrix of the transformation with respect to the standard basisP2(R).
Solution:
Given, T ( f ( x))  f(x)  xf ( x)
Let,
  {1, x, x 2} be the basis of P2 (R).
T (1)  1
T ( x)  x  x(1)  x  x  2 x
T ( x2 )  x2  x(2x)  x 2  2x 2  3x 2
1 0 0 
[T ]  0 2 0 
0 0 3 
Example 3
Let 𝛃 and 𝛄 be the standard ordered bases for 𝐑𝟐 and 𝐑𝟑 respectively. Let T:𝐑𝟐 → 𝐑𝟑 defined by
T(𝐚𝟏 , 𝐚𝟐 )= (𝟐𝐚𝟏 − 𝐚𝟐 , 𝟑𝐚𝟏 + 𝟒𝐚𝟐 , 𝐚𝟏 ),compute T 

Solution:
Given, T(a1 , a2 )= (2a1 − a2 , 3a1 + 4a2 , a1 )
T(1, 0 ) = (2, 3,1)
T(0, 1) = (-1,4,0)
 2 1
T    3 4 
 1 0 
Example 4
Let 𝛃and 𝛄 be the standard ordered bases for 𝐑𝟑 and 𝐑𝟐 respectively.Let T:𝐑𝟑 → 𝐑𝟐 defined by,

T(𝐚𝟏 , 𝐚𝟐, 𝐚𝟑, )= (𝟐𝐚𝟏 + 𝟑𝐚𝟐 − 𝐚𝟑, , 𝐚𝟏 + 𝐚𝟑 ),compute T 
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Regulation: IFETCE R-2019
Academic Year: 2022-2023
Solution:
Given T:R3 → R2 by
T  a1 , a2 , a3    2a1  3a2  a3 , a1  a3 
T 1, 0, 0    2,1
T  0,1, 0    3, 0 
T  0, 0,1   1,1
 2 3 1

Therefore T   

1 0 1 
Example 5
Let 𝛃 and 𝛄 be the standard ordered bases for 𝐑𝟐 and 𝐑𝟑 respectively. Let T:𝐑𝟐 → 𝐑𝟑 defined by,

T(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 + 𝟑𝐚𝟐 , 𝟎, 𝟐𝐚𝟏 − 𝟒𝐚𝟐 ),compute T 
Solution:
Given T:R2 → R3 by T(a1 , a2 )= (a1 + 3a2 , 0,2a1 − 4a2 ),
T(1, 0) = (1, 0, 2) = 1e1 + 0e2 +2e3
And T(0, 1) = ( 3, 0, -4) = 3e1 + 0e2 + 2e3
1 3 



Therefore T    0 0 
 2 4 


Example 6
Let 𝛃 and 𝛄 be the standard ordered bases for 𝐑𝐧 and R respectively. Let T:𝐑𝐧 → 𝐑 defined by

T(𝐚𝟏 , 𝐚𝟐,…, 𝐚𝐧 ) = 𝐚𝟏 + 𝐚𝐧 ,compute T 
Solution:
Given T:Rn → R defined by T(a1 , a2,…, an ) = a1 + an
2
n
Let,   {1, x, x ,........ x } be the basis of 𝐑𝐧
T 1, 0,...0   1  0   1
T  0,1,...0    0  0   0
T  0, 0,1...0    0  0   0
....
....
T  0, 0,...1   0  1  1
T 

1 
0
 
0
 
 .
.
 
. 
1 
  n1
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Regulation: IFETCE R-2019
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Example 7
Let T: 𝐏𝟑(R) → 𝐏𝟐(R) defined by T ( f ( x))  f (x). Let 𝛃 and 𝛄 be the standard ordered bases for
𝐏𝟑(R) → 𝐏𝟐 (R) respectively. Then compute T 

Solution:
Given, T ( f ( x ))  f (x).
T (1)  0
T ( x)  1
T ( x2 )  2 x
0 0 0 
[T ]  0 2 0 
0 0 3
Example 8
 1 0   0 1   0 0   0 0  
Let T : M 22  R   M 22  F  defined by T ( A)  AT   
,
,
,
 ,
0
0
0
0
1
0
0
1









then compute T 
Solution:
1
T
0
0
T
0
0 1 0


0 0 0
1 0 0


0 1 0
0
T
1
0
T
0
0 0 1


0 0 0
0 0 0


1 0 1
T 
1
0

0

0
0
0
1
0
0
1
0
0
0
0 
0

1
Example 9
 1 0   0 1   0 0   0 0  
Let T : M 22  R   F defined by T ( A)  tra( A) ,   
,
,
,
  and   {1}
 0 0   0 0   1 0   0 1  

compute T 
Solution:
1 0
T
 1
0 0
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Regulation: IFETCE R-2019
0
T
0
0
T
1
0
T
0
Academic Year: 2022-2023
1
0
0
0
0
0
0
 1
1
 T   1 0 0 1

Example 10
Let T : P2  R   R defined by T (f(x))  f (2) ,   {1, x, x 2 }and   {1},
compute T 

Solution:
Given   {1, x, x 2 }
T 1  1
T  x  2
T  x2   4

T   1
2 4
If (x)  3  6 x  x ,   {1, x, x }compute  f ( x) 
2
Example 11
2
Solution:
Given f (x)  3  6x  x2 and   {1, x, x 2 }
f (x)  3  6 x  x2  3(1)  6( x)  1( x2 )
 3
 
 f ( x)   6 
1
 
Worked Examples 3.3(B)
Example 1
Let V and W be finite-dimensional vector spaces with ordered bases 𝛃 𝐚𝐧𝐝 𝛄 respectively, and let
T,U:𝐕 → 𝐖 be linear transformations.Then
(i) T  U   T   U  and



(ii) T    U  for allscalars α.


Solution:
Let β= {β1, β2...βn}
And γ ={γ 1, γ 2...γm}
Let  a ij  mxn be the matrix [T:β:γ]
 T   T   j    a ij i, j  1, 2...n
m

i 1
Let  bij 
mxn
be the matrix[U :  : ]
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Regulation: IFETCE R-2019
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 U   U   j    bij i, j  1, 2...n
m

i 1

T  U   T    U 
(i) To prove


T  U   (T  U) ( j )  T( j )  U( j )

m
m
i 1
m
i 1
j  1, 2...n
  a ij i,   bij i, , j  1, 2,3...n
  (a ij  bij ) i
i 1
 Matrix of T  U   [a ij  bij ]m x n

 [a ij ]m x n  [bij ]m x n
Hence T  U   T   U 



(ii ) To prove  cT   c U  for allscalars c.


T   (cT)( j )  cT( j )

, j  1, 2...n
m
 c a ij i,
i 1
m
  (c a ij ) i,
i 1
 The matrix of c T relative to  and  is
cT   [c a ij ]m x n

 c[a ij ]m x n
 c  T 

Hence  cT   c U  for allscalars c.


Example 2
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 − 𝐚𝟐 , 𝐚𝟏 , 𝟐𝐚𝟏 + 𝐚𝟐 ). Let 𝛃 be standard ordered basis for 𝐑𝟐


and 𝛄 ={(1,1,0),(0,1,1),(2,2,3)}. compute T  . If 𝛂 ={(1,2),(2,3)},compute T 
𝟐
𝟑
Solution:
Given T (a1 , a2 )= (a1 − a2 , a1 , 2a1 + a2 ).
γ ={(1,1,0),(0,1,1),(2,2,3)} and α ={(1,2),(2,3)},
 1 
2
T (1, 0)  (1,1, 2)    (1,1, 0)  (0)(0,1,1)    (2, 2,3)
 3 
3
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T (0,1)  (1, 0,1)  (1)(1,1, 0)  (1)(0,1,1)  (0)(2, 2, 3)
T 

 1/ 3
  0
 2 / 3
 1
1
0 
 1 
2
T (1, 2)  T (1, 0)  2T (0,1)    (1,1, 0)  (0)(0,1,1)    (2, 2,3)  2((1)(1,1, 0)  (1)(0,1,1)  (0)(2, 2,3)
 3 
3
 7 
2
   (1,1, 0)  (2)(0,1,1)    (2, 2,3)
 3 
3
Similarly,
 11 
4
T (2,3)    (1,1, 0)  (3)(0,1,1)    (2, 2,3)
 3 
3
 7 / 3  11/ 3 

3 
T    2
 2 / 3
4 / 3
Example 3
Let T: 𝐑𝟐 → 𝐑𝟑 and U: 𝐑𝟐 → 𝐑𝟑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 + 𝟑𝐚𝟐 𝟎, 𝟐𝐚𝟏 − 𝟒𝐚𝟐 ) and
U(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 − 𝐚𝟐 , 𝟐𝐚𝟏 , 𝟑𝐚𝟏 + 𝟐𝐚𝟐 ) Let 𝛃 and 𝛄 be standard ordered basis of 𝐑𝟐 and 𝐑𝟑 respectively.

compute T  U 
Solution:
Let T: R2R3 and U: R2R3 be the linear transformation respectively defined by
T(a1,a2) = ( a1+3a2, 0, 2a1-4a2) and U(a1,a2) = ( a1-a2, 2 a1, 3a1+2a2)
Let β and γ be the standard ordered bases of R2 and R3 , respectively. Then T: R 2R3,
U: R2R3
T(a1,a2) = ( a1+3a2, 0, 2a1-4a2)
U(a1,a2) = ( a1-a2, 2 a1, 3a1+2a2)
β＝{(1,0), (0,1)}
γ = {(1, 0, 0),(0, 1, 0),(0, 0, 1)}
T(1,0) = (1, 0, 2), T(0,1) = (3, 0, -4) U(1,0) = (1, 2, 3),
U(0,1) = (-1, 0, 2)
T 

3
1
1



 0
0  U    2

 2  4 
 3
1 
0 
2
If we compute T+U using the preceding definitions, we obtain
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(T+U) (a1,a2)=( a1+3a2,0,2a1-4a2) + ( a1-a2, 2 a1, 3a1+2a2)
= ( 2a1+2a2, 2a1, 5a1-2a2)
2
  2
5
T  U 

2
0 
 2 
Example 4
x
Let T: 𝐏𝟐(R) → 𝐏𝟑(R) defined by T ( f ( x ))  f (x)   2 f (t )dt find bases for N  T  and R  T  and hence
0
hence verify the dimension theorem. Is T one-to-one?is T onto? Justify your answer.
Solution:
(i)To find the matrix representation of T:
2
Let,   {1, x, x } be the basis of P2(R).
x
T (1)  0  2  dt  0  2[t ]0x  2 x
0
x
t2 
T (1)  1  2  tdt  1  2    1  x 2
 2 0
0
x
x
 t3 
2 x3
T ( x )  2 x  2 t dt  2 x  2    2 x 
3
 3 0
0
0 1 0 
2 0 2 

[T ]  
0 1 0 


 0 0 2 / 3
(ii) To find the R(T):
0 1 0 
2 0 2 

[T ]  
0 1 0 


 0 0 2 / 3
0 1 0 
0 
 2 0 2   x1  0 
  x2    
 
 0 1 0    0 

  x3   
 0 0 2 / 3
0 
x
2
0

2
[T ]  
0

0
2
1 0 0

0 2 0
1 0 0

0 2 / 3 0
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0

2
[T ]  
0

0
Academic Year: 2022-2023
1 0 0

0 2 0
 R3  R3  R1 
0 0 0

0 2 / 3 0
R[T]=3
(iii)To find N(T):
x2  0
x1  x3  0
x3  0
 x1    x3 
x    0 
 2  
 x3   0 
N(T)= span of {}
N(T) = 0
(iv)To verify dimension theorem:
Here, dimension(n)= no.ofcolums of the matrix = 3
By the dimension theorem,
Dimension of T = Rank(T) + Nullity(T)
⟹3=3+0
⟹3=3
Hence verified dimension theorem.
(v)To check one-to-one and onto:
T is one to one if and only if N(T)=0
T is onto if and only if R(T)=dimension of T.
Here N(T)=0, so T is one to one
Here, Rank(T)= Dim(R3), that is 3=3
∴ T is onto.
Therefore, the given transformation is one to one but not onto.
Example 2
𝟑
𝟐
Let T: 𝐑 → 𝐑 be defined by T(𝐱, 𝐲, 𝐳)= (𝟐𝐱 − 𝐲 , 𝟑𝐳). Compute the matrix of transformation with
respect to the standard bases of 𝐑𝟑 𝐚𝐧𝐝 𝐑𝟐 .Find N(T) and R(T).Is T is one-to-one,onto? Justify your
answer.
Solution:
(i)To find the matrix representation of T:
Let,   {1, x, x 2} be the basis of P2(R).
T(1,0,0)=(2,0)
T(0,1,0)=(-1,0)
T(0,0,1)=(0,3)
 2 1 0 
[T ]  

 0 0 3
(ii) To find the R(T):
 2 1 0 
[T ]  

 0 0 3
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 x
 2 1 0    0 

  y   
 0 0 3   z  0 
 
 2 1 0 0 
[T ]  

0 0 3 0
R(T )  2
(iii) To find N(T):
2x  y  0
3z  0
 x   y / 2
1\ 2 
 y   y   y  1 
  



 z   0 
 0 
N(T)= span of {1/2,1,0}
N(T) =1
(iv) To check one-to-one and onto:
T is one to one if and only if N(T)=0
T is onto if and only if R(T)=dimension of T.
Here N(T)=1, so T is not a one to one.
Here, Rank(T)≠ Dim(R3), that is 2≠3
∴ T is not a onto.
Example 3
x
Let T: 𝐏𝟐(R) → 𝐏𝟑(R) defined by T ( f ( x))  2 f (x)   3 f (t )dt find bases for N  T  and R  T  and find bases
0
for N(T) and R(T) and hence verify the dimension theorem. Is T one-to-one?is T onto? Justify your
answer.
Solution:
(i)To find the matrix representation of T:
Let,   {1, x, x 2} be the basis of P2(R).
x
T (1)  2(0)  3 dt  0  3[t ]0x  3 x
0
x
t2 
3x 2
T ( x)  2(1)  3 tdt  2  3    2 
2
 2 0
0
x
x
t3 
T ( x )  2(2 x)  3 t dt  4 x  3    4 x  x 3
 3 0
0
0 2 0 
3 0 4

[T ]  
0 3 / 2 0 


0 0 1 
(ii) To find the R(T):
x
2
2
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0 2
3 0
[T ]  
0 3 / 2

0 0

0 2
3 0

0 3 / 2

0 0
Academic Year: 2022-2023
0
4

0

1
0
 0
 x1   

4   0
 x2   
0    0
  x3   
1    0
0 2

3 0
[T ]  
0 3 / 2

0 0
0

3
[T ]  
0

0
2
0
0
0
0 0

4 0
0 0

1 0 
0 0

4 0 
3 
R3  2 R3  R1 


00 
2 

1 0 
⇒ R[T]=3
(iii) To find the N(T):
2 x2  0
3x1  4 x3  0
x3  0
 x1   4 x3 / 3
x    0 
 2 

 x3   0 
N(T)= span of {}
N(T) = 0
(iv) To verify dimension theorem:
Here, dimension(n)= no.ofcolums of the matrix = 3
By the dimension theorem,
Dimension of T = Rank(T) + Nullity(T)
⟹3=3+0
⟹3=3
Hence verified dimension theorem.
(v) To check the one-to-one and onto.
T is one to one if and only if N(T)=0
T is onto if and only if R(T)=dimension of T.
Here N(T)=0, so T is one to one
Here, Rank(T)= Dim(R3), that is 3=3
∴ T is onto.
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Example 4
Let T: 𝐑𝟐 → 𝐑𝟑 be defined by T(𝐱, 𝐲)= (𝟐𝐱 − 𝐲 , 𝟑𝐱 + 𝟒𝐲, 𝐱). Compute the matrix of transformation with
respect to the standard bases of 𝐑𝟑 𝐚𝐧𝐝 𝐑𝟐 .Find N(T) and R(T).Is T is one-to-one,onto? Justify your
answer.
Solution:
(i)To find the matrix representation of T:
T(1,0,)=(2,3,1)
T(0,1,)=(-1,4,0)
 2 1
[T ]   3 4 
 1 0 
(ii) To find the R(T):
 2 1
[T ]   3 4 
 1 0 
 2 1
0
x


  3 4     0 
y
1 0    0 
 2 1 0 


[T ]   3 4 0 
1 0 0 
1 0.5 0 


[T ] 3
4 0   R1  R1 / 2
1
0 0 
1 0.5 0 

  R2  3R1  R2 
[T ] 0 5.5 0  
R  R3  R1 
0 0.5 0   3
1 0.5 0 


[T ] 0 5.5 0   R3  0.5 R2  5.5 R3 
0
0 0 
R(T )  2
(iii)To find N(T)
x  0.5 y  0
5.5 y  0
 x   y / 2
 y   0 
  

N(T)= span of {}
N(T) = 0
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(iv)To verify dimension theorem:
Here, dimension(n)= no.ofcolums of the matrix = 2
By the dimension theorem,
Dimension of T = Rank(T) + Nullity(T)
⟹2=2+0
⟹2=2
Hence verified dimension theorem.
(v)To check one-to-one and onto:
T is one to one if and only if N(T)=0
T is onto if and only if R(T)=dimension of T.
Here N(T)=0, so T is one to one
Here, Rank(T)= Dim(R3), that is 2=2
∴ T is onto.
Example 5
Let T: 𝐑 → 𝐑 be defined by T(𝐱, 𝐲)= (𝐱 + 𝟑𝐲 , 𝟎, 𝟐𝐱 − 𝟒𝐲). Compute the matrix of
transformation
𝟐
𝟑
with respect to the standard bases of 𝐑 𝐚𝐧𝐝 𝐑 . Find N(T) and R(T). Is T is one-to-one, onto? Justify
your answer.
Solution:
(i)To find the matrix representation of T:
T(1,0,)=(1,0,2)
T(0,1,)=(3,0,-4)
1 3 
[T ]   0 0 
 2 4 
𝟐
𝟑
(ii) To find the R(T):
1 3 
[T ]   0 0 
 2 4 
1 3 
0
 x  


  0 0     0 
 y  0 
 2 4 
 
1 3 0 


[T ]  0 0 0 
 2 4 0 
[T ]
1 3 0 


0 0 0   R3  R3  2 R1 
0 10 0 
R(T )  2
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(iii)To find N(T)
x  3y  0
 10 y  0
 x   3 y 
 y   0 
  

N(T)= span of {}
N(T) = 0
(iv)To verify dimension theorem:
Here, dimension(n)= no.ofcolums of the matrix = 2
By the dimension theorem,
Dimension of T = Rank(T) + Nullity(T)
⟹2=2+0
⟹2=2
Hence verified dimension theorem.
(v)To check one-to-one and onto:
T is one to one if and only if N(T)=0
T is onto if and only if R(T)=dimension of T.
Here N(T)=0, so T is one to one
Here, Rank(T)= Dim(R3), that is 2=2
∴ T is onto.
Example 6
2
2
Let T: R → R be defined by T(a1 , a2 )= (a1 + a2 , 2a1 − a2 ). Prove that T is linear . Examine whether N(T)
and R(T) are subspaces of R2 and R2 respectively. Find the bases of N(T) and R(T).Compute nullity and rank
of T. Determine whether T is one-to-one or onto. (8)
Solution:
Given T: R2 → R2 be defined by T(a1 , a2 )= (a1 + a2 , 2a1 − a2 ).
(i)To prove T is linear
let 𝑢 = (a1 , a2 , ) ∈ R2 and 𝑣 = (a1 ′, a2 ′) ∈ R2
For c ∈ R
T(u  v)  T (a1 ,a 2 )  y(a 1 ',a 2 ') 
 T(a1  a1 ',a 2  a 2 ')
  (a1  a1 ')  (a 2  a 2 '), 2 (a1  a1 ')  (a2  a2 ')
  (a1  a 2 )  (a1 ' a 2 '),(2a1  a 2 )  (2a1 ' a2 ')
 (a1  a 2 , 2a1  a 2 )  (a1 ' a 2 ', 2a1 ' a 2 ')
T(u  v)  T(u)  T(v)
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T(c u)  T c(a1,a 2 ) 
 T (c a1 ,c a 2 ) 
 (c a1  c a 2 , 2 c a 1  ca 2 ) 
 c(a1  a 2 , 2a1  a 2 ) 
 cT(u)
This implies T is linear
(ii)To find basis for N(T):
Let N (T )  x V , T ( x)  0
let x  R 2 , such that x  a1 , a2 
Then T(x)=0
T (a1 , a2 )  0
 a1  a2 ,0,2a1  a2   (0,0,0)
 a1  a2  0,2a1  a2  0
 a1  a2 and 2a1  a2 ……. (3)
From above equation (3) we conclude  a1  0 and a2  0
Clearly the kernel (null space) consists of everything that is transformed to zero.
Here T (a1 , a2 )  (0,0,0) if a1  0 , a2  0
So any element in the null space of the from (0,0)
In general, N (T )  (0,0)  R 2 , T (0,0)  0
Thus, the null space of T is zero dimensional
Therefore, N(T) =0 with the basis 
(iii)To find basis for R(T):
Let R(T )  T ( x), x V 
In general, if T from V To W is a linear transformation,
  v1 , v 2 .....v n  is a basis for V then
R(T )  spanT  
R(T )  spanT v1 , T v2 .......T vn 
Here   (1,0), (0,1) is a standard basis for Therefore
R(T )  spanT  
R(T )  spanT (1,0), T (0,1)…….(4)
Now to find T (1,0), T(0,1)
T (1,0) = (1 + 0,0,2(1) - 0) = (1,0,2)
T (0,1) = (0 + 1,0,2(0))
Then substitute in (4)
R(T)=span ((1,0,2), (1,0, -1))
Also, clearly (1,0,2) and (1,0, -1) are linearly independent
So, ((1,0,2), (1,0, -1)) is a basis of R(T) and the rank R(T)=2
(iv)To determine T is one to one
W.k.t in general if T : V  W is a linear transformation and N(T)
then we have,T is one to one if and only if N(T)=0
T is onto if and only if R(T)=W
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and R(T) are null space and range of T,
Regulation: IFETCE R-2019
Academic Year: 2022-2023
Here N(T)=0, so T is one to one
And T is not onto, because T will never map all R3 also Rank(T)  Dim(R2)
Therefore, the given transformation is one to one but not onto.
(v) To verify dimension theorem
Here, dimension(n)= no.ofcolums of the matrix = 2
By the dimension theorem,
Dimension of T = Rank(T) + Nullity(T)
⟹2=2+0
⟹2=2
Hence verified dimension theorem.
Example 7
𝟐
𝟑
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 )= (𝐚𝟏 + 𝐚𝟐 , 𝟎, 𝟐𝐚𝟏 − 𝐚𝟐 ). Prove that T is linear . Find the bases of
N(T) and R(T).Compute nullity and rank of T. Determine whether T is one-to-one or onto.
Solution:
Given,
T(a1 , a2 )= (a1 + a2 , 0, 2a1 − a2 ).
(i)To proveT is linear
let 𝑢 = (a1 , a2 , )ℇ ∈and 𝑣 = (a1 ′, a2 ′) ∈ R2
For c ∈ R
 T(a1  a1 ',a 2  a 2 ')
  (a1  a1 ')  (a 2  a 2 '),0, 2(a1  a1 ')  (a 2  a 2 ') 
  (a1  a 2 )  (a1 ' a 2 '), 0, (2a1  a 2 )  (2a1 ' a 2 ') 
 (a1  a 2 , 0, 2a1  a 2 )  (a1 ' a 2 ', 0, 2a1 ' a 2 ')
T(u  v)  T(u)  T(v)
T(c u)  T  c(a1 , a 2 ) 
 T  (c a1 , c a 2 ) 
  (c a1  c a 2 , 0, 2 c a1  ca 2 ) 
  c(a1  a 2 , 0, 2a1  a 2 ) 
 cT(u)
This implies T is linear
(ii) To find the basis for N(T)
Let x  R 2 , such that x  (a1 , a2 )
Then
T (x) = 0
 T (a1 , a2 )  0
 (a1  a2 , 0, 2a1  a2 )  (0, 0, 0)
 a1  a2  0, 2a1  a2  0
 a1  a2 ...(4), 2a1  a2 ...(5)
From (4) & (5)  a1  0, a2  0
Clearly, the kernel (null space) consists of everything that is transformed to zero.
Here, T (a1 , a2 )  (0, 0, 0) if a1  0, a2  0
So, any element in the null space will be of the form (0, 0).
In general,
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N (T )  (0, 0) R 2 , T (0, 0)  0
Then, the null space of T is zero dimensional.
Therefore, N(T) = 0 with basis 
Given: T: R2 → R3
T(a1 , a2 )= (a1 + a2 , 0, 2a1 − a2 ). …(6)
R(T )  {T ( x) : x V }
(iii)To find a basis of range space R(T):
In general, if T from V to W is a linear transformation, and
  {v1 , v2 ,..., vn } is a
basis for V,
then, R(T) = span(T(β))
 span{T (v1 ), T ( v2 ),...,T (vn )}
Here, β = {(1, 0), (0, 1)} is a standard basis for R3, so,
R(T) = span(T(β))
= span (T(1, 0), T(0, 1))
Find T(1, 0) , T(0, 1) :
T (1, 0)  (1  0, 0, 2(1)  0) by (6)
T (0,1)  (0  1, 0, 2(0)  1) by (6)
= (1, 0, 2)
= (1, 0, - 1)
Then,
R(T) = span(T(β))
= span (T(1, 0), T(0, 1))
= span ((1, 0, 2), (1, 0, - 1))
Also, clearly (1, 0, 2) and (1, 0, - 1) are linearly independent.
So, ((1, 0, 2), (1, 0, - 1)) is a basis for R(T) and the rank of T.
Therefore, R(T)=2
(iv) To determine whether the given transformation is one-to-one or onto:
In general, if T : V W is a linear transformation, and N(T), and
R(T) are null space and range of T, then,
1.T is one-to-one if and only if N(T) = 0
2.T is onto if and only if R(T) = W.
Here, N(T) = 0, so T is not one-to-one by the above nullity proof.
Clearly, T is not onto, because, T will never map all of R3.
For example T will never map (1, 1, 1).
Also,rank (T) ≠ dim(R3)
2 ≠ 3
Therefore, the given transformation T is one-to-one, but not onto.
(v) To verify dimension theorem
Here, dimension(n)= no.of colums of the matrix = 2
By the dimension theorem,
Dimension of T = Rank(T) + Nullity(T)
⟹2=2+0
⟹2=2
Hence verified dimension theorem.
Example 8
0 
 f (1)  f (2)
Define operator T: 𝐏𝟐(R) → 𝐌𝟐×𝟐(R) by T ( f )  

0
f (0)  .Show that T is linear .Find dim

R(T) and basis for R(T).
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Regulation: IFETCE R-2019
Solution:
 f (1)  f (2)
T( f )  
0

Academic Year: 2022-2023
0 

f (0) 
To prove T is linear.
  f  g  (1)   f  g  (2)
(i)Let T ( f  g )  
0

0   g(1)  g (2) 0 
 f (1)  f (2)
Let T ( f  g )  


0
f (0)  
0
g(0) 

 T ( f  g )  T(f)  T(g)
Let f ,g M 22 ( R)and c  R

0
 f  g  (0) 
0 
 cf(1)  cf (2)
(ii)T (c f )  

0
cf(0) 

 f(1)  f (2) 0 
 c

0
f(0) 

 T (c f )  cT ( f )
Therefore T is linear.
Since   {1, x, x 2 } is a basis for P2 (R), we have
0 0
T (1)  

0 1
 1 0 
T (x)  

 0 0
 3 0 
T (x 2 )  

 0 0
R(T )  span(T ( ))  span({T(1),T(x), T ( x2 )})
  0 0   1 0   3 0   
 span  
,
,
  0 1   0 0   0 0   


  0 0   1 0   
 span  
,
.
  0 1   0 0   


Thus we have found a basis for R(T).
0 0
 1 0 
Therefore the basis of n R(T) are , 
 and 
.
0 1
 0 0
and so dim(R(T))=2.
Example 9
 a11
Define operator 𝐌𝟐×𝟑(F) → 𝐌𝟐×𝟐(F) defined by T 
 a21
linear. Find dim R(T) and basis for R(T).
Solution:
Given T: M2×3(F) → M2×2(F) defined by
29
a12
a22
a13   2a11  a12

a23  
0
a13  2a12 
 Show that T is
0

Regulation: IFETCE R-2019
Academic Year: 2022-2023
a
a   2a  a
a  2a12 
a
T  11 12 13    11 12 13

0
0

 a21 a22 a23  
b
b 
 a11 a12 a13 
b
 ; y   11 12 13   M 23 ( F ) and   F
Let x  
 a 21 a 22 a 23 
 b21 b22 b23 
a
a  b b
b 
a
x  y    11 12 13    11 12 13 
 a 21 a 22 a 23   b21 b22 b23 
Then,
 a a12 a13   b11 b12 b13 
  

  11
 a 21 a 22 a 23   b21 b22 b23 
 a  b a12  b12 a13  b13 

  11 11
 a 21  b21 a 22  b22 a 23  b23 
We have
 2(a11  b11 )  (a12  b12 ) (a13  b13 )  2(a12  b12 ) 
 ...(1)
T (x  y )  
0
0


Also
a  2a12   2b11  b12 b13  2b12 
 2a  a
  

T ( x)  T ( y )    11 12 13
0
0
0
0

 

 2 a11   a12  a13  2 a12   2b11  b12 b13  2b12 

 

0
0
0
0

 

 2a11  a12  2b11  b12 a13  2a12  b13  2b12 

 
0
0


 2(a11  b11 )  (a12  b12 ) (a13  b13 )  2(a12  b12 ) 
 ...(2)
 
0
0


From (1) & (2), we get
T ( x  y)  T ( x)  T ( y)
Hence, T is linear.
We know that the standard basis of the vector space M 23 ( F ) is
1 0 0   0 1 0   0 0 1 0 0 0 0 0 0
0 0 0
,
,
,
,
 and 

0 0 0 0 0 0 0 0 0 1 0 0   0 1 0 
 0 0 1
 
Since,  is a basis of M 23 ( F ) , we have
R(T )  span(T ( ))
 1
T
   0
 span 
  T  0
 1
 
0 1 
, 
0 0   

0 0 

0 1   
  2 0    1 2   0 1   0 0   0 0   0 0   
,
,
,
 ,
,
 
 span 
  0 0   0 0   0 0   0 0   0 0   0 0   


0 0 0
 ,T 
0 0   0
0 0  0
, T 
0 0   0
1 0 0
 ,T 
0 0   0
0 0 0
 ,T 
1 0   0
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  2 0   0 1   
, 
 
 span 
  0 0   0 0   


 2 0
0 1
 and 
 .
Therefore R(T) is spanned by two matrices 
 0 0
0 0
Since these two matrices are linearly independent, therefore the basis elements of
 2 0

R(T) are , 
 0 0
0 1
 .
and 
0 0
The number of basis for R(T) is 2
Therefore, dim R(T)=2.
Example 10
T ( f ( x))  xf ( x)  f (x) find basis for N T  and R T  . And hence verify the dimension theorem.
And hence verify the dimension theorem. Is T one-to-one?is T onto? Justify your answer.
Solution:
Given: T: P2 (R) → P3 (R) defined by T ( f ( x))  xf ( x)  f (x)
(i)To find bases for N(T):
Domain is P2 (R) .So ,its standard basis is 1, x, x 2 
Co-domain is P3 (R) .So ,its standard basis is 1, x, x 2 , x 3 
Suppose T ( f ( x))  0( x)
i.e., xf ( x)  f (x)  0
 f  (x)
 f ( x) 
x
We know that, if f is an nth degree polynomial, then f  is an (n-1)th degree polynomial and consequently,
 f  (x)
is an (n-2)th degree polynomial.
x
 f  (x)
so f ( x) 
implies that nth degree polynomial = (n-2)th degree polynomial, which is not possible in
x
P2 (R) except f ( x)  0
Therefore N(T) = {0} with  .
(ii)To find basis for R(T)
T a0  a1 x  a2 x 2  a1   a0  2a2  x  a1 x 2  a2 x 3


 a1 1  x 2   a0 x  a2  2 x  x3  ,
So the basis for R(T) is given by {1  x2 , x, 2 x  x3}
(clearly we see that it is a generator of {T(a0 + a1x + a2x2)} and is linear).
To check whether T is one to one:
We know that
T a0  a1 x  a2 x 2  a0 x  a1 x 2  a2 x3   a1  2a2 x 


 a1   a0  2a2  x  a1 x 2  a2 x3 ,
with f  x   a0  a1 x  a2 x 2 .
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f ( x)  N (T ) if T  f  x    0 for any x,
That is if a1 = 0 and a0 + 2a2 = 0 then a1  0, a0  0 and a2  0,
i.e. a0  a1  a2  0,
 N(T) = {0}.
 T is one-to-one (Since by a theorem that a linear transformation
T: V → W is one-one iff
N(T) = {0})
(iii) To check whether T is onto.:
By the Dimension Theorem, we deduce that
dim(R(T)) = dim(P2(R))−dim(N(T)) =dim(P2(R)) = 3.
However, dim (R) = 4,
Therefore R(T )  P3 (T )
 T is not onto.
Here, dimension(n)= no.ofcolums of the matrix = 3
By the dimension theorem,
Dimension of T = Rank(T) + Nullity(T)
⟹3=3+0
⟹3=3
Hence verified dimension theorem.
Example 11
𝟑
𝟐
Let T: 𝐑 → 𝐑 be defined by T(𝐚𝟏 , 𝐚𝟐 , 𝐚𝟑 )= (𝐚𝟏 − 𝐚𝟐 , 𝟐𝐚𝟑 ). Prove that T is linear. Examine whether N(T)
and R(T) are subspaces of 𝐑𝟑 𝐚𝐧𝐝 𝐑𝟐 respectively. Find the bases of N(T) and R(T).Compute nullity and
rank of T. Determine whether T is one-to-one or onto and hence verify the dimension theorem.
Solution:
(i)To prove T is linear
let 𝑢 = (a1 , a2 , a3 )ϵ R3 and 𝑣 = (a1 ′, a2 ′, a3 ′)ϵ R2
For c ∈ R
T(u  v)  T (a1 ,a 2 ,a 3 )  (a 1 ',a 2 ',a 3 ') 
 T(a1  a1 ',a 2  a 2 ',a 3  a 3 ')
  a1  a1 '    a 2  a 2 '  , 2  a 3  a 3 ' 
[since T(a1 , a2 , a3 )= (a1 − a2 , 2a3 ). ]
  a1  a 2 , 2a 3    a1 ' a 2 '  , 2a 3 ' 
  a1  a 2 , 2a 3    a1 ' a 2 ', 2a 3 ' 
T(u  v)  T(u)  T(v)
This implies T is linear
T(c u)  T c(a1 ,a 2 ,a 3 ) 
 T (c a1 ,c a 2 ,ca 3 ) 
 (c a1  c a 2 , 2 c a 3 )  This implies T is linear
 c(a1  a 2 , 2a 3 ) 
 cT(u)
(ii)To examine whether N(T) and R(T) are subspaces of R3 and R2
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respectively:
Since T(0) = 0, 0 ∈ N(T); so N(T) is a nonempty subset of V.
Let u, v ∈ N(T); α a scalar.
Then T(u) = T(v) = 0.
Consequently, T(u + αv) = T(u) + αT(v) = 0.
That is, u + αv ∈ N(T). Therefore, N(T) is a subspace of V.
Again, since T(0) = 0, R(T) is a nonempty subset of W.
Let x, y ∈ R(T); β a scalar.
Then there exist u, v ∈ V such that x = T(u) and y = T(v).
Now, T(u+βv) = T(u)+βT(v) = x+βy.
That is, x + βy ∈ R(T).
Therefore, R(T) is a subspace of W.
(iii)To find N(T) and a basis of N(T):
let a1 , a2 , a3 ∈ R3 be such that
T(a1 , a2 , a3 )= (a1 − a2 , 2a3 ) = (0,0).
Then a1 − a2 = 0 and 2a3 = 0. or a1 = a2 and a3 = 0
Therefore N(T)  {(a 2 ,a 2 ,0),a 2  R} and dim(N(T)) = 1,
A basis of N(T) is given by {(1, 1, 0)},
(iv)To find R(T) and Basis of R(T):
Also by the Dimension Theorem, we obtained that ,
dim(R(T)) = 3−dim(N(T)) = 2.
This implies that dimension of range of T coincides with dimension of R2, i.e. R(T) = R2,
A basis of R(T) is any basis of R2, for instance of standard basis
{(1, 0), (0, 1)}.
(v)To find Nullity(T):
The nullity of T, denoted nullity (T), is defined as dim(N(T))
Therefore nullity (T)= dim(N(T)) = 1.
(vi)To find Rank (T):
The rank of T, denoted rank (T), is defined as dim(R(T)) = 2
To Prove T is one to one:
By the definition we know that there are distinct vectors a, b in R 3
such that T(a)  T(b )
with a  b
Indeed T (2, 2, 1) = T(1, 1, 1) = (0, 2),
therefore T is not one-to-one.
Also, since N(T)  {0} we see that T is not one-to-one.
To Prove T is onto:
let (x, y)  R 2 be arbitrary, and find( a1 , a2 , a3 ), if any, such that
T( a1 , a2 , a3 )= (x, y).
This would imply a1 − a2 = x and 2a3 = y.
We see that (x, y) is always in the image of T, by T (a, a − x, y/2) = (x, y).
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This implies T is onto
(vii)To verify the dimension theorem:
The dimension theorem states that ”If V is finite dimensional, then N(T)  R(T)  dim(V)
Dimension of V= dim(V)  R 3  3
Dimension of N(T)  1
Dimension of R(T)  2
Therefore N(T)  R(T)  1  2  3
Hence dimension theorem is verified
3.4 EIGEN VALUES AND EIGEN VECTORS
CHARACTERISTIC POLYNOMIAL
Let T be a linear operator on an n-dimensional vector space V with ordered basis β. We
define the characteristic polynomial f(t) of T to be the characteristic polynomial of A  [T ]  .
That is, f (t )  det A  tI n 
Example:
 5 4 
A

 1 2 
The characteristic polynomial of A is A   I  0
  2  s1  s2  0
s1 Sum of the main diagonal element.
s1 = -5-2 = -7
s2  A  10  4  6
2  7  6  0
METHODS TO FIND CHARACTERISTIC POLYNOMIAL
 a11 a12 a13 


Method I: Let A  a 21 a 22 a 23  be 3 x 3 matrix, then the characteristic equation of A is given by
 a 31 a 32 a 33 
A  I  0
Method II
Case (i) If A is a 3 x 3 matrix, the then the characteristic equation of A is given by λ -S1λ +S2 λ-S3 =0
3
2
S1 =sum of the main diagonal elements = a11  a 22  a 33
S2 =sum of the minors of the main diagonal elements

a11 a12 a 22

a 21 a 22 a 32
a 23 a11 a13

a 33 a 31 a 33
S3 =determinant of the matrix A
Case (ii) If A be a 2 x 2 matrix, the then the characteristic equation of A is given by λ -S1λ+S2 =0
2
S1 =sum of the main diagonal elements = a11  a 22
S2 =determinant of the matrix A
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EIGEN VALUES
Let T be a linear operator on a vector space V. A non zero vector v  V is called an Eigen vector of T
if there exists a scalar  such that T (v)  v . The scalar  is called the eigen value corresponding to the
eigenvector v.
EIGEN VECTORS
Let A be in M nn (F ) . A nonzero vector v  F n is called an eigenvector of A if v is an eigen value of LA ; that
is, if Av  v for some scalar  . The scalar  is called the eigen value of A corresponding to the
eigenvector v.
EIGEN SPACE.
Let T be a linear operator on a vector space V, and let  be an eigen value of T.
Define E  {x V ::T ( x)  x}  N (T  I V ) .
The set E  is called the eigen space of T corresponding to the eigen value  . Analogously, we define the
eigen space of a square matrix A to be the eigen space of LA .
Worked Examples 3.4(A)
Example 1
If 𝛌 is an eigen value of a linear operator T: 𝐕 → 𝐕 then compute the eigen value of the linear Operator
kT.
Solution:
Given, 𝜆 be the eigen value of T.
To find the eigen value of kT.
Let B =kT, then (B  I )  0
 (kT   I )  0
 (T 

I)  0
k
𝛼
This equation shows that is a eigen value of T. 𝜆 is a eigen
𝑘
value of T.
Therefore, 


k
   k
Eigen value of the linear operator kT is k𝜆.
Example 2
Show that similar matrix are same eigen value.
Solution:
Let A and B be the similar matrices.
There exists some invertible matrix P such that P 1 AP  B.
If Av  v , we have
B( P 1v)  P 1v.
And is Bv  v , we have A( pv)  Pv.
Hence A and B have same eigen values.
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Example 3
Prove that similar matrix are same characteristic polynomial.
Solution:
Let A and B are two similar matrices, and p(t) denote kth-degree polynomial.
To show P(A) and P(B) are similar, so
B  P1 AP, for some(n n)invertyible matrix P.
Let det( B)  det( P 1 AP)  det(P 1 ) det( A) det( P)
 det( A) det(P 1 ) det( P)  det( A) det(PP 1 )
 det( A) det(I)  det( A)
now P( B)  B   I
 P 1 AP   I
 P 1 AP   ( P 1 AP )
 P 1 ( A   I ) P
Since determinant of B-λI is same as A-λI.
Therefore, P(A) and P(B) are similar.
Therefore, similar matrix are same characteristic polynomial.
Example 4
Let 𝝀 be an eigen value of an invertible operator T on a vector space V (F). Prove that 𝝀−𝟏 is an eigen
value of 𝑻−𝟏
Solution:
Suppose T is invertible and λ is an eigenvalue of T.
Let λ ≠ 0. Then there exists a nonzero vector v so that
T v = λv.
Applying T-1 to both sides and dividing the resulting equation by λ, we have
and it follows that
1

1

v  T 1v
is an eigenvalue of T-1 .
The converse is proven exactly the same way.
Example 5
The trace and determinate of a 2 ×2 matrix are known to be – 2 and – 35 respectively. Find the eigen values of the
matrix.
Solution:
Gιven, 1  2  2 .........(1)
12  35
(1  2 )2  (1  2 ) 2  412  4  140  144
 1  2  12
take 1  2  12 .......(2)
Solving (1) and (2) we get,
  7 and 2  5.
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Example 6
𝟐 𝟑
] .Then the eigen values of A are 4 and 8. Find the values of X and Y?
If A=[
𝐱 𝐲
Solution:
Given, 1  4, 2  8
We know that, 1  2  2  y
2  y  8  4  12
 y  10
we know that, 12  det A  2 y  3 x
 2 y  3x  (8)(4)  32
(2)(10)  32
x
 4
3
Therefore, x=-4 and y=10.
Example 7
Let A, B, C, D be n × n matrices, each with non-zero determinant, If ABCD = 1 then find 𝑩−𝟏 ?
Solution:
ABCD=1
⟹ ABCDD-1C-1 = D-1C-1
⟹ AB1 = D-1C-1
⟹ A-1AB = A-1 D-1C-1
⟹ B = (CDA)-1
⟹ B-1 = CDA
Example 8
−𝟏 𝟑 𝟓
Find the eigen values of A=[−𝟑 −𝟏 𝟔]
𝟎
𝟎 𝟑
Solution:
The characteristic equation of A is
A  I  0
i, e)  3  S1 2  S2   S3  0 where
S1  sum of the main diagonal elements
S1  1  1  3  1
S 2  sum of the minors of the main diagonl elements \

1 6 1 5 1 3


 ( 3)  ( 3)  (10)  4
0 3 0 3 3  1
1 3 5
S 3  A  3  1 6
0 0 3
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 1(3  0)  3(9  0)  3  27  30
The characteristic equation of A is
 3   2  4  30  0
The eigen values are 3, -1+3i and -1-3i
Example 9
−𝟒
Find the eigen vectors of A=[
𝟒
Solution:
𝟐
]
𝟑
(i) S1  sum of the main diagonal elements
S1  4  3  1
4 2
 20
4 3
The characteristic equation of A is
S2  A 
 2    20  0
The eigen values are
−1±√81
.
2
Worked Examples 3.4(B)
Theorem 1
Let A 𝛜 𝐌𝐧×𝐧(F)Then a scalar 𝛌 is an eigenvalue of Aif and only if det (A - 𝛌𝐈𝐧 )=0.(4)
Proof:
 is an eigen value of A
  v  0 such that Av  v
 v  0, such that Av  v  0
 v  0, such that  A   I n  v  0
 det  A   I n   0
Theorem 2
Let T be a linear operator on a vector space V. and let𝛌𝟏 , 𝛌𝟐,……𝛌𝐤 be distinct eigen values ot'T. For each
i = 1,2,... ,k. let Sbe a finite linearly independent subset of the eigenspace E𝛌𝐢 , Then Prove that
S=𝐒𝟏 𝐔𝐒𝟐 𝐔,…… 𝐔𝐒𝐤 is a linearly independent subset of V.
Proof:
Suppose that for each i, Si  vi1 , vi2 ,......, vin i  . Then S  vij :1  j  n i , and1  i  k .
Consider any scalars a ij  such that
k
ni
 a v
ij ij
i 1 j1
0.
ni
For each i, let w i   a ij vij . Then w i  E  for each i and w1  w 2  ......  w k  0 .
j1
i
Therefore, by the lemma, w i  0 for all i. But such Si is linearly independent and hence a ij  0 for all i.
Thus we conclude that S is linearly independent.
Theorem 3
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Prove that the characteristic polynomial of any diagonalizable linear operator splits.
Proof:
Let T be a diagonalizable linear operator on the dimensional vector space V and let  be an ordered basis for V
such that  T   D is a diagonal matrix.
 1 0

0 2
Suppose that D  


0 0
0

0
and let f(t) be the characteristic polynomial of T. Then


n 
0
 1  t

0
2  t
f  t   det  D  tI   det 


0
 0
  1  t  2  t 
  1
n
0 

0 


n  t 
 n  t 
 t  1  t  2   t  n 
From this it is clear that if T is a diagonalizable linear operator on an n-dimensional vector space that fails
to have distinct eigen values, then the characteristic polynomial of T must have repeated zeros.
Theorem 4
Let T be a linear operator on a vector space V, and let 𝛌𝟏 , 𝛌𝟐,……𝛌𝐤 be distinct eigenvalues of T.
If 𝐯𝟏 , 𝐯𝟐 ,……𝐯𝐤 are eigenvectors of T such that Xi corresponds to 𝐯𝐢 (1< i < k), then{𝐯𝟏 , 𝐯𝟐 ,……𝐯𝐤 } is
linearly independent.
Proof:
The proof is by mathematical induction on k.
Suppose that k =1. Then v1  0 since v1 is an eigen vector and hence v1 is linearly independent. Now
assume that the theorem holds for (k – 1) distinct eigen values, where k  1  1 and that we have k eigen values
v1 , v2 ,......, vk corresponding to the distinct eigen values 1 , 2 ,......, k . We wish to show that v1 , v 2 ,......, v k  is
linearly independent.
Suppose that a1 ,a 2 ,......,a k are scalars such that a1v1  a 2 v 2  ......  a k v k  0
1
Applying T  k to both sides of (1), we obtain
a1  1  k  v1  a 2  2  k  v 2  ......  a k 1  k 1  k  v k 1  0
By induction hypothesis v1 , v 2 ,......, v k 1 is linearly independent and hence
a1  1  k   a 2  2  k   ......  a k 1  k 1  k   0
Since 1 , 2 ,......, k are distinct, it follows that  i  k   0 for 1  i  k  1 .
So a1  a 2  ......  a k 1  0 and (1) therefore reduces to a k vk  0 . But v k  0 and therefore a k  0 .
Consequently a1  a 2  ......  a k  0 and it follows that v1 , v 2 ,......, v k  is linearly independent.
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Theorem 5
Let A be an n x n matrix that is similar to an upper triangular matrix and has the distinct eigenvalues
𝛌𝟏 , 𝛌𝟐 ,……𝛌𝐤 with corresponding multiplicities 𝐦𝟏 , 𝐦𝟐 ,……𝐦𝐤 . Prove the following statements
(i) tra(A)=∑𝐤𝐢=𝟏 𝐦𝐢 𝛌𝐢 (ii) det(A)=𝛌𝟏 𝐦𝟏𝛌𝟐 𝐦𝟐 … … 𝛌𝐤 𝐦𝐤
Proof:
(i) Let M  M n ( F ) be upper triangular matrix,
i.e, M  ( mij ),
where mij  0 whenever i > j. Since the eigen values of M are precisely the roots of its characteristic
polynomial, the roots of

n
p(t )  det( M  tI n )   (mii  t )
i 1
are precisely the eigenvalues. ( * is form the fact the M  tI n is also an upper triangular matrix and determinant
of an upper triangular matrix is just the product of its diagonal entries).
Therefore the eigen values of M are mii ,1  i  n . This completes the proof.
(ii) det( A)  (1 ) m (2 ) m ...(k ) m .
1
k
2
The characteristic polynomial of a is (  1 ) m (  2 ) m ...(  k ) m .
1
2
k
Let B be the upper triangular matrix similar to A, with diagonal entries b1 , b2 ,..., bn .
Because the determinant of an upper triangular matrix is the product of its diagonal entries det( B)  b1b2 ...bn ,
and the characteristic polynomial of B is (  b1 )(  b2 )...(  bn ) Because A and B are similar, they have the
same characteristic polynomial, so
(  1 ) m (  2 ) m ...(  k ) m  (  b1 )(  b2 )...(  bn )
.
1
k
2
Therefore, b1b2 ...bn  (1 ) m (2 ) m ...(k ) m .
1
k
2
Because A and B are similar, they have the same determinant, so
det(A)  det( B)  b1b2 ...bn  (1 ) m (2 ) m ...(k ) m
1
2
k
Theorem 6
Let T be a linear operator on a finite-dimensional vector space V, and let X be an eigen value of T having
multiplicity m Then 1 ≤dim(𝐄𝛌) ≤m.
Proof:
Choose an ordered basis {v1 , v 2 ,..., v p } for E , extended it to an ordered basis
  {v1 , v 2 ,..., v p , v p 1 ,..., v n }
A  [T ]

for V, and let
.
Observe that vi is an eigenvector of T corresponding to  , and therefore
 I B 
A p

 O C
By the characteristic polynomial of T is,
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f (t )  det( A  tI n )
B 
 (  t ) I p
 det 

C  tI n  p 
 O
 det((  t ) I p ) det (C  tI n  p )
 (  t ) p g (t )
Where g(t) is a polynomial.
Thus (  t ) p is a factor of f(t), and hence the multiplicity of  is at least p. But dim(Eλ) = p, and so dim(Eλ)
≤m.
Worked Examples 3.4(C)
Example 1
For the linear operator T: 𝐏𝟐(R) → 𝐏𝟐(R) defined as T ( f ( x))  2f(x)  (2 x  1) f ( x). Find the eigen values
of T and an ordered basis B for P2(R) such that matrix of the given transformation with respect to the
new resultant basis B is a diagonal matrix.
Solution:
T [f(x)]  2f(x)  (2 x  1)f (x)
Let   1, x, x 2 
T (1)  2  (2 x  1)(0)  2
T(x)  2 x  (2 x  1)(1)  4 x  1
T(x 2 )  2 x2  (2 x  1)(2 x)  2 x2  4 x2  2 x  6 x2  2 x
T 
2 1 0
  0 4 2 
 0 0 6 
The characteristic equation of A is T    I  0
  3  s1 2  s2   s3  0
s1 Sum of the main diagonal element.
s1 = 12
s2 Sum of the minors of the main diagonal elements.
s2 
4 2 2 0 2 1


 24  12  8  44
0 6 0 6 0 4
s3  A  48
 3 12 2  44  48  0
  6, 2, 4
1
0   x  0 
2  
 0
4
2   y   0

0
6     z  0
 0
Put   6
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 4 1 0   x 

 
 0 2 2  y   0
 0
0 0   z 
4 x  y  0 z  0
0x  2 y  2z  0
x
1 0
2 2

y

0 4
2 0
z
4 1
0 2
x y z
 
2 8 8
x y z
 
1 4 4
 x  1 
The Eigenvectors vector is  y    4 
 z   4 
Put   4
 2 1 1   x 

 
 0 0 2  y   0
 0 0 2  z 
2 x  y  z  0
0x  0 y  2z  0
0x  0 y  2z  0
x
1 1
0 2

y

1 2
2 0
z
2 1
0 0
x y z
 
2 4 0
x y z
 
1 2 0
 x  1 
The Eigenvectors vector is  y    2 
 z  0 
Put   2
0 1 0   x 

 
0 2 2  y   0
0 0 4  z 
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0 x  y  0 z  0
0x  2 y  z  0
0x  0 y  4z  0
x

y
1 0
0 0
2 1
1 0
x y z
 
1 0 0

z
0 1
0 2
 x  1 
The Eigenvectors vector is  y   0 
 z  0 
Diagonalization  D   Q 1 AQ
0 0
1 
1 1 1 




1

1
1
Where Q   4 2 0  and Q  0
 
2
2

 4 0 0 
1 1
1 

2
4 

0
1 
1 1 1   2 1 0  0






1
1
 D    4 2 0 0 4 2 0 2  2 
 4 0 0  0 0 6  1 1
1 

2
4 
6 0 0 
 D   0 4 0
0 0 2 
Example 2
For the linear operator T: 𝐏𝟐(R) → 𝐏𝟐(R) defined as T ( f ( x))  f(x)  xf ( x)  f (x) find the eigen values of T
and an ordered basis B for P2(R) such that matrix of the given transformation with respect to the new
resultant basis B is a diagonal matrix.
Solution:
T ( f ( x))  f(x)  xf ( x)  f (x)
Let   1, x, x 2 
T (1)  1  (x)(0)  1
T(x)  x  x(1)  0  2 x
T(x 2 )  x2  ( x)2 x  2  3x2  2
T 
1 0 2 
 0 2 0 
0 0 3 
The characteristic equation of A is T    I  0
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  3  s1 2  s2   s3  0
s1 Sum of the main diagonal element.
s1 = 6
s2 Sum of the minors of the main diagonal elements.
2 0 1 2 1 0


 6  3  2  11
0 3 0 3 0 2
s2 
s3  A  6
 3  6 2  11  6  0
  3,1, 2
0
2   x  0
1  
 0
2
0   y   0

 0
0
3     z  0
Put   3
 2 0 2  x 

 
 0 1 0   y   0
 0 0 0   z 
2 x  0 y  2 z  0
0x  2 y  0z  0
x
0 2
2 0

y

2 2
0 0
z
2 0
0 2
x y z
 
4 0 4
x y z
 
1 0 1
 x  1 
The Eigenvectors vector is  y   0 
 z  1 
Put   1
0 0 2  x 

 
0 1 0   y   0
0 0 2  z 
0x  0 y  2z  0
0x  y  0z  0
0x  0 y  2z  0
x
0 1
1 0

y
1 0
0 0

z
0 0
0 1
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Regulation: IFETCE R-2019
Academic Year: 2022-2023
x y z
 
1 0 0
x y z
 
1 0 0
 x  1 
The Eigenvectors vector is  y   0 
 z  0 
Put   2
 1 0 2  x 

 
 0 0 0  y   0
 0 0 1   z 
 x  0 y  2z  0
0x  0 y  z  0
x

0 2
0 1
x y z
 
0 1 0
y

2 1
1 0
z
1 0
0 0
 x  0
The Eigenvectors vector is  y   1 
 z  0 
Diagonalization  D   Q 1 AQ
1 1 0 
Where Q   0 0 1 
1 0 0 
1 1 0  1
 D   0 0 1  0
1 0 0 0
3 0 0 
 D   0 1 0
0 0 2 
0 0 1 
and Q  1 0 1
0 1 0 
0 2  0 0 1 
2 0  1 0 1
0 3  0 1 0 
1
Example 3
For the linear operator T: 𝐏𝟐(R) → 𝐏𝟐(R) defined as T ( f ( x))  f(0)  f (1)(x  x 2 ) Find the eigen values
of T and an ordered basis B for P2(R) such that matrix of the given transformation with respect to the
new resultant basis B is a diagonal matrix.
Solution:
T [f(x)]  f(0)  f (1)(x  x 2 ) f (x), P2 ( R)
Let   1, x, x 2 
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Regulation: IFETCE R-2019
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T (1)  1  1(x  x2 )  1  x  x2
T(x)  0  (x  x2 )  x  x2
T(x 2 )  02  ( x  x2 )  x  x2
T 
1 0 0 
 1 1 1 
1 1 1 
The characteristic equation of A is T    I  0
  3  s1 2  s2   s3  0
s1 Sum of the main diagonal element.
s1 = 2
s2 Sum of the minors of the main diagonal elements.
s2 
1 1 1 0 1 0


 0 11  2
1 1 1 1 1 1
s3  A  0
 3  3 2  12  0
  2,0,1
0
0   x  0
1  
 1 1 
1   y   0

 1
1 1     z  0
Put   0
1 0 0  x 

 
1 1 1   y   0
1 1 1   z 
x  0 y  0z  0
x y z 0
x y z 0
x
0 0
1 1

y
0 1
1 1

z
1 0
1 1
x y z


0 1 1
x y z
 
0 1 1
 x  0 
The Eigenvectors vector is  y   1 
 z   1
Put   1
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Regulation: IFETCE R-2019
Academic Year: 2022-2023
0 0 0  x 

 
1 0 1   y   0
1 1 0  z 
0x  0 y  2z  0
x  0y  z  0
x  y  0z  0
x
0 1
1 0

y
1 1
0 1

z
1 0
1 1
x y z
 
1 1 1
x y
z


1 1 1
 x  1 
The Eigenvectors vector is  y    1
 z   1
Put   2
 1 0 0   x 

 
 1 1 1   y   0
 1 1 1  z 
 x  0 y  0z  0
x y z 0
x yz 0
x

0 0
1 1
x y z
 
0 1 1
y

0 1
1 1
z
1 0
1 1
 x  0
The Eigenvectors vector is  y   1 
 z  1 
1
Diagonalization  D   Q AQ
0
 0 1 0

Where Q   1 1 1  and Q 1  1



 1 1 1 
1
1
2
0
1
2
1 
2
0 

1 
2
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Regulation: IFETCE R-2019
 D
 D

 0 1 0  1 0 0  0
  1 1 1  1 1 1  1

 1 1 1  1 1 1  1

0 0 0 
 0 1 0 
0 0 2 
Academic Year: 2022-2023
1
2
0
1
2
1 
2
0 

1 
2
Example 4
For the linear operator T: 𝐏𝟐(R) → 𝐏𝟐(R) defined as T ( f ( x))  f(x)  ( x  1) f ( x). Find the eigen valuesof T
and an ordered basis B for P2(R) such that matrix of the given transformation with respect to the new
resultant basis B is a diagonal matrix.
Solution:
T ( f ( x))  f(x)  xf ( x)  f (x)
Let   1, x, x 2 
T (1)  1  (x  1)(0)  1
T(x)  x  (x  1)  2 x  1
T(x 2 )  x2  ( x  1)2 x  x2  2 x2  2 x  3x2  2 x
T 
1 1 0 
 0 2 2 
0 0 3 
The characteristic equation of A is T    I  0
  3  s1 2  s2   s3  0
s1 Sum of the main diagonal element.
s1 = 26
s2 Sum of the minors of the main diagonal elements.
s2 
2 2 1 0 1 1


 6  3  2  11
0 3 0 3 0 2
s3  A  6
 3  6 2  11  6  0
  3,1, 2
1
0   x  0
1  
 0
2
2   y   0

0
3     z  0
 0
Put   1
0 1 0   x 

 
0 1 2   y   0
0 0 2  z 
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Regulation: IFETCE R-2019
Academic Year: 2022-2023
0x  y  0z  0
0x  y  z  0
0x  0 y  2z  0
x
y

1 0
0 0
1 1
1 0
x y z
 
1 0 0

z
0 1
0 1
 x  1 
The Eigenvectors vector is  y    0 
 z   0 
Put   2
 1 1 0   x 

 
 0 0 2  y   0
 0 0 1   z 
 x  y  0z  0
0x  0 y  2z  0
0x  0 y  z  0
x

1 0
0 1
x y z
 
1 1 0
y

0 1
1 0
z
1 1
0 0
 x 1 
The Eigenvectors vector is  y    1
 z   0
Put   3
 2 1 0   x 

 
 0 1 2  y   0
 0 0 0   z 
2 x  y  0 z  0
0x  y  2z  0
x yz 0
x

1 0
1 2
x y z
 
2 4 2
y

0 2
2 0
z
2 1
0 1
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Regulation: IFETCE R-2019
Academic Year: 2022-2023
 x  1 
The Eigenvectors vector is  y    2 
 z  1 
Diagonalization  D   Q 1 AQ
1 1 1 
Where Q   0 1 2 
 0 0 1 
1 1 1  1
 D   0 1 2 0
0 0 1  0
1 0 0 
 D   0 2 0
0 0 3
0 1 1 
and Q  0 1 2 
0 0 1 
1 0  0 1 1 
2 2 0 1 2
0 3 0 0 1 
1
50