Linear transformation, Formulas and forms for Mathematics. Daffodil International University

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In mathematics, a linear map (also called a linear mapping, linear transformation or, in some contexts, linear function) is a mapping V → W between two modules (including vector spaces) that preserves (in the sense defin...
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Linear Transformation

Linear Transformation: Let V and U be two vector spaces over the same scalar field F . Then

the mapping :T V U is called a linear transformation or linear mapping if it satisfies the conditions:

1. For any ,u v V ,      T u v T u T v  

2. For any a F and v V ,    T av aT v

Alternatively, A mapping : n mT R R defined by  T v Mv , where M is an m n real matrix; is called a linear transformation.

Transformation matrix: If a linear mapping : n mT R R defined by  T v Mv , where M

is an m n real matrix; then the transformation matrix is  T M .

Problem-01: Test whether the mapping 3 2:T R R defined by    , , 2 , 5T x y z x y y z   is linear or not.

Solution: We have, 3 2:T R R

And    , , 2 , 5T x y z x y y z  

So, we must need a 2 3 matrix consisting of real numbers only.

The given transformation can be written as,

2

5

x x y

T y y z

z

    

        

2 0.

0. 5

x y z

x y z

     

  

1 2 0

0 1 5

x

y

z

    

      

 

Mv

Since,   1 2 0

0 1 5 T M

     

  is a matrix of order 2 3 , so we conclude that T is a linear

transformation.

Problem-02: Test whether the mapping 3 2:T R R defined by    , , 5 ,T x y z x y z yz   is linear or not.

Solution: We have, 3 2:T R R

So, we must need a 2 3 matrix consisting of real numbers only.

The given transformation can be written as,

5 x

x y z T y

yz z

     

      

 

1 1 5

0 0

x

y z

z

           

 

Since the matrix contains variable so M does not exist. We conclude that T is not a linear transformation.

Problem-03: Check the followings for linear transformations and find 1T  if it exists. Also verify for

a non-zero vector.

1) 3 3:T R R is given by    , , 2 , 2 ,T x y z x y z z x   

2) 3 2:T R R is given by    , , 2 ,6 5T x y z x y z x   

3) 3 2:T R R is given by    , , 23 2 5 ,0T x y z x y z  

4) 3 4:T R R is given by    , , 8 , ,4 5 ,5T x y z x z z x  

5) 3 3:T R R is given by    , , 2 ,4 ,3T x y z x y x y z  

6) 5 3:T R R is given by    , , , , ,0,5T p q r s t s t 

7) 3 4:T R R is given by    , , ,0, ,T p q r p q z y  

8) 3 3:T R R is given by    , , 3 , 2 4 3 ,5 4 2T p q r x y x y z x y z      

Solution:1). We have, 3 3:T R R

And    , , 2 , 2 ,T x y z x y z z x   

So, we must need a 3 3 matrix consisting of real numbers only.

The given transformation can be written as,

2

2

x x y

T y z

z z x

                  

2 0.

0. 0. 2

0.

x y z

x y z

x y z

            

1 2 0

0 0 2

1 0 1

x

y

z

               

Mv

Since,  

1 2 0

0 0 2

1 0 1

T M

   

       

is a matrix of order 3 3 , so we conclude that T is a linear

transformation.

Now,     

1 2 0

0 0 2 0 2 0 2 0 4

1 0 1

T

       

Since,   0T  , so   1

T

exists.

Hence, 1T  exists and   1

T

will be the transformation matrix.

The cofactors are, 11 12 13 21 22 23 31 32 330, 2, 0, 2, 1, 2, 4, 2, 0A A A A A A A A A          

The matrix of cofactors is,

0 2 0

2 1 2

4 2 0

   

     

   0 2 0 0 2 4

. 2 1 2 2 1 2

4 2 0 0 2 0

t

       

              

      

1

10 1 20 2 4

. 1 1 1 12 1 2 2 4 24

0 2 0 10 0 2

T

                       

 

The inverse transformation is,

1

10 1 2

1 1 1 2 4 2

10 0 2

a a

T b b

c c

                        

 

2

2 4 2

2

b c

a b c

b

   

               

 1 , , , , 2 2 4 2 2

b a b c b T a b c c

         

 

Verification: Verify for the non-zero vector  1,0,1 .

     1,0,1 1 2.0, 2.1,1 1 1, 2,2T      

And      1 2 22 1 21, 2,2 2, ,

2 2 4 2 2 T

          

   1,0,1 (Verified).

2). We have, 3 2:T R R

And    , , 2 ,6 5T x y z x y z x   

So, we must need a 2 3 matrix consisting of real numbers only

The given transformation can be written as,

2

6 5

x x y

T y z x

z

     

        

2 0.

5 0. 6

x y z

x y z

      

  

1 2 0

5 0 6

x

y

z

     

      

 

Mv

Since,   1 2 0

5 0 6 T M

      

  is a matrix of order 2 3 , so we conclude that T is a linear

transformation.

Again, since the transformation matrix  T is not square, so  T is not possible. So that 1T  does not exist. (Ans.)

3). We have, 3 2:T R R

And    , , 23 2 5 ,0T x y z x y z  

So, we must need a 2 3 matrix consisting of real numbers only.

The given transformation can be written as,

23 2 5

0

x x y z

T y

z

     

      

 

23 2 5

0. 0. 0.

x y z

x y z

     

  

23 2 5

0 0 0

x

y

z

    

      

 

Mv

Since,   23 2 5

0 0 0 T M

     

  is a matrix of order 2 3 , so we conclude that T is a linear

transformation.

Again, since the transformation matrix  T is not square, so  T is not possible. So that 1T  does not exist. (Ans.)

4). We have, 3 4:T R R

And    , , 8 , ,4 5 ,5T x y z x z z x  

So, we must need a 4 3 matrix consisting of real numbers only.

The given transformation can be written as,

8

4 5

5

x x

z T y

z x z

     

            

8 0. 0.

0. 0.

5 0. 4

5

x y z

x y z

x y z

       

      

Since, the transformation matrix  T M does not exist. So, we conclude that T is not a linear transformation. (Ans.)

5). We have, 3 3:T R R

And    , , 2 ,4 ,3T x y z x y x y z  

So, we must need a 3 3 matrix consisting of real numbers only.

The given transformation can be written as,

2

4

3

x x y

T y x y

z z

                  

2 0.

4 0.

0. 0. 3

x y z

x y z

x y z

            

2 1 0

4 1 0

0 0 3

x

y

z

               

Mv

Since,  

2 1 0

4 1 0

0 0 3

T M

   

       

is a matrix of order 3 3 , so we conclude that T is a linear

transformation. (Ans.)

Now,      

2 1 0

4 1 0 2 3 0 1 12 0 0 18

0 0 3

T          

Since,   0T  , so   1

T

exists.

Hence, 1T  exists and   1

T

will be the transformation matrix.

The cofactors are, 11 12 13 21 22 23 31 323, 12, 0, 3, 6, 0, 0, 0,A A A A A A A A          

33 6A   .

The matrix of cofactors is,

3 12 0

3 6 0

0 0 6

         

   3 12 0 3 3 0

. 3 6 0 12 6 0

0 0 6 0 0 6

t

          

               

      

1

1 1 0 6 63 3 0

. 1 2 112 6 0 0 3 318

0 0 6 10 0 3

T

                          

 

The inverse transformation is,

1

1 1 0 6 6

2 1 0 3 3

10 0 3

a a

T b b

c c

                         

 

6 6

2

3 3

3

a b

a b

c

   

              

     1 2, , , ,

6 3 3

a b a b c T a b c

     

 

Verification: Verify for the non-zero vector  1,0,1 .

     1,0,1 2.1 0,4.1 0,3.1 2,4,3T    

And      

 1 2 4 2.2 4 3

2,4,3 , , 1,0,1 6 3 3

T    

    

(verified)

6). We have, 5 3:T R R

And    , , , , ,0,5T p q r s t s t 

So, we must need a 3 5 matrix consisting of real numbers only.

The given transformation can be written as,

0

5

p

s tq

T r

s

t

   

              

   

0. 0. 0.

0. 0. 0. 0. 0.

5

p q r s t

p q r s t

                

Since, the transformation matrix  T M does not exist. So, we conclude that T is not a linear transformation. (Ans.)

7). We have, 3 4:T R R

And    , , ,0, ,T p q r p q z y  

So, we must need a 4 3 matrix consisting of real numbers only.

The given transformation can be written as,

0

p q p

T q z

r y

             

    

0.

0. 0. 0.

p q r

p q r

z

y

    

        

Since, the transformation matrix  T M does not exist. So, we conclude that T is not a linear transformation. (Ans.)

8). We have, 3 3:T R R

And    , , 3 , 2 4 3 ,5 4 2T x y z x y x y z x y z      

So, we must need a 3 3 matrix consisting of real numbers only.

The given transformation can be written as,

3

2 4 3

5 4 2

x x y

T y x y z

z x y z

                     

3 0.

2 4 3

5 4 2

x y z

x y z

x y z

             

3 1 0

2 4 3

5 4 2

x

y

z

               

Mv

Since,  

3 1 0

2 4 3

5 4 2

T M

   

       

is a matrix of order 3 3 , so we conclude that T is a linear

transformation. (Ans.)

Now,      

3 1 0

2 4 3 3 8 12 1 4 15 0 12 11 1

5 4 2

T             

.

Since,   0T  , so   1

T

exists.

Hence, 1T  exists and   1

T

will be the transformation matrix.

The cofactors are, 11 12 13 21 22 23 31 324, 11, 12, 2, 6, 7, 3, 9,A A A A A A A A           

33 10A   .

The matrix of cofactors is,

4 11 12

2 6 7

3 9 10

   

      

   4 11 12 4 2 3

. 2 6 7 11 6 9

3 9 10 12 7 10

t

        

                   

      

1

4 2 3 4 2 3 .

1 11 6 9 11 6 9

12 7 10 12 7 10

T

         

                   

The inverse transformation is,

1

4 2 3

11 6 9

12 7 10

a a

T b b

c c

                

        

4 2 3

11 6 9

12 7 10

a b c

a b c

a b c

              

   1 , , 4 2 3 , 11 6 9 , 12 7 10T a b c a b c a b c a b c         

Verification: Verify for the non-zero vector  1,1,1 .

     1,1,1 3.1 1, 2.1 4.1 3.1,5.1 4.1 2.1 4, 3,7T         

And           1 4, 3,7 4.4 2. 2 3.7, 11.4 6. 2 9.7, 12.4 7. 2 10.7 1,1,1T               

(verified)

Exercise: Check the followings for linear transformations and find 1T  if it exists. Also verify for a non-zero vector.

1) 3 4:T R R is given by    , , 9 2 , ,4 , 5 3T x y z x y z z x x y z       .

2) 3 3:T R R is given by    , , 9 2 , ,T x y z x y xyz z x    .

3) 2 3:T R R is given by    , 0, ,T x y y y x   .

4) 3 4:T R R is given by    , , 8 , ,4 5 ,5T x y z x z z x   .

5) 3:T R R is given by    , , 7 9T x y z z x y    .

6) 3 3:T R R is given by    , , 2 8 ,5 , 2T x y z x y y z   .