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Linear transformation
TheStuffPoint.Com
BY: Abu Bakar Soomro
Linear transformation
( ),U F ( )V F
:T U V
1 2For all , , , ;u u u U a F 
( ) ( ).a aT Tu u
1 2 1 2( ) ( ) ( ),u uT T Tu u  


1 2 1 2
1 2
( ) ( ) ( )
, , , ;
uaT Tu u u
u u
T
a FbU
b  
  
3 2
: R RT 
1 2 11 3 32( , , ) ( , )x x xxT xx x  
1 2 1 2 3 1 2 3
1 2 3 1 2 3
1 1 2 2 3 3
1 1 2 2 1 1 3 3
1 2 1 2 3 1
( ) ( ( , , ) ( , , ))
(( , , ) ( , , ))
( , , )
( , )
( ) ( ) ( , , ) ( ,
T au bu T a x x x b y y y
T ax ax ax by by by
T ax by ax by ax by
ax by ax by ax by ax by
aT u bT u aT x x x bT y y
  
 
   
      
   2 3
1 2 1 3 1 2 1 3
1 2 1 3 1 2 1 3
1 2 1 2 1 3 1 3
1 1 2 2 1 1 3 3
, ))
( , ) ( , )
( ( ), ( )) ( ( ), ( ))
( ( ) ( ), ( ) ( ))
( , )
y
a x x x x b y y y y
a x x a x x b y y b y y
a x x b y y a x x b y y
ax by ax by ax by ax by
     
     
      
      
1 2 1 2( ) ( ) ( )T au bu aT u bT u  
Matrix of linear transformation
:T U V
( ) ,T Au u u U 
: n m
R RT 
( ) ,m
n
nA RT x x x 
:r
r
c
c
RA T R  
Q:
2
3
( ),
( )
R
R
F
F
1 2 3 1 21 2 3 33 4 9 5 3 2( , , ) ( , )x x x x xx xT x x    
3 2
: RRT 
Find the matrix of linear transformation with
respect to standard bases for the vector spaces
Standard basis for
the vector space is:
3
( )R F 2
( )R F
Standard basis for
the vector space is:
1
*
1 1 2 3
2
3
{ , , },
( , , )1 0 0
0 1
,
( , , )0
0 0( )1
,
, , .
u u u
u
u
u
B 



1 2 3
1
2
2
*
{ , , },
( , ),
( ).
1 0
0 1,
v v v
v
v
B 


1 2 3 1 21 2 3 33 4 9 5 3 2( , , ) ( , )x x x x xx xT x x    
1 1 0( ) ( , , ) 3( 50 , )T Tu  
2 0 1 0( ) ( , 3, ) ( , )4uT T  
3 0 0 1( ) ( , 2, ) ( , )9uT T  
21 1( ) ( , , ) 3 5( , ) ( , ) (1 0 01 0 )0 ,1aT u b aT v bv     
( , ) ( , )3 5 0 0( , ) ( , )a b a b  
( , ) ( , )3 5a b 
11 2( ) ( , , ) ( ,3 5) ,1 0 0 3 5T v vTu    
12 2( ) ( , , ) ( , )4 30 1 0 ,4 3u vT T v   
3 1 29( ) ( , , ) ( 20 0 , ) 9 2 .1 ( )v vuT T     
11 2( ) ,3 5vuT v 
2 1 24( ) ,3u v vT 
13 29 2( ) ( ) .T v vu   
Hence, the matrix of transformation is
3 4 9
5 3 2
A
 
   
If is linear transformation
given by the matrix ,
find m, n and express T in terms of coordinates.
Q:
: n m
R RT 
6 1
1 2
1 3
 
 
 
  
:
r
r
c
c
R
A
T R


b
6 1
1 2
1 3
A
 
   
  
Solution:
3 2r c  
2 3
: R RT 
.2 3,n m 
1 2 1 2 1 2 1 2( , ) (6 , 2 , 3 )T x x x x x x x x   
1 2( ) , ( , )T x Ax x x x 
1 2
1
1 2
2
1 2
6 1 6
( ) 1 2 2
1 3 3
x x
x
T x x x
x
x x
   
          
        

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Linear transformations-thestuffpoint.com

  • 2. Linear transformation ( ),U F ( )V F :T U V 1 2For all , , , ;u u u U a F  ( ) ( ).a aT Tu u 1 2 1 2( ) ( ) ( ),u uT T Tu u     1 2 1 2 1 2 ( ) ( ) ( ) , , , ; uaT Tu u u u u T a FbU b     
  • 3. 3 2 : R RT  1 2 11 3 32( , , ) ( , )x x xxT xx x   1 2 1 2 3 1 2 3 1 2 3 1 2 3 1 1 2 2 3 3 1 1 2 2 1 1 3 3 1 2 1 2 3 1 ( ) ( ( , , ) ( , , )) (( , , ) ( , , )) ( , , ) ( , ) ( ) ( ) ( , , ) ( , T au bu T a x x x b y y y T ax ax ax by by by T ax by ax by ax by ax by ax by ax by ax by aT u bT u aT x x x bT y y                    2 3 1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 2 1 3 1 3 1 1 2 2 1 1 3 3 , )) ( , ) ( , ) ( ( ), ( )) ( ( ), ( )) ( ( ) ( ), ( ) ( )) ( , ) y a x x x x b y y y y a x x a x x b y y b y y a x x b y y a x x b y y ax by ax by ax by ax by                           1 2 1 2( ) ( ) ( )T au bu aT u bT u  
  • 4. Matrix of linear transformation :T U V ( ) ,T Au u u U  : n m R RT  ( ) ,m n nA RT x x x  :r r c c RA T R  
  • 5. Q: 2 3 ( ), ( ) R R F F 1 2 3 1 21 2 3 33 4 9 5 3 2( , , ) ( , )x x x x xx xT x x     3 2 : RRT  Find the matrix of linear transformation with respect to standard bases for the vector spaces
  • 6. Standard basis for the vector space is: 3 ( )R F 2 ( )R F Standard basis for the vector space is: 1 * 1 1 2 3 2 3 { , , }, ( , , )1 0 0 0 1 , ( , , )0 0 0( )1 , , , . u u u u u u B     1 2 3 1 2 2 * { , , }, ( , ), ( ). 1 0 0 1, v v v v v B    1 2 3 1 21 2 3 33 4 9 5 3 2( , , ) ( , )x x x x xx xT x x     1 1 0( ) ( , , ) 3( 50 , )T Tu   2 0 1 0( ) ( , 3, ) ( , )4uT T   3 0 0 1( ) ( , 2, ) ( , )9uT T  
  • 7. 21 1( ) ( , , ) 3 5( , ) ( , ) (1 0 01 0 )0 ,1aT u b aT v bv      ( , ) ( , )3 5 0 0( , ) ( , )a b a b   ( , ) ( , )3 5a b  11 2( ) ( , , ) ( ,3 5) ,1 0 0 3 5T v vTu     12 2( ) ( , , ) ( , )4 30 1 0 ,4 3u vT T v    3 1 29( ) ( , , ) ( 20 0 , ) 9 2 .1 ( )v vuT T     
  • 8. 11 2( ) ,3 5vuT v  2 1 24( ) ,3u v vT  13 29 2( ) ( ) .T v vu    Hence, the matrix of transformation is 3 4 9 5 3 2 A      
  • 9. If is linear transformation given by the matrix , find m, n and express T in terms of coordinates. Q: : n m R RT  6 1 1 2 1 3          : r r c c R A T R   b 6 1 1 2 1 3 A          Solution: 3 2r c   2 3 : R RT  .2 3,n m 
  • 10. 1 2 1 2 1 2 1 2( , ) (6 , 2 , 3 )T x x x x x x x x    1 2( ) , ( , )T x Ax x x x  1 2 1 1 2 2 1 2 6 1 6 ( ) 1 2 2 1 3 3 x x x T x x x x x x                        