Matrices and linear transformations For grade 1, undergraduate students For grade 1, undergraduate...

Matrices and linear transformationsMatrices and linear transformations

For grade 1, undergraduate studentsFor grade 1, undergraduate students

Made by Department of Math. ,Anqing Teachers collegeMade by Department of Math. ,Anqing Teachers college

1. Representing a linear transformation by a matrix Let us suppose given a linear transformation

between the finite-dimensional vector spaces V and W.

:T V W

Let be an ordered basis for V and

1, n

1, m

an ordered basis for W. It is then possible to fine unique numbers

, 1, , ; 1, ,ija i n j m

such that

1 11 1 21 2 1 ,m mT a a a

2 12 1 21 2 2 ,m mT a a a

1 1 2 2n n n mn mT a a a

which we have seen may be written more compactly as

1

1,2, , .m

j ij ii

T a j n

1, n 1, m

( )ijA am nThe matrix

is called the matrix of T relative to the ordered bases

EXAMPLE 1. Let be the linear transformation given by

Calculate the matrix of T relative to the standard basis of .

2 2:T R R

( , ) ( , )T x y y x

2R

Solution. By the “standard basis of ”, we of course mean the basis

and that we are to use this ordered basis in both the domain and range of T. We have

2R

1 21,0 , 0,1 ,E E

1,0 0,1 , 0,1 1,0 ,T T

so the matrix we seek is

0 1.

1 0A

EXAMPLE2. With T as in Example 1, find the matrix of T relative to the pair of ordered bases and , where 1, 2E E 1, 2F F

1 2(0,1), (1,0).F F

Solution. We still have

1,0 0,1 , 0,1 1,0 ,T T

but now we must write these equation as

1 1 2 2 2 2( ) 1 0 ( ) 0T E F F T E F F

so that 0 1

1 0B

is the matrix that we now seek .  EXAMPLE3. Calculate the matrix of the differentiation operator D: ( ) ( )n nP PR R

relative to the usual basis

for . 21, , , nx x x

( )nP R

Solution . We have for 1,2,m 1 10 0 0m m m nDx mx x mx x

and (1) 0.D

Thus the matrix that we seek is 0 1 0 0

0 0 2 0

.

0 0 0

0 0 0 0

N

n

Remark. There is of course no reason to restrict us to square matrices. For example we have the linear transformation

1: ( ) ( )n nD P P R R

and we can ask for its matrix relative to the standard bases of and . ( )nP R 1( )nP R

Question. What size is this matrix?

How to calculate the matrix of linear transformation 1: ( ) ( )?n nD P P R R

2. An isomorphism and its matrix

Proposition. is an isomorphism iff its matrix A is invertible.

:T V V

PROOF. Suppose that T is an isomorphism Let:S V V

be the linear transformation inverse to T. Let B be the matrix of S relative to the basis pair 1 1,n n (Note that we have interchanged the role of the bases ). 1 1,n n

Thus if B=( ) then . ijb ( )j ij jS b Since the matrix product AB is the matrix of the linear transformation

:TS V Vrelative to the basis pair

1 ,n 1 .n

But for all since T and S are inverse isomorphisms.

( )TS V

In particular

1 2 1 1( ) 0 0 0 1 0 0j j j j nT S B B

and hence the matrix of relative to the bases

is

T S

1 ,n 1 n

1 0

1

0 1

I

Therefore AB=I.

Likewise, since the matrix product BA is the matrix of the linear transformation

:ST V V

relative to the basis pair-

1 1, .n n

But

( )ST

for all in V because S and T are inverse isomorphisms, and hence as before we find BA=I.

This shows that if :T V V

is an isomorphism then a matrix S for T is always invertible.

To prove the converse, we suppose that the matrix A of T is invertible. Let B be a matrix such that AB=I=BA.

Let be the linear transformation whose matrix relative to the ordered bases

:S V V

1 1,n n

is B . Then the matrix of T relative to the ordered bases

is

:S V V

1 ,n 1 n

1 0

1.

0 1

I

Therefore and I have the same matrix relative to the bases

S T

1 ,n 1 n

so that by , that is S T I ( )S T for all in V.

Likewise we see that

( )TS for all in V,

so that S and T are inverse isomorphisms.

EXAMPLE 4. Find the matrix of the identity linear transformation

relative to the ordered bases

3 3:I R R

1 2 3

1 2 3

1,0,0 , 0,1,0 , 0,0,1

1,1,1 , 1,1,0 , 1,0,0

Solution. We have

1 1 2 3

2 1 2 3

3 1 2 3

1 1 1,0,0 , 1,0,0 0 0 1 ,

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

1 0,0,1 0,0,1 1 ( 1) 0 .

So the matrix we seek is

0 0 1

0 1 1 .

1 1 0

A

Remark. In view of Example above, it is reasonable to expect that when we calculate with matrices of transformations

, we insist upon using the same ordered basis

twice to do the calculation, rather than work with distinct ordered bases

:T V V

1 n

1 .n 1 ,n

EXAMPLE 5. Let be the linear transformation

given by

3 3:T R R

( , , ) ( , , )T x y z y z x z y z

Calculate the matrix of T relative to (a) the standard basis of 3R

(b) the basis

used twice.

(1,1,1), (1, 1,0), (1,1, 2) ,

Solution .To do Part (a) we compute as follows:

: (1,0,0) (0,1,1)

: (0,1,0) (1,0,1)

: (0,0,1) (1,1,0).

T

T

T

Thus the desired matrix is 0 1 1

1 0 1 .

0 1 0

A

To so the computations of Part (b) , let us set

123 1,1,1,1,1,0,1,1,2.

Then we find

1 1 2 3

2 1 2 3

3 1 2 3

( ) 1,1,1 2,2,2 2 0 0

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

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

T T

T T

T T

So the matrix for Part (b) is

2 0 0

0 1 0 .

0 0 1

B

3. Matrices relative to different bases  

Theorem. Let A and B be

matrices, V and n-dimensional vector space and W

an m-dimensional vector space. Then A and Brepresent the same linear transformation

m n

:T V W

relative to (perhaps) different pairs of ordered bases iff there exist nonsingular matrices P and Q such that

1A PBQwhere P is and Q is . m n n n

PROOF. There are two things we must prove. First, if A and B represent the same linear transformation relative to different bases of V and B represent the same linear transformation relative to different bases of V and W we must construct invertible matrices P and Q such that

1A PBQ

we must construct a linear transformation:T V W

and pairs of ordered bases for V and W such that A represents T relative to one pair and S relative to the other. Consider the first of these. We suppose given bases

1 ,n 1 n

such that the matrix of T relative to these bases is A, and bases

1 ,n 1 n

, such that the matrix of T relative to these bases is B . Let P be the matrix of

:I W Wrelative to the bases

1 ,n 1 .n

Then by the proposition above, P is invertible.

,

1Q

relative to the bases

Then is also invertible and represent the matrix of

:I V V

1 ,n 1 .n

Let Q be the matrix of

relative to he bases

1 ,n 1 .n

:I V V

Therefore PB is the matrix of

relative to the bases

:T V W

1 .n 1 ,n

If we apply it again we see that

is the matrix of T relative to the bases

1PBQ

1 .n 1 ,n

But Q is also the matrix of T relative to the bases

1 .n 1 ,n

To prove the converse, suppose given invertible matrices P and Q such that

1A PBQ

so that

as required .

1A PBQ

:T V W

Choose bases

1 ,n 1 n

for V and W respectively . Let be the linear transformation whose matrix is A relative to these bases. Let

1 1

1 1

( ), , ( )

( ), , ( )n n

n m

P P

Q Q

,

since P and A are isomorphisms, the collections

1 ,n 1 n

are bases for V and W respectively. A brute force computation now shows that B is the matrix of T relative to the bases

1 ,n 1 .n

EXAMPLE 6. Recall that we are given the linear transformation

defined by

3 3:T R R

( , , ) ( , , )T x y z y z x z y z

and

is the matrix of T relative to the standard basisof , while

0 1 1

1 0 1

0 1 0

A

3R2 0 0

0 1 0

0 0 1

B

is the matrix of relative to the ordered basis

(1,1,1), (1, 1,0), (1,1, 2) ,

3.Rof

Since there are invertible matrices P, and

such that

, our task is to

1Q

1A PBQ

compute P and . We compute them as follows.1Q

(1) P is the matrix of

relative to the basis pair

and

3 3:I R R

(1,1,1), (1, 1,0), (1,1, 2) ,

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

1Q

(2) is the matrix of

relative to the basis pair

and

3 3:I R R

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

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

The computation of P is easy and gives us 1 1 1

1 1 1 .

1 0 2

P

The computation of is not hard and depends on the following equations

1Q

1 1 1(1,0,0) (1,1,1) (1, 1,0) (1,1, 2)

3 2 61 1 1

(0,1,0) (1,1,1) (1, 1,0) (1,1, 2)3 2 61 1

(0,0,1) (1,1,1) 0(1, 1,0) (1,1, 2).3 3

so that

1

1 1 1

2 2 61 1 1

.3 2 61 1

03 3

Q

A tedious computation shows that .

1A PBQ

That is 1 1 1

2 2 60 1 1 0 1 1 2 0 01 1 1

1 0 1 1 0 1 0 1 0 .3 2 6

0 1 0 0 1 0 0 0 11 1

03 3

4. Some exercises

1. Find the matrix of following linear transformations relative to the stand are bases for

3R

(a) given by 3 5:T R R

1 2 3 1 1 3 1 3 1 2, , , , , .T a a a a a a a a a a

(b) given by 4 4:T R R

1 2 3 4 1 1 2 1 2 3 1 2 3 4, , , , , , .T a a a a a a a a a a a a a a

(c) given by 3:T R R

1 2 3 1 2 3, ,T a a a a a a

(d) given by 2 4:T R R

1 2 1 2 1 2 1 1 2, , 2 ,3 ,2T a a a a a a a a a

2. Let be the linear transformation whose matrix relative to the standard bases is

4 7:T R R

1 2 1 0

0 1 4 1

2 0 2 0

.0 1 3 1

3 2 5 0

0 0 7 1

4 0 1 0

Find

(1,2,3,4).T

3. Let be the linear transformations with matrices

3 2, :S T R R

6 1 2,

2 4 1

5 0 7

2 1 9

respectively. What is the matrix of the linear transformation

3 23 7 : .S T R R

Find (3S-7T)(1,2,3).

Thanks !!!