The Matrix of a Linear Transformation

The Matrix of a Linear Transformation

Linear AlgebraMATH 2076

Section 4.7 The Matrix of an LT 27 March 2017 1 / 7


Recall that every LT Rn T−→ Rm is a matrix transformation; i.e.,

there is anm × n matrix A so that T (~x) = A~x . In fact, Colj(A) = T (~ej).

Suppose V T−→W is a LT. Can we view T as a matrix transformation?Yes, if we use coordinate vectors. Let B,A be bases for V,W resp.

Consider the coordinate maps V [·]B−−→ Rn and W [·]A−−→ Rm. Given ~v in V,we get

[~v]B in Rn, and given ~w in W, we get

[~w]A in Rm.

Consider ~x =[~v]B and ~y =

[~w]A where ~v is in V and ~w = T (~v). By

linearity props of coord vectors, the map ~x 7→ ~y (from Rn to Rm) is a lineartransformation. So ~x 7→ ~y is given by multiplication by some matrix M:

[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.

We call M the matrix for T relative to B and Aand we write

[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .



Recall that every LT Rn T−→ Rm is a matrix transformation; i.e., there is anm × n matrix A so that T (~x) = A~x .

In fact, Colj(A) = T (~ej).




[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .



Recall that every LT Rn T−→ Rm is a matrix transformation; i.e., there is anm × n matrix A so that T (~x) = A~x . In fact, Colj(A) = T (~ej).




[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .




Suppose V T−→W is a LT. Can we view T as a matrix transformation?

Yes, if we use coordinate vectors. Let B,A be bases for V,W resp.



[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .




Suppose V T−→W is a LT. Can we view T as a matrix transformation?Yes, if we use coordinate vectors.

Let B,A be bases for V,W resp.



[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .





Consider the coordinate maps V [·]B−−→ Rn and W [·]A−−→ Rm.

Given ~v in V,we get


[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .






[~v]B in Rn, and

given ~w in W, we get[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.


[~w]A where ~v is in V and ~w = T (~v).

Bylinearity props of coord vectors, the map ~x 7→ ~y (from Rn to Rm) is a lineartransformation. So ~x 7→ ~y is given by multiplication by some matrix M:

[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.



linearity props of coord vectors, the map ~x 7→ ~y (from Rn to Rm) is a lineartransformation.

So ~x 7→ ~y is given by multiplication by some matrix M:

[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




[T (~v)

]A =

[~w]A =

~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M

[~v]B. We call M the matrix for T relative to B and A

and we write[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




[T (~v)

]A =

[~w]A = ~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M



[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.



linearity props of coord vectors, the map ~x 7→ ~y (from Rn to Rm) is a lineartransformation. So ~x 7→ ~y is given by multiplication by some matrix M:[

T (~v)]A =

[~w]A = ~y = M~x

= M[~v]B;

i.e.,[T (~v)

]A = M



[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




T (~v)]A =

[~w]A = ~y = M~x = M

[~v]B;

i.e.,[T (~v)

]A = M



[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




T (~v)]A =

[~w]A = ~y = M~x = M

[~v]B;

i.e.,[T (~v)

]A = M

[~v]B.


[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




T (~v)]A =

[~w]A = ~y = M~x = M

[~v]B;

i.e.,[T (~v)

]A = M


and

we write[T]AB= M, so

[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




T (~v)]A =

[~w]A = ~y = M~x = M

[~v]B;

i.e.,[T (~v)

]A = M



[T (~v)

]A =

[T]AB[~v]B .







[~w]A in Rm.




T (~v)]A =

[~w]A = ~y = M~x = M

[~v]B;

i.e.,[T (~v)

]A = M



[T (~v)

]A =

[T]AB[~v]B .


Picture for the matrix for T relative to B and A

We have a linear transformation V T−→W and bases B,A for V,W resp.

Consider the B and A coord maps V→ Rn

and W→ Rm. Given ~v in V, ~w in W, let~x =

[~v]B and ~y =

[~w]A.

Consider the LT Rn → Rm given by ~x 7→ ~y .This is a matrix transformation, and

[T (~v)

]A =

[~w]A =

~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y





and W→ Rm.

Given ~v in V, ~w in W, let~x =

[~v]B and ~y =

[~w]A.


[T (~v)

]A =

[~w]A =

~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y





and W→ Rm. Given ~v in V, ~w in W, let

~x =[~v]B and ~y =

[~w]A.


[T (~v)

]A =

[~w]A =

~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.


[T (~v)

]A =

[~w]A =

~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.

Consider the LT Rn → Rm given by ~x 7→ ~y .

This is a matrix transformation, and

[T (~v)

]A =

[~w]A =

~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.


[T (~v)

]A =

[~w]A =

~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.


[T (~v)

]A =

[~w]A =

~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.


[T (~v)

]A =

[~w]A = ~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.

Consider the LT Rn → Rm given by ~x 7→ ~y .This is a matrix transformation, and[T (~v)

]A =

[~w]A = ~y = M~x

= A[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.


]A =

[~w]A = ~y = M~x = A

[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y






[~v]B and ~y =

[~w]A.


]A =

[~w]A = ~y = M~x = A

[~v]B

where M =[T]AB.

V WT

Rn Rm

[·]B [·]A

~v ~w

~x ~y


Finding the Matrix for a Linear Transformation

Suppose V T−→W is a linear transformation and B,A are bases for V,Wresp. Then

[T (~v)

]A =

[T]AB[~v]B where

[T]AB is the matrix for T

relative to B and A.

To find[T]AB, we need to know B. Suppose B = {~b1, ~b2, . . . , ~bn}. Then

Colj

([T]AB

)=[T (~bj)

]A .

When V = Rn,W = Rm and B,A are the standard bases, this is the usualformula for the standard matrix for T .

You can remember this as[T]AB=

[T (B)

]A, but this abuses notation!




[T (~v)

]A =

[T]AB[~v]B

where[T]AB is the matrix for T



Colj

([T]AB

)=[T (~bj)

]A .



[T (B)





[T (~v)

]A =

[T]AB[~v]B where




Colj

([T]AB

)=[T (~bj)

]A .



[T (B)





[T (~v)

]A =

[T]AB[~v]B where



To find[T]AB, we need to know B.

Suppose B = {~b1, ~b2, . . . , ~bn}. Then

Colj

([T]AB

)=[T (~bj)

]A .



[T (B)





[T (~v)

]A =

[T]AB[~v]B where




Colj

([T]AB

)=[T (~bj)

]A .



[T (B)





[T (~v)

]A =

[T]AB[~v]B where




Colj

([T]AB

)=[T (~bj)

]A .



[T (B)





[T (~v)

]A =

[T]AB[~v]B where




Colj

([T]AB

)=[T (~bj)

]A .



[T (B)





[T (~v)

]A =

[T]AB[~v]B where




Colj

([T]AB

)=[T (~bj)

]A .



[T (B)



The B-Matrix for a Linear Transformation

Suppose V = Rn = W and B = A, so Rn T−→ Rn is a linear transformationand B is some basis for Rn.

Here we write[T]B=[T]BB for the matrix for

T relative to B and B, and call this the the B-matrix for T .

Consider the B coord map Rn → Rn. Given ~xand ~y = T (~x) = A~x , look at[

~x]B and

[~y]B =

[A~x]B.

Recall that ~x = P[~x]B where P = PEB =

[B].

So[~y]B = P−1~y and we see that[

T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B

It follows that the B-matrix for T is given by[T]B = P−1AP, so

A = P[T]BP−1. Thus A =

[T]E and

[T]B are similar matrices!



Suppose V = Rn = W and B = A, so Rn T−→ Rn is a linear transformationand B is some basis for Rn. Here we write

[T]B=[T]BB for the matrix for

T relative to B and B,

and call this the the B-matrix for T .


~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and







Consider the B coord map Rn → Rn.

Given ~xand ~y = T (~x) = A~x , look at[

~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and







Consider the B coord map Rn → Rn. Given ~xand ~y = T (~x) = A~x , look at

[~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].

So[~y]B = P−1~y and we see that

[T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B =

P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B = P−1~y

=

P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B = P−1~y

= P−1A~x =

P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B = P−1~y

= P−1A~x = P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B = P−1~y

= P−1A~x = P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B = P−1~y

= P−1A~x = P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B


A = P[T]BP−1.

Thus A =[T]E and








~x]B and

[~y]B =

[A~x]B.


[B].


T]B[~x]B =

[T (~x)

]B =

[~y]B = P−1~y

= P−1A~x = P−1AP[~x]B.

Rn RnT

Rn Rn

[·]B [·]B

~x ~y = A~x

[~x]B

[~y]B



[T]E and



Connection with Diagonalization

Let A be a diagonalizable n × n matrix.

Define Rn T−→ Rn by T (~x) = A~x .

Since A is diagonalizable, there is an eigenbasis assoc’d with A; that is,there is a basis B = {~v1, ~v2, . . . , ~vn} for Rn such that each vector ~vi is aneigenvector for A. Assume A~vi = λi~vi . Then

A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.

Recall that P = PEB; so, P−1 = PBE which means that[~y]B = P−1~y .

Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.

This says that D =[T]B =

[T]BB.



Let A be a diagonalizable n × n matrix. Define Rn T−→ Rn by T (~x) = A~x .


A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.




Since A is diagonalizable, there is an eigenbasis assoc’d with A; that is,

there is a basis B = {~v1, ~v2, . . . , ~vn} for Rn such that each vector ~vi is aneigenvector for A. Assume A~vi = λi~vi . Then

A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.




Since A is diagonalizable, there is an eigenbasis assoc’d with A; that is,there is a basis B = {~v1, ~v2, . . . , ~vn} for Rn such that each vector ~vi is aneigenvector for A.

Assume A~vi = λi~vi . Then

A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.

Recall that P = PEB; so,

P−1 = PBE which means that[~y]B = P−1~y .

Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.

Recall that P = PEB; so, P−1 = PBE which means that

[~y]B = P−1~y .

Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=

[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=[A~x]B

=

P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=[A~x]B

= P−1(A~x)

=

P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)=

D(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)=

D[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.





A = PDP−1 where P =[~v1 ~v2 . . . ~vn

]and D =

λ1 0 . . . 00 λ2 . . . 0...

...... 0

0 0 . . . λn

.


Thus[T (~x)

]B

=[A~x]B

= P−1(A~x)

= P−1(PDP−1~x

)= D

(P−1~x

)= D

[~x]B.


[T]BB.


A 3× 3 Example

The matrix A =

4 −1 0−1 5 −10 −1 4

has simple eigenvalues 3, 4, 6 with

associated eigenvectors

~v1 =

111

, ~v2 =

−101

, ~v3 =

1−21

.

Since B = {~v1, ~v2, ~v3} is a basis for R3, A is diagonalizable with

A = PDP−1 where P =

1 −1 11 0 −21 1 1

and D =

3 0 00 4 00 0 6

.

Here D is the B-matrix for the LT ~x 7→ A~x .


A 3× 3 Example

The matrix A =

4 −1 0−1 5 −10 −1 4


associated eigenvectors ~v1 =

111

, ~v2 =

−101

, ~v3 =

1−21

.



1 −1 11 0 −21 1 1

and D =

3 0 00 4 00 0 6

.



A 3× 3 Example

The matrix A =

4 −1 0−1 5 −10 −1 4



111

, ~v2 =

−101

, ~v3 =

1−21

.



1 −1 11 0 −21 1 1

and D =

3 0 00 4 00 0 6

.



A 3× 3 Example

The matrix A =

4 −1 0−1 5 −10 −1 4



111

, ~v2 =

−101

, ~v3 =

1−21

.



1 −1 11 0 −21 1 1

and D =

3 0 00 4 00 0 6

.Here D is the B-matrix for the LT ~x 7→ A~x .


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The Matrix of a Linear Transformation