2Linear maps

IB Linear Algebra

2.4 Change of basis

Suppose we have a linear map

α

:

U → V

. Given a basis

{e

i

}

for

U

, and a basis

{f

i

} for V , we can obtain a matrix A.

U V

F

m

F

n

α

A

(e

i

) (f

i

)

We now want to consider what happens when we have two different basis

{u

i

}

and

{e

i

}

of

U

. These will then give rise to two different maps from

F

m

to our

space

U

, and the two basis can be related by a change-of-basis map

P

. We can

put them in the following diagram:

U U

F

m

F

m

ι

U

P

(u

i

) (e

i

)

where

ι

U

is the identity map. If we perform a change of basis for both

U

and

V

,

we can stitch the diagrams together as

U U V V

F

m

F

m

F

n

F

n

ι

U α

ι

V

P

(u

i

)

B

A

(e

i

) (f

i

)

Q

(v

i

)

Then if we want a matrix representing the map

U → V

with respect to bases

(u

i

) and (v

i

), we can write it as the composition

B = Q

−1

AP.

We can write this as a theorem:

Theorem.

Suppose that (

e

1

, ··· , e

m

) and (

u

1

, ··· , u

m

) are basis for a finite-

dimensional vector space

U

over

F

, and (

f

1

, ··· , f

n

) and (

v

1

, ··· , v

n

) are basis

of a finite-dimensional vector space V over F.

Let

α

:

U → V

be a linear map represented by a matrix

A

with respect to

(e

i

) and (f

i

) and by B with respect to (u

i

) and (v

i

). Then

B = Q

−1

AP,

where P and Q are given by

u

i

=

m

X

k=1

P

ki

e

k

, v

i

=

n

X

k=1

Q

ki

f

k

.

Note that one can view

P

as the matrix representing the identity map

i

U

from

U

with basis (

u

i

) to

U

with basis (

e

i

), and similarly for

Q

. So both are

invertible.

Proof. On the one hand, we have

α(u

i

) =

n

X

j=1

B

ji

v

j

=

X

j

X

`

B

ji

Q

`j

f

`

=

X

`

[QB]

`i

f

`

.

On the other hand, we can write

α(u

i

) = α

m

X

k=1

P

ki

e

k

!

=

m

X

k=1

P

ki

X

`

A

`k

f

`

=

X

`

[AP ]

`i

f

`

.

Since the f

`

are linearly independent, we conclude that

QB = AP.

Since Q is invertible, we get B = Q

−1

AP .

Definition

(Equivalent matrices)

.

We say

A, B ∈ Mat

n,m

(

F

) are equivalent if

there are invertible matrices

P ∈ Mat

m

(

F

),

Q ∈ Mat

n

(

F

) such that

B

=

Q

−1

AP

.

Since

GL

K

(

F

) =

{A ∈ Mat

k

(

F

) :

A is invertible}

is a group, for each

k ≥

1,

this is indeed an equivalence relation. The equivalence classes are orbits under

the action of GL

m

(F) ×GL

n

(F), given by

GL

m

(F) ×GL

n

(F) ×Mat

n,m

(F) → Mat(F)

(P, Q, A) 7→ QAP

−1

.

Two matrices are equivalent if and only if they represent the same linear map

with respect to different basis.

Corollary.

If

A ∈ Mat

n,m

(

F

), then there exists invertible matrices

P ∈

GL

m

(F), Q ∈ GL

n

(F) so that

Q

−1

AP =

I

r

0

0 0

for some 0 ≤ r ≤ min(m, n).

This is just a rephrasing of the proposition we had last time. But this tells

us there are min(m, n) + 1 orbits of the action above parametrized by r.

Definition (Column and row rank). If A ∈ Mat

n,m

(F), then

–

The column rank of

A

, written

r

(

A

), is the dimension of the subspace of

F

n

spanned by the columns of A.

–

The row rank of

A

, written

r

(

A

), is the dimension of the subspace of

F

m

spanned by the rows of A. Alternatively, it is the column rank of A

T

.

There is no a priori reason why these should be equal to each other. However,

it turns out they are always equal.

Note that if

α

:

F

m

→ F

n

is the linear map represented by

A

(with respect to

the standard basis), then

r

(

A

) =

r

(

α

), i.e. the column rank is the rank. Moreover,

since the rank of a map is independent of the basis, equivalent matrices have

the same column rank.

Theorem.

If

A ∈ Mat

n,m

(

F

), then

r

(

A

) =

r

(

A

T

), i.e. the row rank is equivalent

to the column rank.

Proof. We know that there are some invertible P, Q such that

Q

−1

AP =

I

r

0

0 0

,

where r = r(A). We can transpose this whole equation to obtain

(Q

−1

AP )

T

= P

T

A

T

(Q

T

)

−1

=

I

r

0

0 0

So r(A

T

) = r.