2Linear maps

IB Linear Algebra

2.2 Linear maps and matrices

Recall that our first example of linear maps is matrices acting on

F

n

. We will

show that in fact, all linear maps come from matrices. Since we know that all

vector spaces are isomorphic to

F

n

, this means we can represent arbitrary linear

maps on vector spaces by matrices.

This is a useful result, since it is sometimes easier to argue about matrices

than linear maps.

Proposition.

Suppose

U, V

are vector spaces over

F

and

S

=

{e

1

, ··· , e

n

}

is a

basis for

U

. Then every function

f

:

S → V

extends uniquely to a linear map

U → V .

The slogan is “to define a linear map, it suffices to define its values on a

basis”.

Proof.

For uniqueness, first suppose

α, β

:

U → V

are linear and extend

f

:

S →

V . We have sort-of proved this already just now.

If u ∈ U, we can write u =

P

n

i=1

u

i

e

i

with u

i

∈ F since S spans. Then

α(u) = α

X

u

i

e

i

=

X

u

i

α(e

i

) =

X

u

i

f(e

i

).

Similarly,

β(u) =

X

u

i

f(e

i

).

So α(u) = β(u) for every u. So α = β.

For existence, if

u ∈ U

, we can write

u

=

P

u

i

e

i

in a unique way. So defining

α(u) =

X

u

i

f(e

i

)

is unambiguous. To show linearity, let λ, µ ∈ F, u, v ∈ U. Then

α(λu + µv) = α

X

(λu

i

+ µv

i

)e

i

=

X

(λu

i

+ µv

i

)f(e

i

)

= λ

X

u

i

f(e

i

)

+ µ

X

v

i

f(e

i

)

= λα(u) + µα(v).

Moreover, α does extend f .

Corollary.

If

U

and

V

are finite-dimensional vector spaces over

F

with bases

(e

1

, ··· , e

m

) and (f

1

, ··· , f

n

) respectively, then there is a bijection

Mat

n,m

(F) → L(U, V ),

sending A to the unique linear map α such that α(e

i

) =

P

a

ji

f

j

.

We can interpret this as follows: the

i

th column of

A

tells us how to write

α(e

i

) in terms of the f

j

.

We can also draw a fancy diagram to display this result. Given a basis

e

1

, ··· , e

m

, by our bijection, we get an isomorphism

s

(

e

i

) :

U → F

m

. Similarly,

we get an isomorphism s(f

i

) : V → F

n

.

Since a matrix is a linear map

A

:

F

m

→ F

n

, given a matrix

A

, we can

produce a linear map α : U → V via the following composition

U F

m

F

n

V.

s(e

i

)

A

s(f

i

)

−1

We can put this into a square:

F

m

F

n

U V

A

s(e

i

)

α

s(f

i

)

Then the corollary tells us that every

A

gives rise to an

α

, and every

α

corresponds

to an A that fit into this diagram.

Proof.

If

α

is a linear map

U → V

, then for each 1

≤ i ≤ m

, we can write

α

(

e

i

)

uniquely as

α(e

i

) =

n

X

j=1

a

ji

f

j

for some

a

ji

∈ F

. This gives a matrix

A

= (

a

ij

). The previous proposition tells

us that every matrix A arises in this way, and α is determined by A.

Definition

(Matrix representation)

.

We call the matrix corresponding to a

linear map

α ∈ L

(

U, V

) under the corollary the matrix representing

α

with

respect to the bases (e

1

, ··· , e

m

) and (f

1

, ··· , f

n

).

It is an exercise to show that the bijection

Mat

n,m

(

F

)

→ L

(

U, V

) is an iso-

morphism of the vector spaces and deduce that

dim L

(

U, V

) = (

dim U

)(

dim V

).

Proposition.

Suppose

U, V, W

are finite-dimensional vector spaces over

F

with

bases R = (u

1

, ··· , u

r

) , S = (v

1

, .., v

s

) and T = (w

1

, ··· , w

t

) respectively.

If

α

:

U → V

and

β

:

V → W

are linear maps represented by

A

and

B

respectively (with respect to

R

,

S

and

T

), then

βα

is linear and represented by

BA with respect to R and T .

F

r

F

s

F

t

U V W

A B

s(R)

α

s(S)

β

s(T )

Proof.

Verifying

βα

is linear is straightforward. Next we write

βα

(

u

i

) as a linear

combination of w

1

, ··· , w

t

:

βα(u

i

) = β

X

k

A

ki

v

k

!

=

X

k

A

ki

β(v

k

)

=

X

k

A

ki

X

j

B

jk

w

j

=

X

j

X

k

B

jk

A

ki

!

w

j

=

X

j

(BA)

ji

w

j