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