2Linear maps
IB Linear Algebra
2.1 Definitions and examples
Definition
(Linear map)
.
Let
U, V
be vector spaces over
F
. Then
α
:
U → V
is a linear map if
(i) α(u
1
+ u
2
) = α(u
1
) + α(u
2
) for all u
i
∈ U.
(ii) α(λu) = λα(u) for all λ ∈ F, u ∈ U.
We write L(U, V ) for the set of linear maps U → V .
There are a few things we should take note of:
–
If we are lazy, we can combine the two requirements to the single require-
ment that
α(λu
1
+ µu
2
) = λα(u
1
) + µα(u
2
).
–
It is easy to see that if
α
is linear, then it is a group homomorphism (if we
view vector spaces as groups). In particular, α(0) = 0.
–
If we want to stress the field
F
, we say that
α
is
F
-linear. For example,
complex conjugation is a map C → C that is R-linear but not C-linear.
Example.
(i)
Let
A
be an
n×m
matrix with coefficients in
F
. We will write
A ∈ M
n,m
(
F
).
Then α : F
m
→ F
n
defined by v → Av is linear.
Recall matrix multiplication is defined by: if A
ij
is the ijth coefficient of
A, then the ith coefficient of Av is A
ij
v
j
. So we have
α(λu + µv)
i
=
m
X
j=1
A
ij
(λu + µv)
j
= λ
m
X
j=1
A
ij
u
j
+ µ
m
X
j=1
A
ij
v
j
= λα(u)
i
+ µα(v)
i
.
So α is linear.
(ii)
Let
X
be a set and
g ∈ F
X
. Then we define
m
g
:
F
X
→ F
X
by
m
g
(
f
)(
x
) =
g(x)f(x). Then m
g
is linear. For example, f(x) 7→ 2x
2
f(x) is linear.
(iii)
Integration
I
: (
C
([
a, b
])
, R
)
→
(
C
([
a, b
])
, R
) defined by
f 7→
R
x
a
f
(
t
) d
t
is
linear.
(iv) Differentiation D : (C
∞
([a, b]), R) → (C
∞
([a, b]), R) by f 7→ f
0
is linear.
(v)
If
α, β ∈ L
(
U, V
), then
α
+
β
defined by (
α
+
β
)(
u
) =
α
(
u
) +
β
(
u
) is linear.
Also, if λ ∈ F, then λα defined by (λα)(u) = λ(α(u)) is also linear.
In this way, L(U, V ) is also a vector space over F.
(vi)
Composition of linear maps is linear. Using this, we can show that many
things are linear, like differentiating twice, or adding and then multiplying
linear maps.
Just like everything else, we want to define isomorphisms.
Definition
(Isomorphism)
.
We say a linear map
α
:
U → V
is an isomorphism
if there is some β : V → U (also linear) such that α ◦ β = id
V
and β ◦ α = id
U
.
If there exists an isomorphism
U → V
, we say
U
and
V
are isomorphic, and
write U
∼
=
V .
Lemma.
If
U
and
V
are vector spaces over
F
and
α
:
U → V
, then
α
is an
isomorphism iff α is a bijective linear map.
Proof.
If
α
is an isomorphism, then it is clearly bijective since it has an inverse
function.
Suppose
α
is a linear bijection. Then as a function, it has an inverse
β
:
V → U
. We want to show that this is linear. Let
v
1
, v
2
∈ V
,
λ, µ ∈ F
. We
have
αβ(λv
1
+ µv
2
) = λv
1
+ µv
2
= λαβ(v
1
) + µαβ(v
2
) = α(λβ(v
1
) + µβ(v
2
)).
Since α is injective, we have
β(λv
1
+ µv
2
) = λβ(v
1
) + µβ(v
2
).
So β is linear.
Definition
(Image and kernel)
.
Let
α
:
U → V
be a linear map. Then the
image of α is
im α = {α(u) : u ∈ U}.
The kernel of α is
ker α = {u : α(u) = 0}.
It is easy to show that these are subspaces of V and U respectively.
Example.
(i)
Let
A ∈ M
m,n
(
F
) and
α
:
F
n
→ F
m
be the linear map
v 7→ Av
. Then the
system of linear equations
m
X
j=1
A
ij
x
j
= b
i
, 1 ≤ i ≤ n
has a solution iff (b
1
, ··· , b
n
) ∈ im α.
The kernel of α contains all solutions to
P
j
A
ij
x
j
= 0.
(ii) Let β : C
∞
(R, R) → C
∞
(R, R) that sends
β(f)(t) = f
00
(t) + p(t)f
0
(t) + q(t)f(t).
for some p, q ∈ C
∞
(R, R).
Then if
y
(
t
)
∈ im β
, then there is a solution (in
C
∞
(
R, R
)) to the differential
equation
f
00
(t) + p(t)f
0
(t) + q(t)f(t) = y(t).
Similarly, ker β contains the solutions to the homogeneous equation
f
00
(t) + p(t)f
0
(t) + q(t)f(t) = 0.
If two vector spaces are isomorphic, then it is not too surprising that they
have the same dimension, since isomorphic spaces are “the same”. Indeed this is
what we are going to show.
Proposition. Let α : U → V be an F-linear map. Then
(i)
If
α
is injective and
S ⊆ U
is linearly independent, then
α
(
S
) is linearly
independent in V .
(ii) If α is surjective and S ⊆ U spans U, then α(S) spans V .
(iii) If α is an isomorphism and S ⊆ U is a basis, then α(S) is a basis for V .
Here (iii) immediately shows that two isomorphic spaces have the same
dimension.
Proof.
(i)
We prove the contrapositive. Suppose that
α
is injective and
α
(
S
) is linearly
dependent. So there are
s
0
, ··· , s
n
∈ S
distinct and
λ
1
, ··· , λ
n
∈ F
not all
zero such that
α(s
0
) =
n
X
i=1
λ
i
α(s
i
) = α
n
X
i=1
λ
i
s
i
!
.
Since α is injective, we must have
s
0
=
n
X
i=1
λ
i
s
i
.
This is a non-trivial relation of the s
i
in U. So S is linearly dependent.
(ii)
Suppose
α
is surjective and
S
spans
U
. Pick
v ∈ V
. Then there is some
u ∈ U
such that
α
(
u
) =
v
. Since
S
spans
U
, there is some
s
1
, ··· , s
n
∈ S
and λ
1
, ··· , λ
n
∈ F such that
u =
n
X
I=1
λ
i
s
i
.
Then
v = α(u) =
n
X
i=1
λ
i
α(s
i
).
So α(S) spans V .
(iii) Follows immediately from (i) and (ii).
Corollary.
If
U
and
V
are finite-dimensional vector spaces over
F
and
α
:
U → V
is an isomorphism, then dim U = dim V .
Note that we restrict it to finite-dimensional spaces since we’ve only shown
that dimensions are well-defined for finite dimensional spaces. Otherwise, the
proof works just fine for infinite dimensional spaces.
Proof.
Let
S
be a basis for
U
. Then
α
(
S
) is a basis for
V
. Since
α
is injective,
|S| = |α(S)|. So done.
How about the other way round? If two vector spaces have the same di-
mension, are they necessarily isomorphic? The answer is yes, at least for
finite-dimensional ones.
However, we will not just prove that they are isomorphic. We will show that
they are isomorphic in many ways.
Proposition.
Suppose
V
is a
F
-vector space of dimension
n < ∞
. Then writing
e
1
, ··· , e
n
for the standard basis of F
n
, there is a bijection
Φ : {isomorphisms F
n
→ V } → {(ordered) basis(v
1
, ··· , v
n
) for V },
defined by
α 7→ (α(e
1
), ··· , α(e
n
)).
Proof.
We first make sure this is indeed a function — if
α
is an isomorphism,
then from our previous proposition, we know that it sends a basis to a basis. So
(α(e
1
), ··· , α(e
n
)) is indeed a basis for V .
We now have to prove surjectivity and injectivity.
Suppose
α, β
:
F
n
→ V
are isomorphism such that Φ(
α
) = Φ(
β
). In other
words, α(e
i
) = β(e
i
) for all i. We want to show that α = β. We have
α
x
1
.
.
.
x
n
= α
n
X
i=1
x
i
e
i
!
=
X
x
i
α(e
i
) =
X
x
i
β(e
i
) = β
x
1
.
.
.
x
n
.
Hence α = β.
Next, suppose that (v
1
, ··· , v
n
) is an ordered basis for V . Then define
α
x
1
.
.
.
x
n
=
X
x
i
v
i
.
It is easy to check that this is well-defined and linear. We also know that
α
is
injective since (
v
1
, ··· , v
n
) is linearly independent. So if
P
x
i
v
i
=
P
y
i
v
i
, then
x
i
=
y
i
. Also,
α
is surjective since (
v
1
, ··· , v
n
) spans
V
. So
α
is an isomorphism,
and by construction Φ(α) = (v
1
, ··· , v
n
).