8Inner product spaces
IB Linear Algebra
8.3 Adjoints, orthogonal and unitary maps
Adjoints
Lemma.
Let
V
and
W
be finite-dimensional inner product spaces and
α
:
V →
W
is a linear map. Then there exists a unique linear map
α
∗
:
W → V
such that
(αv, w) = (v, α
∗
w) (∗)
for all v ∈ V , w ∈ W.
Proof.
There are two parts. We have to prove existence and uniqueness. We’ll
first prove it concretely using matrices, and then provide a conceptual reason of
what this means.
Let (
v
1
, ··· , v
n
) and (
w
1
, ··· , w
m
) be orthonormal basis for
V
and
W
.
Suppose α is represented by A.
To show uniqueness, suppose
α
∗
:
W → V
satisfies (
αv, w
) = (
v, α
∗
w
) for
all v ∈ V , w ∈ W , then for all i, j, by definition, we know
(v
i
, α
∗
(w
j
)) = (α(v
i
), w
j
)
=
X
k
A
ki
w
k
, w
j
!
=
X
k
¯
A
ki
(w
k
, w
j
) =
¯
A
ji
.
So we get
α
∗
(w
j
) =
X
i
(v
i
, α
∗
(w
j
))v
i
=
X
i
¯
A
ji
v
i
.
Hence α
∗
must be represented by A
†
. So α
∗
is unique.
To show existence, all we have to do is to show
A
†
indeed works. Now let
α
∗
be represented by
A
†
. We can compute the two sides of (
∗
) for arbitrary
v, w
.
We have
α
X
λ
i
v
i
,
X
µ
j
w
j
=
X
i,j
¯
λ
i
µ
j
(α(v
i
), w
j
)
=
X
i,j
¯
λ
i
µ
j
X
k
A
ki
w
k
, w
j
!
=
X
i,j
¯
λ
i
¯
A
ji
µ
j
.
We can compute the other side and get
X
λ
i
v
i
, α
∗
X
µ
j
w
j
=
X
i,j
¯
λ
i
µ
j
v
i
,
X
k
A
†
kj
v
k
!
=
X
i,j
¯
λ
i
¯
A
ji
µ
j
.
So done.
What does this mean, conceptually? Note that the inner product
V
defines
an isomorphism
V →
¯
V
∗
by
v 7→
(
·, v
). Similarly, we have an isomorphism
W →
¯
W
∗
. We can then put them in the following diagram:
V W
¯
V
∗
¯
W
∗
α
∼
=
∼
=
α
∗
Then
α
∗
is what fills in the dashed arrow. So
α
∗
is in some sense the “dual” of
the map α.
Definition (Adjoint). We call the map α
∗
the adjoint of α.
We have just seen that if
α
is represented by
A
with respect to some or-
thonormal bases, then α
∗
is represented by A
†
.
Definition
(Self-adjoint)
.
Let
V
be an inner product space, and
α ∈ End
(
V
).
Then α is self-adjoint if α = α
∗
, i.e.
(α(v), w) = (v, α(w))
for all v, w.
Thus if
V
=
R
n
with the usual inner product, then
A ∈ Mat
n
(
R
) is self-
adjoint if and only if it is symmetric, i.e.
A
=
A
T
. If
V
=
C
n
with the usual
inner product, then
A ∈ Mat
n
(
C
) is self-adjoint if and only if
A
is Hermitian,
i.e. A = A
†
.
Self-adjoint endomorphisms are important, as you may have noticed from IB
Quantum Mechanics. We will later see that these have real eigenvalues with an
orthonormal basis of eigenvectors.
Orthogonal maps
Another important class of endomorphisms is those that preserve lengths. We
will first do this for real vector spaces, since the real and complex versions have
different names.
Definition
(Orthogonal endomorphism)
.
Let
V
be a real inner product space.
Then α ∈ End(V ) is orthogonal if
(α(v), α(w)) = (v, w)
for all v, w ∈ V .
By the polarization identity,
α
is orthogonal if and only if
kα
(
v
)
k
=
kvk
for
all v ∈ V .
A real square matrix (as an endomorphism of
R
n
with the usual inner product)
is orthogonal if and only if its columns are an orthonormal set.
There is also an alternative way of characterizing these orthogonal maps.
Lemma.
Let
V
be a finite-dimensional space and
α ∈ End
(
V
). Then
α
is
orthogonal if and only if α
−1
= α
∗
.
Proof. (⇐) Suppose α
−1
= α
∗
. If α
−1
= α
∗
, then
(αv, αv) = (v, α
∗
αv) = (v, α
−1
αv) = (v, v).
(
⇒
) If
α
is orthogonal and (
v
1
, ··· , v
n
) is an orthonormal basis for
V
, then for
1 ≤ i, j ≤ n, we have
δ
ij
= (v
i
, v
j
) = (αv
i
, αv
j
) = (v
i
, α
∗
αv
j
).
So we know
α
∗
α(v
j
) =
n
X
i=1
(v
i
, α
∗
αv
j
))v
i
= v
j
.
So by linearity of α
∗
α, we know α
∗
α = id
V
. So α
∗
= α
−1
.
Corollary. α ∈ End
(
V
) is orthogonal if and only if
α
is represented by an
orthogonal matrix, i.e. a matrix
A
such that
A
T
A
=
AA
T
=
I
, with respect to
any orthonormal basis.
Proof.
Let (
e
1
, ··· , e
n
) be an orthonormal basis for
V
. Then suppose
α
is
represented by
A
. So
α
∗
is represented by
A
T
. Then
A
∗
=
A
−1
if and only if
AA
T
= A
T
A = I.
Definition
(Orthogonal group)
.
Let
V
be a real inner product space. Then the
orthogonal group of V is
O(V ) = {α ∈ End(V ) : α is orthogonal}.
It follows from the fact that
α
∗
=
α
−1
that
α
is invertible, and it is clear
from definition that O(
V
) is closed under multiplication and inverses. So this is
indeed a group.
Proposition.
Let
V
be a finite-dimensional real inner product space and
(e
1
, ··· , e
n
) is an orthonormal basis of V . Then there is a bijection
O(V ) → {orthonormal basis for V }
α 7→ (α(e
1
, ··· , e
n
)).
This is analogous to our result for general vector spaces and general bases,
where we replace O(V ) with GL(V ).
Proof. Same as the case for general vector spaces and general bases.
Unitary maps
We are going to study the complex version of orthogonal maps, known as unitary
maps. The proofs are almost always identical to the real case, and we will not
write the proofs again.
Definition
(Unitary map)
.
Let
V
be a finite-dimensional complex vector space.
Then α ∈ End(V ) is unitary if
(α(v), α(w)) = (v, w)
for all v, w ∈ V .
By the polarization identity,
α
is unitary if and only if
kα
(
v
)
k
=
kvk
for all
v ∈ V .
Lemma.
Let
V
be a finite dimensional complex inner product space and
α ∈
End(V ). Then α is unitary if and only if α is invertible and α
∗
= α
−1
.
Corollary. α ∈ End
(
V
) is unitary if and only if
α
is represented by a unitary
matrix A with respect to any orthonormal basis, i.e. A
−1
= A
†
.
Definition
(Unitary group)
.
Let
V
be a finite-dimensional complex inner prod-
uct space. Then the unitary group of V is
U(V ) = {α ∈ End(V ) : α is unitary}.
Proposition.
Let
V
be a finite-dimensional complex inner product space. Then
there is a bijection
U(V ) → {orthonormal basis of V }
α 7→ {α(e
1
), ··· , α(e
n
)}.