4Hilbert spaces
II Linear Analysis
4.6 Self-adjoint operators
We have just looked at compact operators. This time, we are going to add a
condition of self-adjointness.
Definition
(Self-adjoint operator)
.
Let
H
be a Hilbert space,
T ∈ B
(
H
). Then
T is self-adjoint or Hermitian if for all x, y ∈ H, we have
hT x, yi = hx, T yi.
It is important to note that we defined the term for bounded linear operators
T
. If we have unbounded operators instead, Hermitian means something different
from self-adjoint, and we have to be careful.
Recall that we defined the adjoint of a linear map to be a map of the dual
spaces. However, we will often abuse notation and call
T
∗
:
H → H
the adjoint,
which is the (unique) operator such that for all x, y ∈ H,
hT x, yi = hx, T
∗
yi.
It is an exercise to show that this is well-defined.
How is this related to the usual adjoint? Let
˜
T
∗
:
H
∗
→ H
∗
be the usual
adjoint. Then we have
T
∗
= φ
−1
◦
˜
T
∗
◦ φ,
where φ : H → H
∗
is defined by
φ(v)(w) = hw, vi
as in the Reisz representation theorem.
The main result regarding self-adjoint operators is the spectral theorem:
Theorem
(Spectral theorem)
.
Let
H
be an infinite dimensional Hilbert space
and T : H → H a compact self-adjoint operator.
(i) σ
p
(T ) = {λ
i
}
N
i=1
is at most countable.
(ii) σ
p
(T ) ⊆ R.
(iii) σ(T ) = {0} ∪ σ
p
(T ).
(iv) If E
λ
i
are the eigenspaces, then dim E
λ
i
is finite if λ
i
6= 0.
(v) E
λ
i
⊥ E
λ
j
if λ
i
6= λ
j
.
(vi) If {λ
i
} is infinite, then λ
i
→ 0.
(vii)
T =
N
X
i=1
λ
i
P
E
λ
i
.
We have already shown (i), (iii), (iv) and (vi). The parts (ii) and (v) we
already did in IA Vectors and Matrices, but for completeness, we will do the
proof again. They do not require compactness. The only non-trivial bit left is
the last part (vii).
We first do the two easy bits.
Proposition.
Let
H
be a Hilbert space and
T ∈ B
(
H
) self-adjoint. Then
σ
p
(T ) ⊆ R.
Proof.
Let
λ ∈ σ
p
(
T
) and
v ∈ ker
(
T − λI
)
\ {
0
}
. Then by definition of
v
, we
have
λ =
hT v, vi
kvk
2
H
=
hv, T vi
kvk
2
H
=
¯
λ.
So λ ∈ R.
Proposition.
Let
H
be a Hilbert space and
T ∈ B
(
H
) self-adjoint. If
λ, µ ∈
σ
p
(T ) and λ 6= µ, then E
λ
⊥ E
µ
.
Proof. Let v ∈ ker(T −λI) \ {0} and w ∈ ker(T −µI) \ {0}. Then
λhv, wi = hT v, wi = hv, T wi = ¯µhv, wi = µhv, wi,
using the fact that eigenvalues are real. Since
λ 6
=
µ
by assumption, we must
have hv, wi = 0.
To prove the final part, we need the following proposition:
Proposition.
Let
H
be a Hilbert space and
T ∈ B
(
H
) a compact self-adjoint
operator. If T 6= 0, then T has a non-zero eigenvalue.
This is consistent with our spectral theorem, since if
T
is non-zero, then
something in the sum
λ
i
P
E
λ
i
has to be non-zero. It turns out this is most of
the work we need.
However, to prove this, we need the following lemma:
Lemma.
Let
H
be a Hilbert space, and
T ∈ B
(
H
) a compact self-adjoint
operator. Then
kT k
B(H)
= sup
kxk
H
=1
|hx, T xi|
Proof. Write
λ = sup
kxk
H
=1
|hx, T xi|.
Note that one direction is easy, since for all x, Cauchy-Schwarz gives
|hx, T xi| ≤ kT xk
H
kxk
H
= kT k
B(H)
kxk
2
H
.
So it suffices to show the inequality in the other direction. We now claim that
kT k
B(H)
= sup
kxk
H
=1,kyk
H
=1
|hT x, yi|.
To show this, recall that
φ
:
H → H
∗
defined by
v 7→ h·, vi
is an isometry. By
definition, we have
kT k
B(H)
= sup
kxk
H
=1
kT xk
H
= sup
kxk
H
=1
kφ(T x)k
H
∗
= sup
kxk
H
=1
sup
kyk
H
=1
|hy, T xi|.
Hence, it suffices to show that
sup
kxk
H
=1,kyk
H
=1
|hT x, yi| ≤ λ.
Take
x, y ∈ H
such that
kxk
H
=
kyk
H
= 1. We first perform a trick similar to
the polarization identity. First, by multiplying
y
by an appropriate scalar, we
can wlog assume hT x, yi is real. Then we have
|hT (x + y), x + yi−hT (x −y), x − yi| = 2|hT x, yi+ hT y, xi|
= 4|hT x, yi|.
Hence we have
|hT x, yi| =
1
4
|hT (x + y), x + yi−hT (x −y), x − yi|
≤
1
4
(λkx + yk
2
H
+ λkx − yk
2
H
)
=
λ
4
(2kxk
2
H
+ 2kyk
2
H
)
= λ,
where we used the parallelogram law. So we have kT k
B(H)
≤ λ.
Finally, we can prove our proposition.
Proposition.
Let
H
be a Hilbert space and
T ∈ B
(
H
) a compact self-adjoint
operator. If T 6= 0, then T has a non-zero eigenvalue.
Proof.
Since
T 6
= 0, then
kT k
B(H)
6
= 0. Let
kT k
B(H)
=
λ
. We now claim that
either λ or −λ is an eigenvalue of T .
By the previous lemma, there exists a sequence
{x
n
}
∞
n=1
⊆ H
such that
kx
n
k
H
= 1 and hx
n
, T x
n
i → ±λ.
We consider the two cases separately. Suppose
hx
n
, T x
n
i → λ
. Consider
T x
n
−λx
n
. Since
T
is compact, there exists a subsequence such that
T x
n
k
→ y
for some
y ∈ H
. For simplicity of notation, we assume
T x
n
→ y
itself. We have
0 ≤ kT x
n
− λx
n
k
2
H
= hT x
n
− λx
n
, T x
n
− λx
n
i
= kT x
n
k
2
H
− 2λhT x
n
, x
n
i + λ
2
kx
n
k
2
→ λ
2
− 2λ
2
+ λ
2
= 0
as
n → ∞
. Note that we implicitly used the fact that
hT x
n
, x
n
i
=
hx
n
, T x
n
i
since hT x
n
, x
n
i is real. So we must have
kT x
n
− λx
n
k
2
H
→ 0.
In other words,
x
n
→
1
λ
y.
Finally, we show y is an eigenvector. This is easy, since
T y = lim
n→∞
T (λx
n
) = λy.
The case where
x
n
→ −λ
is entirely analogous. In this case,
−λ
is an eigenvalue.
The proof is exactly the same, apart form some switching of signs.
Finally, we can prove the last part of the spectral theorem.
Proposition.
Let
H
be an infinite dimensional Hilbert space and
T
:
H → H
a compact self-adjoint operator. Then
T =
N
X
i=1
λ
i
P
E
λ
i
.
Proof. Let
U = span{E
λ
1
, E
λ
2
, ···}.
Firstly, we clearly have
T |
U
=
N
X
i=1
λ
i
P
E
λ
i
.
This is since for any x ∈ U can be written as
x =
n
X
i=1
P
E
λ
i
x.
Less trivially, this is also true for
¯
U, i.e.
T |
¯
U
=
N
X
i=1
λ
i
P
E
λ
i
,
but this is also clear from definition once we stare at it hard enough.
We also know that
H =
¯
U ⊕ U
⊥
.
It thus suffices to show that
T |
U
⊥
= 0.
But since
T |
U
⊥
has no non-zero eigenvalues, this follows from our previous
proposition. So done.