This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.7 Spectral Theory
Beware of ghosts2 in this area!
As we saw operators could be added and multiplied each other, in some
sense they behave like numbers, but are much more complicated. In this
lecture we will associate to each operator a set of complex numbers
which reflects certain (unfortunately not all) properties of this operator.
The analogy between operators and numbers become even more deeper
since we could construct functions of operators (called
functional calculus) in a way we build
numeric functions. The most important functions of this sort is called
resolvent (see Definition 5). The methods of
analytical functions are very powerful in operator theory and students
may wish to refresh their knowledge of complex analysis before this
part.
7.1 The spectrum of an operator on a Hilbert space
An eigenvalue of operator
T∈B(H) is a complex number λ such that there
exists a nonzero x∈ H, called
eigenvector with property
Tx=λ x, in other words x∈ker(T−λ I).
In finite
dimensions T−λ I is invertible if and only if λ
is not an eigenvalue.
In infinite dimensions it is not the same:
the right shift operator S is not
invertible but 0 is not its eigenvalue because Sx=0 implies
x=0 (check!).
Definition 1
The resolvent set ρ(
T)
of an operator T is the set
| ρ (T)={λ∈ℂ: T−λ I is invertible}.
|
The spectrum of operator
T∈
B(
H)
, denoted σ(
T)
, is the complement of
the resolvent set ρ(
T)
:
| σ(T)={λ∈ℂ: T−λ I is not invertible}.
|
Example 2
If H is finite dimensional the from previous discussion follows
that σ(
T)
is the set of eigenvalues of T for any T.Even this example demonstrates that spectrum does not provide a
complete description for operator even in finite-dimensional
case. For example, both operators in ℂ2 given by matrices
(
) and
(
) have a single point spectrum {0}, however are
rather different. The situation became even worst in the infinite
dimensional spaces.
Theorem 3
The spectrum σ(
T)
of a bounded
operator T is a nonempty compact (i.e. closed and bounded)
subset of ℂ
.
For the proof we will need several Lemmas.
Lemma 4
Let A∈
B(
H)
. If ||
A||<1
then I−
A is
invertible in B(
H)
and inverse is given by the
Neumann series (C. Neumann, 1877):
|
(I−A)−1=I+A+A2+A3+…= | | Ak.
(40) |
Proof.
Define the sequence of operators
Bn=
I+
A+⋯+
AN—the partial sums
of the infinite series (
40). It is a
Cauchy sequence, indeed:
| = | ⎪⎪
⎪⎪ | Am+1+Am+2+⋯+An | ⎪⎪
⎪⎪ |
(if n<m) |
|
| | ≤ | ⎪⎪
⎪⎪ | Am+1 | ⎪⎪
⎪⎪ | + | ⎪⎪
⎪⎪ | Am+2 | ⎪⎪
⎪⎪ | +⋯+ | ⎪⎪
⎪⎪ | An | ⎪⎪
⎪⎪ |
|
| | ≤ | ⎪⎪
⎪⎪ | A | ⎪⎪
⎪⎪ | m+1+ | ⎪⎪
⎪⎪ | A | ⎪⎪
⎪⎪ | m+2+⋯+ | ⎪⎪
⎪⎪ | A | ⎪⎪
⎪⎪ | n |
|
| | ≤ | |
|
for a large
m. By the
completeness of
B(
H) there is a limit, say
B, of the sequence
Bn. It is a simple algebra to check that
(
I−
A)
Bn=
Bn(
I−
A)=
I−
An+1, passing to the limit in the norm
topology, where
An+1→ 0 and
Bn→
B we
get:
| (I−A)B=B(I−A)=I ⇔ B=(I−A)−1.
|
□
Definition 5
The resolvent
of an operator T is the operator valued
function defined on the resolvent set by the formula:
Corollary 6
-
If | λ |>||T|| then
λ∈ ρ(T), hence the spectrum is
bounded.
-
The resolvent set ρ(T) is open, i.e for any λ ∈
ρ(T) then there exist є>0 such
that all µ with | λ−µ |<є are also
in ρ(T), i.e. the resolvent set is open and the spectrum is
closed.
Both statements together imply that the spectrum is compact.
Proof.
- If | λ |>||T|| then ||λ−1T||<1
and the operator T−λ I=−λ(I−λ−1T)
has the inverse
|
R(λ,T)= (T−λ I)−1=− | | λ−k−1Tk.
(42) |
by the
previous Lemma.
- Indeed:
| T−µ I | = | T−λ I + (λ−µ)I |
| | = | (T−λ I)(I+(λ−µ)(T−λ I)−1).
|
|
The last line is an invertible operator because T−λ I is
invertible by the assumption and I+(λ−µ)(T−λ
I)−1 is invertible by the
previous Lemma, since
||(λ−µ)(T−λ I)−1||<1 if
є<||(T−λ I)−1||.
□
Exercise 7
-
Prove the first resolvent
identity:
|
R(λ,T)−R(µ,T)=(λ−µ)R(λ,T)R(µ,T)
(43) |
- Use the identity (43) to show that
(T−µ I)−1→
(T−λ I)−1 as µ→ λ.
-
Use the identity (43) to show that
for z∈ρ(t) the complex derivative d/dz R(z,T)
of the resolvent R(z,T) is well defined, i.e. the resolvent is
an analytic function operator valued function of z.
Lemma 8
The spectrum is non-empty.
Proof.
Let us assume the opposite, σ(
T)=∅ then the resolvent
function
R(λ,
T) is well defined for all
λ∈ℂ. As could be seen from the von Neumann
series (
42) ||
R(λ,
T)||→ 0
as λ→ ∞. Thus for any vectors
x,
y∈
H the function
f(λ)=⟨
R(λ,
T)
x,
y)
⟩ is
analytic (see Exercise
3)
function tensing to zero at infinity. Then by the Liouville theorem from
complex analysis
R(λ,
T)=0, which is impossible. Thus the
spectrum is not empty.
□
Proof.[Proof of Theorem
3]
Spectrum is nonempty by Lemma
8 and compact
by Corollary
6.
□
7.2 The spectral radius formula
The following definition is of interest.
Definition 10
The spectral radius of T is
| r(T)=sup{ | ⎪
⎪ | λ | ⎪
⎪ | : λ∈ σ(T)}.
|
From the Lemma 1 immediately follows that
r(T)≤||T||. The more accurate estimation is given by the
following theorem.
Theorem 11
For a bounded operator T we have
|
r(T)= | | ⎪⎪
⎪⎪ | Tn | ⎪⎪
⎪⎪ | 1/n.
(44) |
We start from the following general lemma:
Lemma 12
Let a sequence (
an)
of positive real numbers satisfies
inequalities: 0≤
am+n≤
am+
an for all m and
n. Then there is a limit
lim
n→∞(
an/
n)
and its equal to
inf
n(
an/
n)
.
Proof.
The statements follows from the observation that for any n and
m=nk+l with 0≤ l≤ n we have am≤ kan+la1
thus, for big m we got am/m≤ an/n +la1/m ≤
an/n+є.
□
Proof.[Proof of Theorem
11]
The existence of the limit
lim
n→∞||
Tn||
1/n
in (
44) follows from the previous Lemma
since by the Lemma
12
log||
Tn+m||≤ log||
Tn||+log||
Tm||. Now we
are using some results from the complex analysis. The Laurent series
for the resolvent
R(λ,
T) in the neighbourhood of infinity
is given by the von Neumann series (
42). The
radius of its convergence (which is equal, obviously, to
r(
T))
by the Hadamard theorem is exactly
lim
n→∞||
Tn||
1/n.
□
Corollary 13
There exists λ∈σ(
T)
such that
| λ |=
r(
T)
.
Proof.
Indeed, as its known from the complex analysis the boundary of the
convergence circle of a Laurent (or Taylor) series contain a
singular point, the singular point of the resolvent is obviously
belongs to the spectrum.
□
Example 14
Let us consider the left shift
operator S*, for any λ∈ℂ
such that
| λ | <1
the vector
(1,λ,λ
2,λ
3,…)
is in l2
and is an eigenvector of S* with eigenvalue λ
, so the
open unit disk | λ |<1
belongs to σ(
S*)
. On
the other hand spectrum of S* belongs to the closed unit disk
| λ |≤ 1
since r(
S*)≤
||
S*||=1
. Because spectrum is closed it should coincide with
the closed unit disk, since the open unit disk is dense in
it. Particularly 1∈σ(
S*)
, but it is easy to see that 1
is not an eigenvalue of S*.
Proposition 15
For any T∈B(H) the spectrum of the adjoint operator
is σ(T*)={λ: λ∈ σ(T)}.
Proof.
If (T−λ I)V=V(T−λ I)=I the by taking adjoints
V*(T*−λI)=(T*−λI)V*=I. So λ
∈ ρ(T) implies λ∈ρ(T*), using the property
T**=T we could invert the implication and get the statement
of proposition.
□
Example 16
In continuation of Example 14 using the
previous Proposition we conclude that σ(
S)
is also the
closed unit disk, but S does not have eigenvalues at all!
7.3 Spectrum of Special Operators
Theorem 17
-
If U is a unitary operator then σ(U)⊆
{| z |=1}.
-
If T is Hermitian then σ(T)⊆ ℝ.
Proof.
- If | λ |>1 then ||λ−1U||<1 and
then λ I−U=λ(I−λ−1U) is invertible, thus
λ∉σ(U).
If | λ |<1 then ||λ U*||<1 and
then λ I−U=U (λ U*−I) is invertible, thus
λ∉σ(U). The remaining set is exactly
{z:| z |=1}.
- Without lost of generality we
could assume that ||T||<1, otherwise we could multiply
T by a small real scalar. Let us consider the Cayley
transform which maps real axis to the unit circle:
Straightforward calculations show that U is unitary if
T is Hermitian. Let us take λ∉ℝ and
λ≠ −i (this case could be checked directly by
Lemma 4). Then the Cayley transform
µ=(λ−i)(λ+i)−1 of
λ is not on the unit circle and thus the operator
| U−µ I=(T−iI)(T+iI)−1−(λ−i)(λ+i)−1I=
2i(λ+i)−1(T−λ I)(T+iI)−1,
|
is invertible, which implies invertibility of T−λ I. So
λ∉ℝ.
□
The above reduction of a self-adjoint operator to a unitary one (it
can be done on the opposite direction as well!) is an important tool
which can be applied in other questions as well, e.g. in the following
exercise.
Exercise 18
-
Show that an operator U: f(t) ↦ eitf(t) on
L2[0,2π] is unitary and has the
entire unit circle {| z |=1} as its spectrum .
- Find a self-adjoint operator T with the entire real line
as its spectrum.
Last modified: November 6, 2024.