# The spectral theory for compact self adjoint operators

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```					   • The spectral theory for compact, self-adjoint operators

– Spectrum
– Compact operators
– The spectral theorem
– Hilbert-Schmidt operators
– Functions of operators

Deﬁnitions:
In ﬁnite dimensions:

• eigenvalue, eigenvector, diagonalizable, spectrum, normal matrix (n ×
n matrix is normal iﬀ Cn has an orthonormal basis of eigenvectors),
multiplicity
Bounded linear operators on inﬁnite-dimensional Hilbert spaces:

• Resolvant Set: A ∈ B(H), ρ(A) = {λ ∈ C|(A−λI) : H → H is one-to-one and onto.}

• Spectrum: A ∈ B(H), σ(A) = C\ρ(A).

• A ∈ B(H),

– The point spectrum of A is all λ ∈ σ(A) such that A − λI is not
one-to-one, and λ is an eigenvalue of A.
– The continuous spectrum of A is all λ ∈ σ(A) such that A − λI is
one-to-one but not onto and ran(A − λI) is dense in H.
– The residual spectrum is λ ∈ σ(A) such that A − λI is one-to-one
but not onto and ran(A − λI) is not dense in H.

• If λ ∈ ρ(A) then A−λI has an everywhere deﬁned and bounded inverse,
Rλ = (λI−A)−1 is called the resolvant of A, which is an operator-valued
function deﬁned on ρ(A) ∈ C.

• The spectral radius of A, r(A), is the radius of the smallest disk centered
at zero which contains σ(A),

r(A) = sup{|λ| | λ ∈ σ(A)}.
• A linear subspace M of H is called an invariant subspace of a linear
operator A on H if Ax ∈ M for all x ∈ M.

• A bounded linear operator A on a separable Hilbert space H is Hilbert-
Schmidt if there is an ONB {en | n ∈ N} such that
∞
2
Aen       < ∞.
n=1

Furthermore, if A is a Hilbert-Schmidt operator then

∞
A         =           Aen    2
HS
n=1

is called the Hilbert-Schmidt norm of A.

Theorems:

• If A ∈ B(H), then ρ(A) is an open subset of C that contains the
exterior disc {λ ∈ C | |λ| > A }. The resolvant Rλ is an operator
valued analytic function of λ deﬁned on ρ(A).
For any λ0 ∈ ρ(A) write

λI − A = (λ0 I − A)[I − (λ0 − λ)( 0I − A)−1 ].

Then for small enough |λ0 − λ| invert the operator on RHS above by
the Neumann series and the resolvant is given by an operator-norm
convergent Taylor series in the disc so it is analytic where deﬁned. If
| > A then the same Neumann series shows that Rλ = λ(I − A/λ)
is invertible so λ ∈ ρ(A).

• If A ∈ B(H) then
r(A) = lim An          1/n
n→∞

and if A is self-adjoint then r(A) = A .

• The spectrum of a bounded linear operator on a Hilbert space is nonempty
(may just be zero, if r(A) = 0 then A is called nilpotent.)
Proof uses Liouville’s theorem from complex analysis which states that
a bounded entire function is constant. If A ∈ B(H) then Rλ the re-
solvant is an analytic function on the resolvant set. Deﬁne f (λ) =
x, Rλ y , for all x, y ∈ H. then f is analytic in ρ(A) and limλ→∞ f (λ) =
0. If σ(A) is empty f is a constant function but by it’s limiting value
it must be zero. This would imply Rλ = 0 for all λ ∈ C which is a
contradiction, so σ(A) is not empty.

• The eigenvalues of a bounded, self-adjoint operator are real, and eigen-
vectors associated with diﬀerent eigenvalues are orthogonal.

λ x, x = x, Ax = Ax, x = λ x, x

and

λ x, y = Ax, y = x, Ay = µx, y

So if λ = µ then x, y = 0

• If A is a bounded, self-adjoint operator on H and M is an invariant
subspace of A then M⊥ is also invariant.
If x ∈ M⊥ and y ∈ M then

y, Ax = Ay, x = 0

so Ax ∈ M⊥ .

• If λ is in the residual spectrum of a bdd operator A on a Hilbert space
then λ is an eigenvalue of A∗ .
This follows from a general theorem about bounded linear operators,
namely
ranA = (kerA∗ )⊥ , kerA = (ranA∗ )⊥ .

• If A is bdd, self-adjoint on a Hilbert space then the spectrum is real
and contained in the interval [− A , A ].

• The residual spectrum of a bounded, self-adjoint operator is empty.
• A non-zero eigenvalue of a compact, self-adjoint operator has ﬁnite
multiplicity. A countably inﬁnite set of nonzero eigenvalues has zero as
an accumulation point and no other accumulation points.
If λ = 0 has inﬁnite multiplicity then there is a sequence (en ) of or-
thonormal eigenvectors, which is bounded but (Aen ) has no convergent
subsequence since Aen = λen and this contradicts that A is compact.
Furthermore if A has a countably inﬁnite set {λn } of non-zero eigen-
values then since they are bdd by A there is a convergent subseq.
(λnk ). If λnk → λ = 0 then the orthogonal sequence of eigenvectors
((fnk ) where fnk = λ−1 enk with enk = 1) would be bdd. But (Afnk )
nk
has no convgt subsequence since Afnk = enk .

• Spectral theorem for compact, self-adjoint operators: Let A : H → H
be a compact, self-adjoint operator on a Hilbert space H. There is an
orthonormal basis of H consisting of eigenvectors of A. The nonzero
eigenvalues of A form a ﬁnite or countably inﬁnite set {λk } or real
numbers and
A=       λk Pk
k

where Pk is the orthogonal projection onto the ﬁnite-dimensional eigenspace
of eigenvectors with eigenvalue λk . If the number of nonzero eigenval-
ues is countably inﬁnte then the series above converges to A in the
operator norm .
Idea: First prove that if A is compact and self-adjoint then λ = A
or − A (or both) is an eigenvalue of A. Do this by writing

A = sup | x, Ax |,
x =1

and look at sequence leading to the sup. By self-adjoint xn , Axn is
real so taking a further subsequence have

lim xn , Axn = λ
n→∞

where λ = ± A . The sequence is bounded and so by compactness of A
there is a subsequence such that (Axn ) converges. Let y = limn→∞ Axn .
Now show that y is an eigenvector of A with eigenvalue λ.
Compute:
2                                 2
(A − λI)y          = lim (A − λI)Axn
n→∞
2                                 2
≤ A      lim (A − λI)xn
n→∞
2
= A lim        Axn           + λ 2 xn      2
− 2λ xn , Axn
n→∞
2              2            2
≤ A      lim    A        xn           + λ 2 xn   2
− 2λ xn , Axn   .
n→∞
2
= A       2
λ + λ2 − 2λ2 .

Finish the proof by using an orthogonal decomposition of H into in-
variant subspaces and apply the above result to smaller and smaller
subspaces. Using previous results the eigenvalues have ﬁnite multiplic-
ities and limn→∞ λn = 0 so can construct the series
n
A=          λk Pk + An+1 ,
k=1

where An+1 is zero on the subspace spanned by the previously con-
structed {e1 , ...., en }. Furthermore An+1 = |λn+1 | → 0.
Generalizations for compact, normal operators hold, eigenvectors are
orthogonal but may be complex. Also can generalize for bounded,
self-adjoint or normal operators. There the sum must be replaced by
an integral with respect to an appropriate measure to account for the
possibility of a continuous spectrum.
• To show an operator is compact, one can verify directly that if E is
bounded in H then A(E) is precompact (using Arzela-Ascoli or Rellich’s
theorem or... )
other criteria:
• Let E be a subset of an inﬁnite dimensional separable Hilbert space.
If E is precompact, then for every orthonormal set {en | n ∈ N]} and
every ε > 0 there is an N such that

sum∞ +1 | en , x |2 < ε for allx ∈ E.
n=N

Also if E is bounded and there is an ONB such with the abover property
then E is precompact.
ﬁrst show if E is bounded and the condition doesn’t hold then E is not
precompact. Construct an explicit subsequence using Parseval. For the
second part use a diagonalization argument to contruct a convergent
subsequence showing precompactness.

• Hilbert-Schmidt operators are compact.
2             2
Diagonal operators A :        (N) →         (N) deﬁned by

A(x1 , x2 , ...) = (λ1 x1 , λ2 x2 , ...)

with λn ∈ C are compact iﬀ λn → 0 as n → ∞. Any compact, normal
operator on a separable Hilbert space is unitarily equivalent to such a
diagonal operator. A diagonal operator is H-S iﬀ

sum∞ |λn |2 < ∞.
n=1

An operator A is called trace class if
∞
|λn | < ∞.
n=1

A trace class operator is Hilbert-Schmidt and a Hilbert-Schmidt oper-
ator is compact.

• A BLO on a H-space is compact iﬀ it maps weakly convergent sequences
into strongly convergent sequences.

• Spectral Mapping: If A is a compact, self-adkoint operator on a Hilbert
space and f : σ(A) → C is continuous then

σ(f (A)) = f (σ(A)).
Problems:

1. Suppose that P : H → H is a bounded, not necessarily self-adjoint
projection (meaning that P 2 = P ) and P = 0, I. For λ ∈ C, compute
(λI − P )−1 explicitly when it exists, and show that the spectrum of P
is the set {0, 1}. HINT. Consider the series expansion of (λI − P )−1 .

2. Let U = 2 (Z) → L2 (T) be the unitary map from a complex sequence
c = (cn )n∈Z to the function f : T → C whose sequence of Fourier
coeﬃcients is c :
1
(U c)(x) = √        cn einx .
2π n∈Z
2           2
Deﬁne the right-shift map S :       (Z) →       (Z) by

Sc = b, where c = (cn ), b = (bn ), and bn = cn−1 .

Deﬁne the multiplication operator M : L2 (T) → L2 (T) by

(M f )(x) = eix f (x).

(a) Show that S = U −1 M U.
(b) Determine the spectrum of M, and deduce the spectrum of S.
Classify the spectrum.

3. (W’08) Suppose that A : H → H is a bounded linear operator on a
(complex) Hilbert space H with spectrum σ(A) ⊂ C and resolvant set
ρ(A) = C\σ(A). For µ ∈ ρ(A), let

R(µ, A) = (µI − A)−1

denote the resolvant operator of A.

(a) If µ ∈ ρ(A) and
1
|ν − µ| <           ,
R(µ, A)
prove that ν ∈ ρ(A) and

R(ν, A) = [I − (µ − ν)R(µ, A)]−1 R(µ, A).
(b) If µ ∈ ρ(A), prove that
1
R(µ, A) ≥
d(µ, σ(A))

where
d(µ, σ(A)) = inf |µ − λ|
λ∈σ(A)

is the distance of µ from the spectrum of A.

4. (W ’07) Let U be a unitary operator on a Hilbert space. Prove that
the spectrum of U lies on the unit circle.

5. (W’05) Let 2 (Z) denote the complex Hilbert space of sequences xn ∈ C,
n ∈ Z, such that
∞
|xn |2 < ∞.
n=−∞
2           2
Deﬁne the shift operator S :       (Z) →       (Z) by

S((xn )) = (xn+1 ).

Show that S has no eigenvalues.

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Description: The spectral theory for compact self adjoint operators