Lecture 7: Consequences of the Spectral Theorem; the Fredholm Alternative

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We begin by recording some nice consequences of the spectral theorem for s.a. compact operators proved in the last lecture. We will use the notation

$\sigma (T)$ for the set of eigenvalues of a compact operator

$T$.

Functional Calculus. If

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$f:\sigma (T)\to \mathbb{R}$ is any bounded function and

$T\in K(H)$ is selfadjoint with eigendecomposition

${\lambda }_n,{\varphi }_n$, define

$$f(T):=\sum _{n}f({\lambda }_n){\varphi }_n{\varphi }_n^{\ast }.$$
Then the above definition is an algebra homomorphism, in that

$$af(T)+bg(T)=(af+bg)(T)$$

$$f(T)g(T)=g(T)f(T)=(fg)(T)$$
for such functions

$f,g$.

The proof follows immediately from orthogonality of the eigenvectors

${\varphi }_n$. Since the norm of a diagonal operator is its largest (in magnitude) entry, we have for compact s.a.

$T$:

$$\Vert f(T)\Vert =\underset{\lambda \in \sigma (T)}{max}|f(\lambda )|.$$ Some interesting consequences, by considering

$f(t)=\sqrt{t}$ and

$f(t)=|t|$ are that for a compact s.a. operator

$T$,

$\sqrt{T}$ and

$|T|$ may be calculated "eigenvalue-wise" using the above calculus, and these are also are compact since

$f(t)\to 0$ as

$t\to 0$ for these functions.

Theorem 1. (Singular Value Decomposition) If

$T\in K(H)$, there are orthnormal sets

${\psi }_n$ and

${\varphi }_n$ and nonnegative numbers

${\sigma }_n\to 0$ such that

$$T=\sum _n{\sigma }_n{\psi }_n{\varphi }_n^{\ast }.$$

Proof. By the polar decomposition, we have

$T=U|T|$, and

$|T|$ is compact by the functional calculus. Since it is also selfadjoint, the spectral theorem yields

$$|T|=\sum _n{\sigma }_n{\psi }_n{\psi }_n^{\ast }$$
for some nonzero eigenvalues

${\sigma }_n$ which must be positive since

$|T|\ge 0$. Moreover

$U$ is an isometry on

$\ker (|T|{)}^{\perp }$, which contains all the

${\varphi }_n$; so

${\psi }_n:=U{\varphi }_n$ is an orthonormal set. This gives the desired result.

$◻$

The second consequence is a variational characterization of the eigenvalues, often useful in proving bounds when it is hard to calculate them exactly (which it often is).

Theorem 2. (Courant-Fisher) If

$T\ge 0$ is compact, then its eigenvalues may be ordered

${\lambda }_1\ge {\lambda }_2,\dots {\lambda }_k\dots$. We then have:

$${\lambda }_k=\underset{\dim (V)=k}{max}\underset{x\in V-\{0\}}{min}(x,Tx)/\Vert x{\Vert }^2$$

$$=\underset{\dim (V)=k}{min}\underset{x\in V^{\perp }-\{0\}}{max}(x,Tx)/\Vert x^2\Vert .$$
where the max/min is over subspaces of

$H$.

This theorem is also true and extremely useful in finite dimensions. The proof is left to the homework. Two of its important conceptual consequences, immediate since the right hand sides are expressed entirely in terms of quadratic forms, are:

1. If
   $0\le A\le B$ for compact operators, then
   ${\lambda }_k(A)\le {\lambda }_k(B)$ for every
   $k$.
2. If
   $\Vert A-B\Vert \le ϵ$ for positive compact operators, then
   $|{\lambda }_k(A)-{\lambda }_k(B)|\le ϵ$.

The second part implies in particular that if finite rank

$T_n\to T\ge 0$, then for every

$k$

${\lambda }_k(T_n)\to {\lambda }_k(T)$. The result may be extended to nonpositive (but still selfadjoint) operators by writing

$A=A_{+}-A_{-}$ for

$A_{+},A_{-}\ge 0$ using the spectral theorem.

Trace-Class and Hilbert-Schmidt Operators

Definition. If

$A\in L(H)$ is positive and

${\varphi }_n$ is an ONB of

$H$ then

$$\mathrm{tr}(A):=\sum _n({\varphi }_n,A{\varphi }_n)\in [0,\mathrm{\infty }].$$

The trace is well-defined as a function of

$A$ because the above definition is independent of the choice of ONB; see RS Theorem VI.18 for a proof. Note that the sum above may not necessarily converge, though it converges absolutely whenever it does due to positivity; there can be more subtle convergence issues when dealing with operators which are not positive. The trace is well-behaved on a restricted class of operators defined below.

An operator is called trace class (denoted

$S_1$ for "Schatten-1") if

$$\mathrm{tr}(|A|)<\mathrm{\infty }$$
and Hilbert-Schmidt (denoted

$S_2$ for "Schatten-2") if

$$\mathrm{tr}(|A{|}^2)<\mathrm{\infty }.$$

Theorem 3. If

$\mathrm{tr}(|A{|}^p)<\mathrm{\infty }$ for

$p=1$ (resp.

$p=2$) then

$A$ is compact.
Proof. Choose any ONB

${\varphi }_n$ and notice that

$$\sum _n({\varphi }_n,|A{|}^p{\varphi }_n)=\sum _n\Vert |A{|}^{p/2}{\varphi }_n|{|}^2<\mathrm{\infty }.$$
Let

$P_N$ be the projection onto the span of the first

$N$ basis vectors. For any

$\Vert x\Vert =1$:

$$\Vert (I-P_N)|A{|}^{p/2}x\Vert =\Vert \sum _{n>N}{\varphi }_n({\varphi }_n,|A{|}^{p/2}x)\Vert =\sum _{n>N}|((|A{|}^{p/2}{)}^{\ast }{\varphi }_n,x){|}^2$$

$$\le \sum _{n>N}\Vert |A{|}^{p/2}{\varphi }_n{\Vert }^2\Vert \Vert x{\Vert }^2\to 0,$$
uniformly in

$x$, where the last inequality is Cauchy-Schwartz and we used

$|A|=|A{|}^{\ast }$.
Thus,

$|A{|}^{p/2}$ is a limit of finite rank operators and must be compact; so

$\sqrt{|A|}$ (resp.

$|A|$) is compact. Since compact operators are an ideal,

$|A|$ is compact in both cases, whence

$A=U|A|$ is compact.

$◻$

Remark. The above proof actually shows that

$\mathrm{tr}(|A{|}^p)<\mathrm{\infty }$ for any

$p>0$ is sufficient for compactness, since

$|A{|}^{p/2}$ compact implies

$|A|$ compact by the functional calculus with

$f(t)=t^{2/p}$.

Corollary. A trace class operator is Hilbert-Schmidt.
Proof. If

$T\in S_1$ then

$|T|$ is compact, so by the SVD and invariance of the trace

$\mathrm{tr}(|T|)=\sum _n{\sigma }_n$, where

${\sigma }_n$ are the singular values of

$T$. But since this is convergent,

$\mathrm{tr}(|T{|}^2)=\sum _n{\sigma }_n^2$ must also be convergent.

$◻$

The punch line is that the definition (and orthogonal invariance) of the trace extends to all of

$S_1$ (including not necessarily positive operators). The following theorem is proven on the homework.

Theorem 4. If

$A$ is trace-class, then for any ONB

${\varphi }_n$,

$$\mathrm{tr}(A):=\sum _n({\varphi }_n,A{\varphi }_n)$$
is independent of the choice of ONB.

This implies in particular that

$\mathrm{tr}(A)=\sum _n{\lambda }_n$ for any any selfadjoint trace class

$A$.

Remark. There is a deep theorem of Lidskii which shows that

$\mathrm{tr}(A)=\sum _n{\lambda }_n$ for any trace class

$A$ (not necessarily self adjoint!).

The Fredholm Alternative

In finite dimensional linear algebra, the solvability of the equation

$Ax=b$ is nicely characterized by the fact that

$$\mathrm{ran}(A)=\ker (A^{\ast }{)}^{\perp },$$
but this is no longer true in infinite dimensions, since the range may not be closed as we saw with the example of the Volterra operator. The Fredholm alternative states that the finite-dimensional relation can be recovered for operators of type

$\lambda I-T$ where

$T$ is compact and

$\lambda \ne 0$. We will denote

$\lambda I$ by

$\lambda$ as is customary.

Theorem 5. (Fredholm Alternative) If

$T\in K(H)$ and

$\lambda \in \mathbb{C}-\{0\}$ then for

$A=\lambda -T$:

1. $\mathrm{ran}(A)=\overline{\mathrm{ran}(A)}=\ker (A^{\ast }{)}^{\perp }.$
2. $\dim (\ker (A))=\dim (\ker (A^{\ast }))<\mathrm{\infty }.$

This theorem has the following important conceptual consequences:

1. $\lambda$ is an eigenvalue of
   $T$ iff
   $\bar{\lambda }$ is an eigenvalue of
   $T^{\ast }$ (since these correspond to
   $\ker (A)\ne \{0\}$ and
   $\ker (A^{\ast })\ne \{0\}$.
2. $(\lambda -T)x=b$ has a solution for every
   $b$ iff
   $\lambda$ is not an eigenvalue of
   $T$. Otherwise, the set of
   $b$ for which a solution exists is precisely
   $\ker (A^{\ast }{)}^{\perp }$.

Proof of (1).
Suppose

$$y_n=(\lambda -T)x_n(\ast )$$
is a Cauchy sequence in

$R:=\mathrm{ran}(\lambda -T)$, converging to some

$y\in H$. We want to show that

$y\in R$.

Let

$V=\ker (\lambda -T)$, and note that

$V$ is closed. Decompose each vector

$x_n=u_n+v_n$ for

$v_n\in V^{\perp }$. Assume for now that

$v_n$ is a bounded sequence. By compactness,

$T{v}_n$ must have a convergent subsequence; passing to this subsequence, let

$T{v}_n\to x$. Since

$u_n\in \ker (\lambda -T)$, equation (*) now implies

$$y_n=\lambda v_n-T{v}_n\Rightarrow \lambda v_n=y_n+T{v}_n\to y+x,$$
as

$n\to \mathrm{\infty }$. Since

$\lambda \ne 0$ we have

$v_n\to (x+y)/\lambda$, which satisfies

$$(\lambda -T)v=\underset{n}{lim}(\lambda -T)v_n=y,$$
as desired.

To see that

$v_n$ is bounded, assume for contradiction that

$\Vert v_n\Vert \to \mathrm{\infty }.$ The normalized sequence

$v_n/\Vert v_n\Vert$ is bounded, so

$T{v}_n/\Vert v_n\Vert$ must have a convergent subsequence by compactness; let

$z$ be the limit of this sequence.

Observe that again by (*):

$$T{v}_n/\Vert v_n\Vert =\lambda v_n/\Vert v_n\Vert -y_n/\Vert v_n\Vert .$$
Since the LHS converges to

$z$, the RHS must also converge to

$z$, revealing that

$\Vert z\Vert =|\lambda |\ne 0$ and

$z\in V^{\perp }$ since

$v_n\in V^{\perp }$ for all

$n$ and that subspace is closed.

On the other hand, we also have

$$(\lambda -T)z=\underset{n}{lim}(\lambda -T)T{v}_n/\Vert v_n\Vert =\underset{n}{lim}T{y}_n/\Vert v_n\Vert =0$$
since

$\Vert v_n\Vert \to \mathrm{\infty }$ and

$T{y}_n$ is bounded, so

$z\in V$, a contradiction.

$◻$

Proof of (2). Given

$\lambda \ne 0$, decompose

$T=T_0+T_1$ where

$\Vert T_0\Vert <|\lambda |$ and

$T_1$ is finite rank (this is possible because

$T$ is a norm limit of finite rank operators). Observe that since the power series

$$(1-x{)}^{-1}=\sum _{k\ge 0}{x}^k=:S(x)$$
is absolutely geometrically convergent in

$\{|x|<1\}$ and satisfies

$(1-x)S(x)=1$ for such

$x$, we have that

$$S(T_0/\lambda )=\sum _{k\ge 0}(T_0/\lambda {)}^k$$
is convergent in norm (by applying the triangle inequality to

$S$ minus its partial sums). Thus,

$(1/\lambda )S(T_0/\lambda )=(\lambda -T_0{)}^{-1}\in L(H)$. We now have

$$(\lambda -T_0{)}^{-1}(\lambda -T)=I-(\lambda -T_0{)}^{-1}{T}_1=:I-T_2,$$
where

$T_2$ has finite rank. As left or right multiplying by an invertible transformation doesn't change the dimension of the kernel,

$$\dim (\ker (\lambda -T))=\dim (\ker (I-T_2))$$
and

$$\dim (\ker ({\lambda }^{\ast }-T^{\ast }))=\dim (\ker (I-T_2^{\ast }))$$
the latter can be shown to be equal by elementary linear algebra (homework) since

$T_2$ is finite rank.

$◻$