# Lecture 7: Consequences of the Spectral Theorem; Hilbert-Schmidt and Trace-Class Operators
###### tags: `224a 2020`
$\newcommand{\tr}{\mathrm{tr}}$
$\newcommand{\ran}{\mathrm{ran}}$
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 $\renewcommand{\R}{\mathbb{R}}$ $f:\sigma(T)\rightarrow\R$ is any bounded function and $T\in K(H)$ is selfadjoint with eigendecomposition $\lambda_n,\phi_n$, define
$$f(T):=\sum_n f(\lambda_n)\phi_n\phi_n^*.$$
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 $\phi_n$. Since the norm of a diagonal operator is its largest (in magnitude) entry, we have for compact s.a. $T$:
$$\|f(T)\|=\max_{\lambda\in\sigma(T)} |f(\lambda)|.$$ 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 $f(t)\rightarrow 0$ as $t\rightarrow 0$ for these functions.
**Theorem 1. (Singular Value Decomposition)** If $T\in K(H)$, there are orthnormal sets $\psi_n$ and $\phi_n$ and nonnegative numbers $\sigma_n\rightarrow 0$ such that
$$T = \sum_n \sigma_n \psi_n \phi_n^*.$$
*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
$$|T| = \sum_n \sigma_n \psi_n\psi_n^*$$
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 $\phi_n$; so $\psi_n := U\phi_n$ is an orthonormal set. This gives the desired result. $\square$
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 $\lambda_1\ge \lambda_2,\ldots\lambda_k\ldots$. We then have:
$$ \lambda_k = \max_{\dim(V)=k}\min_{x\in V-\{0\}} (x,Tx)/\|x\|^2$$
$$= \min_{\dim(V)=k}\max_{x\in V^\perp-\{0\}} (x,Tx)/\|x^2\|.$$
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 $\|A-B\|\le \epsilon$ for positive compact operators, then $|\lambda_k(A)-\lambda_k(B)|\le \epsilon$.
The second part implies in particular that if finite rank $T_n\rightarrow T\ge 0$, then for every $k$ $\lambda_k(T_n)\rightarrow \lambda_k(T)$. 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 $\phi_n$ is an ONB of $H$ then
$$ \tr(A):=\sum_n (\phi_n,A\phi_n) \in [0,\infty].$$
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
$$ \tr(|A|)<\infty$$
and Hilbert-Schmidt (denoted $S_2$ for "Schatten-2") if
$$ \tr(|A|^2)<\infty.$$
**Theorem 3.** If $\tr(|A|^p)<\infty$ for $p=1$ (resp. $p=2$) then $A$ is compact.
*Proof.* Choose any ONB $\phi_n$ and notice that
$$\sum_n (\phi_n,|A|^p\phi_n) = \sum_n \||A|^{p/2}\phi_n||^2<\infty.$$
Let $P_N$ be the projection onto the span of the first $N$ basis vectors. For any $\|x\|=1$:
$$ \|(I-P_N)|A|^{p/2}x\| = \|\sum_{n>N} \phi_n (\phi_n, |A|^{p/2}x)\| = \sum_{n>N} |((|A|^{p/2})^*\phi_n, x)|^2$$
$$ \le \sum_{n>N} \||A|^{p/2}\phi_n\|^2\|\|x\|^2\rightarrow 0,$$
uniformly in $x$, where the last inequality is Cauchy-Schwartz and we used $|A|=|A|^*$.
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.$\square$
**Remark.** The above proof actually shows that $\tr(|A|^p)<\infty$ 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 $\tr(|T|)=\sum_n \sigma_n$, where $\sigma_n$ are the singular values of $T$. But since this is convergent, $\tr(|T|^2)=\sum_n \sigma_n^2$ must also be convergent.$\square$
The punch line is that the definition (and orthogonal invariance) of the trace extends to all of $S_1$ (including not necessarily positive operators).
**Theorem 4.** If $A$ is trace-class, then for any ONB $\phi_n$,
$$\tr(A):=\sum_n(\phi_n, A \phi_n)$$
is independent of the choice of ONB.
This implies in particular that $\tr(A)=\sum_n \lambda_n$ for any any selfadjoint trace class $A$.
**Remark.** There is a deep theorem of Lidskii which shows that $\tr(A)=\sum_n \lambda_n$ for *any* trace class $A$ (not necessarily self adjoint!).
See the notes for a proof that $S_1$ and $S_2$ are linear subspaces and ideals in $L(H)$.

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