# Lecture 6: Spectral Theorem for Compact Operators

## More on Compact Operators

In the last lecture we defined compact operators, and stated that they are equivalent to norm limits of finite rank operators. We begin by proving one direction of this theorem (the other direction is left to the homework).

*Proof.* Assume $T\in K(H)$, let $B\subset H$ be the unit ball, and let $K=\overline{T(B)}$. Let $n$ be an integer. The set
$$\cup_{z\in K} \{ y: \|z-y\|<1/n\}$$
is an open cover of $K$ so it must have a finite subcover; call this $S_n$. Let $P_n$ be the orthogonal projection onto $span(S_n)$. Note that $P_nT$ is a finite rank operator since $S_n$ is finite.

Consider a point $x\in H$. Choose $z\in S_n$ such that $\|Tx-z\|<1/n$. Since $P_nTx$ is the closest point to $Tx$ in $span(S_n)$, we must have
$$\|P_nTx-Tx\|\le \|z-Tx\|<1/n,$$
whence $\|P_nT-T\|<1/n$, so that $\|P_nT-T\|\rightarrow 0$ as $n\rightarrow \infty.$. $\square$

Two interesting families of compact operators are the following.

**Integral Kernel Operators.** Suppose $K\in L^2(0,1)^2$. A Cauchy-Schwartz exercise shows that $\|T_K\|\le \|K\|_{L^2(0,1)^2}$, where
$$T_Kf:=\int_0^1 K(x,y)f(y)dy.$$
Let $\{\phi_i\}$ be an ONB of $L^2(0,1)$. It is easy to check that the family of bivariate functions $\{\phi_i(x)\phi_j(y)\}_{i,j}$ is an ONB of $L^2(0,1)^2$, so we may expand
$$K(x,y)=\sum_{ij} c_{ij} \phi_i(x)\phi_j(y).$$
Let $K_n(x,y)=\sum_{i,j\le n} c_{ij}\phi_i(x)\phi_j(y).$ As the difference $T_{K_n}-T_K$ is an integral kernel operator of the same type, we have
$$\|T_K-T_{K_n}\|\le \|K-K_n\|_{L^2}=\sum_{i,j>n}|c_{ij}|^2\rightarrow 0$$
as $n\rightarrow \infty$, showing that $T_k$ is compact by the previous theorem.

**Diagonal Operators.** If $\alpha_n$ is any sequence with $|\alpha_n|\rightarrow 0$, then the diagonal multiplication operator $T(e_i)=\alpha_i e_i$ is compact, since it is approximated by $T_n(e_i)=\alpha_i e_i \{i\le n\}$.

Note that diagonal multiplication operators in $L^2(0,1)$, $T_gf = g(x)f(x)$ for $g\in L^\infty(0,1)$ are *not* compact, as seen by considering $g$ to be the indicator of any set of positive Lebesgue measure, since then the image $\overline{T_g(B)}$ contains infinitely many vectors pairwise separated by a constant distance.

Finally, we mention that the set $K(H)$ of compact operators on $H$ is a 2-sided $*-$ideal in $L(H)$, which means that if $T\in K(H)$ then $T^*, BT, TB\in K(H)$ for any $B\in L(H)$. These properties are easy to verify from the characterization as of $K(H)$ as norm limits of finite rank operators.

## The Spectral Theorem for Compact Operators
We will now show that the diagonal operators above are in a sense the *only* examples of compact operators, up to isomorphism.

**Definition.** If $Tf=\lambda f$ for some $f\in H\setminus\{0\}$ then $f$ is called an eigenvector of $T$ and $\lambda$ is called an eigenvalue.

**Theorem.** Suppose $T=T^*\in K(H)$. Then:
1. The eigenvalues of $T$ are real and may be ordered $\lambda_1,\lambda_2,\ldots \rightarrow 0$.
2. If $\lambda\neq 0$ is an eigenvalue of $T$, the eigenspace $V_\lambda:=\ker(\lambda-T)$ is finite dimensional.
3. There is an orthonormal basis of $H$ consisting of eigenvectors of $T$.

The last property implies that we have the expansion
$$T=\sum_n \lambda_n \phi_n \phi_n^*$$
where $\phi_n^*$ denotes the linear functional $(\phi_n,x)$ dual to $\phi_n$. Alternately, this can be expressed as
$$T = UDU^*$$
where $U(e_n)=\phi_n$ is unitary and $D$ is a diagonal multiplication operator.

*Proof of Theorem.* We begin by observing that for any eigenvectors $Tv=\lambda v$ and $Tw=\mu w$:
$$ \mu(v,w) = (v,Tw) = (T^*v,w)=(Tv,w)=\overline{\lambda}(v,w).$$
Plugging in $v=w$ shows that $\lambda=\overline{\lambda}$ so $\lambda$ must be real. For $\mu\neq \lambda$ we then must have $(v,w)=0$, so eigenvectors from distinct eigenvalues must be orthogonal.

**Lemma 1.** For every $\epsilon>0$, the subspace
$$S_\epsilon := span\{x: Tx=\lambda x, |\lambda|\ge \epsilon\}$$
is finite dimensional.
*Proof of Lemma.* We first show that for every nonzero eigenvalue $\lambda$, $V_\lambda$ is finite dimensional. Assume not, i.e., there is an infinite sequence of orthonormal vectors $\{x_n\}$ such that $Tx_n=\lambda x_n$. Then $Tx_n\in K=\overline{T(B)}$ and we have
$$ \|Tx_n-Tx_m\|=|\lambda|\|x_n-x_m\|=\sqrt{2}|\lambda|$$
whenever $n\neq m$. But compactness implies that every sequence in $K$ must have a convergent subsequence, so this is impossible.

The orthogonality of distinct eigenspaces (which are closed since they are kernels) implies that
$$ S_\epsilon = \bigoplus_{|\lambda|\ge \epsilon} V_\lambda.$$
Assume for contradiction that there are infinitely many direct summands, and choose one unit eigenvector $x_n$ from each eigenspace. By the same argument above, we have
$$\|Tx_n-Tx_m\|=\|\lambda_nx_n-\lambda_mx_m\|\ge \epsilon$$
whenever $n\neq m$, since the hypotenuse of a right triangle is longer than its shorter side. This is impossible by compactness. $\square$

Lemma 1 implies properties (1) and (2) and that there are at most countably many eigenvalues. We now show that there are enough to form an orthonormal basis; the key is to show that we can always find one eigenvector, and the rest will follow by induction.

**Lemma 2.** Either $\|T\|$ or $-\|T\|$ is an eigenvalue of $T$.

*Proof of Lemma.* Let $c=\sup_{\|x\|=1} \|Tx\|$. If $c=0$ then we are done since $T=0$ and any unit vector in the kernel will do. Otherwise assume $c>0$ and let $\{x_n\}\subset B$ be a sequence such that $\|Tx_n\|\rightarrow c$. By compactness of $K$ we may pass to a subsequence such that $Tx_n$ converges to some vector $y\neq 0$.

Let $A=c^2-T^2$, and observe that $(x,Ax)\ge 0$ for every $x\in H$, since $(x,T^2x) = \|Tx\|^2\le c^2$ for every $x$. Observe that
$$ (x_n,Ax_n)=c^2-\|Tx_n\|^2\rightarrow 0.$$
Since $A$ is positive it has a square root $\sqrt{A}$, so we have
$$ (x_n,\sqrt{A}^2x_n)=\|\sqrt{A}x_n\|^2\rightarrow 0.$$
Since $T\sqrt{A}$ is a bounded operator and $TA=AT$, this implies
$$\|TAx_n\|=\|ATx_n\|\rightarrow 0,$$
which by continuity of $A$ implies $Ay=0$. Thus we have
$$(c^2-T^2)y = (c-T)(c+T)y=0.$$
If $(c+T)y=0$ then $-c$ is an eigenvalue; otherwise $c$ is an eigenvalue with eigenvector $(c+T)y\neq 0$.$\square$

To finish the proof of the theorem, let $\gamma_n$ denote the dimension of $V_{\lambda_n}$ and let
$\{\phi_{nj}\}_{n=1,j=1}^{\infty, \gamma_n}$ denote a sequence of orthonormal eigenvectors for all of the (countably many by Lemma 1) *nonzero* eigenvalues $\lambda_n$. Set $M=span\{\phi_{nj}\}_{n,j}$ (meaning the set of finite linear combinations). Observe that $T(M)\subset M$ by construction, and since $T$ is continuous this implies $T(\overline{M})\subset \overline{M}.$ On the other hand, if $y\in M^\perp$ and $x\in M$ we have
$$(Ty,x) = (y,Tx)=0$$
since $Tx\in M$. Thus, both $\overline{M}$ and $M^\perp$ are closed invariant subspaces of $T$.

Observe that if $P=proj(M^\perp)$, the restricted operator $PTP$ is also compact. Since it has no nonzero eigenvectors (by construction), Lemma 2 implies that it must satisfy $\|PTP\|=0$, which means $Tx=0$ whenever $x\in M^\perp$. Thus, every vector in $M^\perp$ is an eigenvector of $T$ with eigenvalue $0$.

To complete the proof, take $\{\psi_k\}$ to be any orthonormal basis of $M^\perp$. Since $H=\overline{M}\oplus M^\perp$ the union $\{\phi_{nj}\}_{n,j}\cup \{\psi_k\}$ is an ONB of $H$ consisting of eigenvectors of $T$, as desired.$\square$

**Remark.** As pointed out by Tarun in class, it is possible to prove Lemma 2 by considering $cI-T$ instead of $c^2I-T^2$ if one assumes $\|T\|=\sup_{\|x\|=1} (x,Tx)$, which can be achieved by possibly replacing $T$ with $-T$.

**Remark.** Yeshwanth pointed out that one can also prove Lemma 2 by showing that $\|y/c-x_n\|^2\rightarrow 0$, which can be seen by expanding the left hand side in inner products. This proof has the advantage of not using a square root.