Compact operator on Hilbert space

If (${\displaystyle m(T)=0}$, then ${\displaystyle T=0}$ by the polarization identity, and this case is clear. Consider the function $${\displaystyle {\begin{cases}f:H\to \mathbf {R} \\f(x)=\langle Tx,x\rangle \end{cases}}}$$

Replacing ${\displaystyle T}$ by ${\displaystyle -T}$ if necessary, one may assume that the supremum of ${\displaystyle f}$ on the closed unit ball ${\displaystyle B\subset H}$ is equal to ${\displaystyle m(T)>0}$.

By the Banach–Alaoglu theorem and the reflexivity of ${\displaystyle H}$, the closed unit ball ${\displaystyle B}$ is weakly compact. Also, the compactness of ${\displaystyle T}$ means (see above) that ${\displaystyle T}$ seen as an operator from ${\displaystyle H}$ with the weak topology to ${\displaystyle H}$ with the norm topology is continuous ^{[disputed – discuss]}. These two facts imply that ${\displaystyle f}$ is continuous on ${\displaystyle B}$ equipped with the weak topology, and ${\displaystyle f}$ therefore attains its maximum ${\displaystyle m}$ on ${\displaystyle B}$ at some ${\displaystyle y\in B}$.

We now show that ${\displaystyle Ty=m(T)y}$. This is equivalent to

${\displaystyle \forall z\in H,\mathrm {Re} (\langle Ty-m(T)y,z\rangle =0}$

Now since y is the maximum of f, it is also the maximum of the Rayleigh quotient: $${\displaystyle g(x)={\frac {\langle Tx,x\rangle }{\|x\|^{2}}},\qquad 0\neq x\in H.}$$ Therefore, the function $${\displaystyle {\begin{cases}h:\mathbf {R} \to \mathbf {R} \\h(t)=g(y+tz)\end{cases}}}$$ achieves its maximum at 0, and thus h′(0) = 0. A computation now shows that $${\displaystyle {\begin{aligned}h'(0)&={\frac {1}{\|y\|^{2}}}\left({\frac {d}{dt}}{\frac {\langle T(y+tz),y+tz\rangle }{\langle y+tz,y+tz\rangle }}\right)(0)\\&=\operatorname {Re} (\langle Ty-m(Y)y,z\rangle )\end{aligned}}}$$

Note. The compactness of ${\displaystyle T}$ is crucial. In general, ${\displaystyle f}$ need not be continuous for the weak topology on the unit ball ${\displaystyle B}$. For example, let ${\displaystyle T}$ be the identity operator, which is not compact when ${\displaystyle H}$ is infinite-dimensional. Take any orthonormal sequence ${\displaystyle (y_{n})_{n\in \mathbb {N} }}$. Then ${\displaystyle y_{n}}$ converges to 0 weakly, but ${\displaystyle \lim f(y_{n})=1\neq 0=f(0)}$.

Let ${\displaystyle T}$ be a compact operator on a Hilbert space ${\displaystyle H}$. A finite (possibly empty) or countably infinite orthonormal sequence ${\displaystyle (e_{n})}$ of eigenvectors of ${\displaystyle T}$, with corresponding non-zero eigenvalues, is constructed by induction as follows. Let ${\displaystyle H_{0}=H}$ and ${\displaystyle T_{0}=T}$. If ${\displaystyle m(T_{0})=0}$, then ${\displaystyle T=0}$ and the construction stops without producing any eigenvector ${\displaystyle e_{n}}$. Suppose that orthonormal eigenvectors ${\displaystyle e_{0},\dots ,e_{n-1}}$ of ${\displaystyle T}$ have been found. Then ${\displaystyle E_{n}:=\mathrm {Span} (e_{0},\dots ,e_{n-1})}$ is invariant under ${\displaystyle T}$, and by self-adjointness, the orthogonal complement ${\displaystyle H_{n}}$ is an invariant subspace of ${\displaystyle T}$. Let ${\displaystyle T_{n}}$ denote the restriction of ${\displaystyle T}$ to ${\displaystyle H_{n}}$. If ${\displaystyle m(T_{n})=0}$, then ${\displaystyle T_{n}=0}$, and the construction stops. Otherwise, applying to ${\displaystyle T_{n}}$ the property of existence of an eigenvector, there is a norm one eigenvector ${\displaystyle e_{n}}$ of ${\displaystyle T}$ in ${\displaystyle H_{n}}$, with corresponding non-zero eigenvalue ${\displaystyle \lambda _{n}=m(T_{n})}$.

Let ${\displaystyle F=\mathrm {Span} ((e_{n})_{n})^{\perp }}$, where ${\displaystyle (e_{n})_{n}}$ is the finite or infinite sequence constructed by the inductive process; by self-adjointness, ${\displaystyle F}$ is invariant under ${\displaystyle T}$. Let ${\displaystyle S}$ denote the restriction of ${\displaystyle T}$ to ${\displaystyle F}$. If the process was stopped after finitely many steps, with the last vector ${\displaystyle e_{m-1}}$, then ${\displaystyle F=H_{m}}$ and ${\displaystyle S=T_{m}=0}$ by construction. In the infinite case, compactness of ${\displaystyle T}$ and the weak-convergence of ${\displaystyle e_{n}}$ to 0 imply that ${\displaystyle Te_{n}=\lambda _{n}e_{n}\to 0}$, therefore ${\displaystyle \lambda _{n}\to 0}$. Since ${\displaystyle F}$ is contained in ${\displaystyle H_{n}}$ for every ${\displaystyle n}$, it follows that ${\displaystyle m(S)\leq m(T_{n})=\vert \lambda _{n}\vert }$ for every ${\displaystyle n}$, hence ${\displaystyle m(S)=0}$. This implies again that ${\displaystyle S=0}$.

The fact that S = 0 means that F is contained in the kernel of ${\displaystyle T}$. Conversely, if x ∈ ker(${\displaystyle T}$) then by self-adjointness, x is orthogonal to every eigenvector {e_n} with non-zero eigenvalue. It follows that F = ker(${\displaystyle T}$), and that {e_n} is an orthonormal basis for the orthogonal complement of the kernel of ${\displaystyle T}$. One can complete the diagonalization of ${\displaystyle T}$ by selecting an orthonormal basis of the kernel. This proves the spectral theorem.

A shorter but more abstract proof goes as follows: by Zorn's lemma, select ${\displaystyle U}$ to be a maximal subset of ${\displaystyle H}$ with the following three properties: all elements of ${\displaystyle U}$ are eigenvectors of ${\displaystyle T}$, they have norm one, and any two distinct elements of ${\displaystyle U}$ are orthogonal. Let ${\displaystyle F}$ be the orthogonal complement of the linear span of ${\displaystyle U}$. If ${\displaystyle F\neq \{0\}}$, it is a non-trivial invariant subspace of ${\displaystyle T}$, and by the existence property of an eigenvector, there exists a norm one eigenvector ${\displaystyle y}$ of ${\displaystyle T}$ in ${\displaystyle F}$. But then ${\displaystyle U\cup \{y\}}$ contradicts the maximality of ${\displaystyle U}$. It follows that ${\displaystyle F=\{0\}}$, hence the linear span of ${\displaystyle U}$ is dense in ${\displaystyle H}$. This shows that ${\displaystyle U}$ is an orthonormal basis of ${\displaystyle H}$ consisting of eigenvectors of ${\displaystyle T}$.

Case I: all the operators have each exactly one eigenvalue on ${\displaystyle S}$. Take any ${\displaystyle s\in S}$ of unit length. It is a common eigenvector.

Case II: there is some operator ${\displaystyle T\in {\mathcal {F}}}$ with at least 2 eigenvalues on ${\displaystyle S}$ and let ${\displaystyle 0\neq \alpha \in \sigma (T\upharpoonright S)}$. Since ${\displaystyle T}$ is compact and α is non-zero, we have ${\displaystyle S':=\ker(T\upharpoonright S-\alpha )}$ is a finite-dimensional (and therefore closed) non-zero ${\displaystyle {\mathcal {F}}}$-invariant sub-space (because the operators all commute with ${\displaystyle T}$, we have for ${\displaystyle T'\in {\mathcal {F}}}$ and ${\displaystyle x\in \ker(T\upharpoonright S-\alpha )}$, that ${\displaystyle (T-\alpha )(T'x)=(T'(T~x)-\alpha T'x)=0}$). In particular, since α is just one of the eigenvalues of ${\displaystyle T}$ on ${\displaystyle S}$, we definitely have ${\displaystyle \dim S'<\dim S}$. Thus we could in principle argue by induction over dimension, yielding that ${\displaystyle S'\subseteq S}$ has a common eigenvector for ${\displaystyle {\mathcal {F}}}$.

The following set $${\displaystyle \mathbf {P} =\{A\subseteq H:A{\text{ is an orthonormal set of common eigenvectors for }}{\mathcal {F}}\},}$$ is partially ordered by inclusion. This clearly has the Zorn property. So taking ${\displaystyle Q}$ a maximal member, if ${\displaystyle Q}$ is a basis for the whole Hilbert space ${\displaystyle H}$, we are done. If this were not the case, then letting ${\displaystyle S=\langle Q\rangle ^{\bot }}$, it is easy to see that this would be an ${\displaystyle {\mathcal {F}}}$-invariant non-trivial closed subspace; and thus by the lemma above, therein would lie a common eigenvector for the operators (necessarily orthogonal to ${\displaystyle Q}$). But then there would then be a proper extension of ${\displaystyle Q}$ within ${\displaystyle \mathbf {P} }$; a contradiction to its maximality.

Fix ${\displaystyle T_{0}\in {\mathcal {F}}}$ compact injective. Then we have, by the spectral theory of compact symmetric operators on Hilbert spaces: $${\displaystyle H={\overline {\bigoplus _{\lambda \in \sigma (T_{0})}\ker(T_{0}-\sigma )}},}$$ where ${\displaystyle \sigma (T_{0})}$ is a discrete, countable subset of positive real numbers, and all the eigenspaces are finite-dimensional. Since ${\displaystyle {\mathcal {F}}}$ a commuting set, we have all the eigenspaces are invariant. Since the operators restricted to the eigenspaces (which are finite-dimensional) are automatically all compact, we can apply Theorem 1 to each of these, and find orthonormal bases Q_σ for the ${\displaystyle \ker(T_{0}-\sigma )}$. Since ${\displaystyle T}$₀ is symmetric, we have that $${\displaystyle Q:=\bigcup _{\sigma \in \sigma (T_{0})}Q_{\sigma }}$$ is a (countable) orthonormal set. It is also, by the decomposition we first stated, a basis for H.

Case I: all operators have exactly one eigenvalue. Then any basis for ${\displaystyle H}$ will do.

Case II: Fix ${\displaystyle T_{0}\in {\mathcal {F}}}$ an operator with at least two eigenvalues, and let ${\displaystyle P\in \operatorname {Hom} (H,H)^{\times }}$ so that ${\displaystyle P^{-1}T_{0}P}$ is a symmetric operator. Now let α be an eigenvalue of ${\displaystyle P^{-1}T_{0}P}$. Then it is easy to see that both: $${\displaystyle \ker \left(P^{-1}~T_{0}(P-\alpha )\right),\quad \ker \left(P^{-1}~T_{0}(P-\alpha )\right)^{\bot }}$$ are non-trivial ${\displaystyle P^{-1}{\mathcal {F}}P}$-invariant subspaces. By induction over dimension we have that there are linearly independent bases Q₁, Q₂ for the subspaces, which demonstrate that the operators in ${\displaystyle P^{-1}{\mathcal {F}}P}$ can be simultaneously diagonalisable on the subspaces. Clearly then ${\displaystyle P(Q_{1}\cup Q_{2})}$ demonstrates that the operators in ${\displaystyle {\mathcal {F}}}$ can be simultaneously diagonalised.