Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 2

The spectral theorem for compact self-adjoint operators says that a compact self-adjoint $A$ has an orthogonal decompositionwith real nonzero eigenvalues of finite multiplicity. Only finitely many eigenvalues, counting multiplicity, have absolute value at least any given $\epsilon >0$. Thus the nonzero eigenvalues form a finite list or a sequence tending to zero, and $A={\sum }_{\lambda \ne 0}\lambda P_{\lambda }$ in operator norm. The kernel may have arbitrary Hilbert dimension.

For existence of an extremal eigenvector, self-adjointness gives $\Vert A\Vert ={\sup }_{\Vert x\Vert =1}|⟨Ax,x⟩|$. One elementary proof of the nontrivial inequality uses polarization: if the quadratic form is bounded in absolute value by $m\Vert x{\Vert }^2$, its polarized real part is bounded by $m$ on two unit vectors; adjusting a phase bounds $|⟨Ax,y⟩|$ by $m$. Choose unit $x_j$ with quadratic values tending to $\lambda =\Vert A\Vert$ or $-\Vert A\Vert$. ThenIf $A\ne 0$, compactness supplies a convergent subsequence of $A{x}_j$, and the last estimate supplies a convergent unit subsequence of $x_j$, with eigenvalue $\lambda$.

Eigenvectors for distinct eigenvalues are orthogonal, since $A=A^{\ast }$. An infinite orthonormal sequence in one nonzero eigenspace would violate compactness. More generally an infinite orthonormal family with eigenvalues bounded away from zero has images separated in norm, again impossible. Take all nonzero eigenspaces and let $M$ be their closed span. Its complement reduces $A$; if the restriction there were nonzero, the extremal-eigenvector argument would produce another nonzero eigenvector, a contradiction. Thus $M^{\perp }=\ker A$. The finite truncations of the eigenvalue sum have remainder norm tending to zero, proving the asserted norm expansion.

Apply part (a) to the compact positive operator $T^{\ast }T$. Its positive square root $B=|T|$ is defined by multiplying each eigenvector of eigenvalue $\lambda$ by $\sqrt{\lambda }$ and vanishing on the kernel. The eigenvalues tend to zero, so $B$ is compact. For every $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a partial isometry with initial space $(\ker T{)}^{\perp }$, final space $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot 0$ with many different unitary $U$.

The Von Neumann double commutant theorem states that for a unital star-subalgebra $A\subseteq B(H)$,The commutant of an operator algebra consists of operators commuting with every element. Each commutation equation is closed in the weak operator topology, so $A^{\prime \prime }$ is weakly closed and contains both indicated closures.

Let $T\in A^{\prime \prime }$ and fix ${\xi }_1,\dots ,{\xi }_m\in H$. On $H^m$ let $A$ act diagonally and put $\xi =({\xi }_1,\dots ,{\xi }_m)$. The subspace $K=\overline{\{(a{\xi }_1,\dots ,a{\xi }_m):a\in A\}}$ reduces the diagonal star-algebra. Its orthogonal projection $P$ therefore commutes with every diagonal $a$. Writing $P$ as an $m\times m$ operator matrix shows that each entry belongs to $A^{\prime }$. Since $T\in A^{\prime \prime }$, the diagonal $T$ commutes with $P$. Unitality gives $\xi \in K$, so $P\xi =\xi$ and $P(T\xi )=T\xi$. Thus $T\xi \in K$. By definition of $K$, for any $\epsilon >0$ a single $a\in A$ satisfies ${\sum }_j\Vert (T-a){\xi }_j{\Vert }^2<{\epsilon }^2$. These are precisely the finite-vector neighbourhoods of the strong operator topology. Hence $T$ belongs to the strong closure, completing the proof. Unitality, or the corresponding nondegeneracy hypothesis, is essential to the theorem's stated form.

Pass to $B={\overline{A}}^{\Vert \cdot \Vert }$. Closed invariant subspaces and intertwining maps are unchanged by this closure, and $BH$ is still dense. Because $B$ is a star-algebra, every closed invariant subspace reduces it.

Every nonzero nondegenerate compact-operator representation contains an irreducible reducing subspace. Indeed choose a nonzero positive compact element $b$; an isolated nonzero eigenvalue gives a nonzero finite-rank spectral projection $e\in B$ by continuous functional calculus, using a function vanishing at zero. The finite-dimensional corner $eBe$ has a minimal nonzero projection $p$, with $pBp=\mathbb{C}p$. Choose a unit vector $v\in pH$ and set $K=\overline{Bv}$. Then $pK=\mathbb{C}v$, since $pbv=pbpv$ is scalar times $v$. If $Q$ is the projection onto a reducing subspace of $K$, it commutes with $B$, and $Qv=pQv\in \mathbb{C}v$. Its projection property makes $Qv$ either zero or $v$. Cyclicity then makes $Q=0$ or $I_K$. Thus $K$ is irreducible.

Take a maximal orthogonal family of such irreducible reducing subspaces, by the axiom of choice. If their orthogonal complement were nonzero, its restricted representation would be nondegenerate: a vector annihilated by all restricted $B$ would, by taking adjoints, be orthogonal to $BH$ and hence zero. The image of $B$ on this complement is a norm-closed C-star algebra of compact operators, so the preceding construction gives another irreducible summand, contradicting maximality. The family therefore exhausts $H$.

Finally, suppose one irreducible equivalence class occurred infinitely often. Choose some $b\in B$ acting nontrivially on that class, and a unit vector $v$ with $\Vert bv\Vert >0$. Transport $v$ by unitary intertwiners into distinct equivalent summands. These vectors are orthonormal; their images under $b$ are orthogonal and all have the same nonzero norm. That contradicts compactness of $b$. The decomposition has finite multiplicity for every irreducible type, as claimed in nondegenerate representations by compact operators.