 Solution

Write $Q_n=P_n=P_n^{\ast }$ for the orthogonal projections and regard the compression $A_n=Q_{n}A{Q}_n$ as acting on $\mathrm{ran}{Q}_n$. The compactness argument from part iii applies to any strongly convergent sequence of orthogonal projections, so

We first rule out spectral pollution. Suppose ${\lambda }_n\in \mathrm{Sp}(A_n)$ and, after taking a subsequence, ${\lambda }_n\to \lambda \ne 0$. Choose unit eigenvectors $x_n\in \mathrm{ran}{Q}_n$:Compactness gives a convergent subsequence of $A{x}_n$. Becausethe relation $x_n={\lambda }_n^{-1}{A}_n{x}_n$ then makes $x_n$ converge to a nonzero vector $x$, and passage to the limit gives $Ax=\lambda x$. Thus every nonzero limit of finite-section spectral points belongs to $\mathrm{Sp}(A)$. The only remaining possible limit is zero, which belongs to the spectrum of a compact operator on an infinite-dimensional space.

Conversely, the Riesz–Schauder theorem says that every nonzero $\lambda \in \mathrm{Sp}(A)$ is an isolated eigenvalue of finite algebraic multiplicity. Put a small contour around $\lambda$ containing no other point of $\mathrm{Sp}(A)$. Norm convergence of $A_n$ gives uniform resolvent convergence on the contour, so the associated Riesz projections converge in norm and eventually have the same positive rank. Hence $\mathrm{Sp}(A_n)$ meets every neighborhood of $\lambda$.

Finally, zero is also approximated. Otherwise some subsequence would have all its eigenvalues bounded away from zero. Outside any small disk, $\mathrm{Sp}(A)$ has only finitely many eigenvalues, and the preceding Riesz-projection argument fixes the total algebraic multiplicity of nearby finite-section eigenvalues. This cannot account forThus finite-section eigenvalues also approach zero. Both directed spectral distances vanish, proving