TITLE: Can we approximate any eigenvalue of an infinite matrix via eigenvalues of some sequence of submatrices which approximates the matrix?
QUESTION [6 upvotes]: Let $T:\ell^2\to\ell^2$ be a compact linear operator. Let $[T]=(a_{i,j})_{i,j=1}^{\infty}$ be the representing infinite matrix of $T$ with respect to the canonical base. Let $T_n$ be the finite rank operator defined by the matrix $(a_{i,j})_{i,j=1}^{n}$ embedded into an infinite matrix. Thus $T_n\to T$ in norm.
Can we approximate the eigenvalues of $T$ with eigenvalues of $T_n$?

REPLY [3 votes]: Today I by chance came across Chapter XI.9 of the Dunford & Schwartz classic "Linear operators. Part II. Spectral theory", and Lemma 5 in said chapter reads as follows:

Let $T_n,T$ be compact operators, and let $T_n\to T$ in the uniform operator topology. Let $\lambda_m(T)$ be an enumeration of the non-zero eigenvalues of $T$, each repeated according to its multiplicity. Then there exist enumerations $\lambda_m(T_n)$ of the non-zero eigenvalues of $T_n$, with repetitions according to multiplicity, such that
  $$
\lim_{n\to\infty}\lambda_m(T_n)=\lambda_m(T),\qquad m\geq 1,
$$
  the limit being uniform in $m$.

As you correctly observed the finite truncations $T_n$ converge to $T$ in norm ${}^1$ so the eigenvalue approximation property you were interested in holds. 

 ${}^1$: For anyone who wants to see why that is: in separable Hilbert spaces, such as $\ell^2$, finite-dimensional projections $\Pi_n:=\sum_{i=1}^n\langle e_i,\cdot\rangle e_i$ converge to the identity in the strong operator topology (simple consequence of Bessel's inequality) so compactness of $T$ ensures $T_n=\Pi_nT\Pi_n\to T$ in norm (see, e.g., Proposition 2.1 in this article by Widom).