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Consider the compact operator $ T: \ell^2(\mathbb{N}) \rightarrow\ell^2(\mathbb{N})$, represented by the infinite matrix $(a_{i j})_{i,j=0}^\infty$ in the canonical Hilbert basis $(\delta_i)_{i=0}^\infty $ of $\ell^2(\mathbb{N})$. Consider its "truncations" $T_n =\pi_n \circ T \circ \pi_n $ where $\pi_n$ is the orthogonal projector on $\rm{Vect}$ $ (\delta_0 , ... ,\delta_n)$ . Those correspond to the finite sub-matrices $(a_{i j})_{i,j=0}^n$ embedded in the aforementioned infinite one.

It is known that $T_n \rightarrow T$ in operator norm. Furthermore, since $T_n$ has finite rank, its has finately many eigenvalues.

Now, suppose $T$ is diagonalizable. In this question, it was said that eigenvalues of $T_n$ approach exactly those of $T$, though no furthermore explanation was given. Could someone explain why this is the case?

Rodrigo de Azevedo
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KRPO
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    The comment you are referring to probably considers operators which are diagonal themselves (or so I think) where this result is of course trivial as the first $n$ eigenvalues of $T_n$ and $T$ match exactly. Either way this eigenvalue approximation result holds for arbitrary compact operators as I stated in this answer. – Frederik vom Ende May 06 '20 at 15:53
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    Thank you, this is exactly what I was looking for! – KRPO May 08 '20 at 11:32

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