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?