Suppose that I have a compact operator on a separable complex Hilbert space $T:H \to H$.
Furthermore suppose that with respect to a suitable orthonormal basis $(e_n)_{n\in\mathbb Z}$, the matrix of $T$ takes the form of a bi-infinite block-upper-triangular matrix:
$$ \begin{pmatrix} \ddots & \vdots & \vdots & \vdots & \vdots & \ddots\\ \cdots & B_{-1} & \ast & \ast & \ast & \cdots\\ \cdots & 0 & B_0 & \ast & \ast & \cdots\\ \cdots & 0 & 0 & B_1 & \ast & \cdots\\ \cdots & 0 & 0 & 0 & B_2 & \cdots \\ \ddots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, $$ where the $B_k$ $(k \in \mathbb Z)$ are some square matrices, the 0's correspond to rectangular 0 matrices and $\ast$ can be anything at all.
Question. Is it true that the eigenvalues of $T$ correspond to the eigenvalues of the $B_k$, as in the finite dimensional case?