Any bounded linear operator $T : \ell^1 \to \ell^1$ can be represented by an infinite matrix $A = (A_{ij})$ with $A_{ij} = e_i^T A e_j = (T e_j)_i$. Assume that there is a decomposition of the index set $\mathbb{N} = \bigcup_{i \in K} I_i$ where $K = \{ 1, \dots, n \}$ or $K = \mathbb{N}$ (and the $I_i$ are pairwise disjoint) and that w.r.t. this decomposition, $A$ has an upper block triangular form
$$ A = \begin{pmatrix} A^{11} & A^{12} & A^{13} & \dots \\ 0 & A^{22} & A^{23} & \dots \\ & & \ddots & \ddots \end{pmatrix} $$
The $I_i \times I_j$-matrices $A^{ij}$ define bounded linear operators $T^{ij} : \ell^1(I_i) \to \ell^1(I_j)$.
Is it true that the spectrum $\sigma(T) = \sigma(A)$ is the closure of the union of all the spectra $\sigma(T^{ii}) = \sigma(A^{ii})$, i.e.
$$ \sigma(A) = \overline{\bigcup_{i \in K} \sigma(A^{ii})} $$
Partial answer: In this general setup, the conjecture is wrong. For instance, the one-sided shift operator on $\ell^1$ has upper triangular representation with $A^{ii} = 0$ on the diagonal and $A^{i, i+1} = 1$ on the first off-diagonal (so the blocks $A^{ij}$ are one-dimensional), but its spectrum is not $0$: $\sigma(A)$ is the closed unit ball around $0$.
So let us reduce the question to 2 cases:
$A$ splits into finitely many (finite or infinite) blocks $A^{ij}$, $1 \leq i \leq j \leq n$. Then $\sigma(A) \subseteq \bigcup_{i=1}^n \sigma(A^{ii})$. In general, equality need not hold, see e.g. [Jin Kyu Han, Hong Youl Lee and Woo Young Lee, "Invertible Completions of 2 × 2 Upper Triangular Operator Matrices" (2000)].
$A$ is a block diagonal matrix with infinitely many blocks $A^{ii}$. I think, in this setup the conjecture is true. How would you prove it?
