If $a,b,c,d$ denote four bounded operators on Banach spaces, let $$ A=\begin{pmatrix} a&b\\ c&d \end{pmatrix} $$
Can the spectral radius of $A$ be linked to those of $a,b,c,d$ ?
For example, if $c=0$, then I would guess that $r(A)=\max(r(a),r(b))$, is that the case ? Also, can this be generalized to a 3x3 matrix ? Or to a $n\times n$ matrix ?
If $c = 0$ then the spectrum of $A$ is contained in the union of the spectra of $a$ and $d$. In fact, if $a-\lambda I$ and $d - \lambda I$ are invertible then $$ \pmatrix{a-\lambda I & b\cr 0 & d - \lambda I}^{-1} = \pmatrix{ (a-\lambda I)^{-1} & - (a - \lambda I)^{-1} b (d - \lambda I)^{-1}\cr 0 & (d-\lambda I)^{-1}}$$ This generalizes to $n \times n$ upper triangular matrices of operators, since e.g. you can write $$ \pmatrix{a & b & c\cr 0 & d & e\cr 0 & 0 & f} = \pmatrix{A & B\cr 0 & f}$$ where $$ A = \pmatrix{a & b\cr 0 & d},\ B = \pmatrix{c\cr e\cr}$$