The maximal eigenvalue of a block matrix.

Question

I have a block matrix \begin{equation} A=\left( \begin{matrix} 0&0&\cdots&0&A_{1,\alpha}&0&\cdots&0\\ 0&0&\cdots&0&A_{2,\alpha}&0&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&A_{\alpha,\alpha}&0&\cdots&0\\ \vdots&\vdots&\cdots&\vdots&\vdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0&A_{s,\alpha}&0&\cdots&0 \end{matrix} \right)_{N\times N}. \end{equation} Here, each $A_{k,\alpha},k=1,2,\cdots,s$ is a matrix of $N_k\times N_\alpha$. Obviously $A_{\alpha,\alpha}$ is a square matrix of $N_{\alpha}\times N_{\alpha}$ and we assume that the diagonal line of $A_{\alpha,\alpha}$ is on the diagonal line of $A$. The main question that I concerned is the relationship of the maximal eigenvalue of $A$ and the maximal eigenvalue of $A_{\alpha,\alpha}$. I have used software to calculate several examples and the results are all $$\lambda_{\max}(A)=\lambda_{\max}(A_{\alpha,\alpha}).$$ Hence I think that it may hold in general but I cannot give a proof. I will highly appreciate if anyone could help me!

Answer 1

The blocks correspond to a direct sum decomposition of $\mathbb R^M$: $$ \mathbb R^M = \mathbb R^{N_1}\oplus \mathbb R^{N_2}\oplus \ldots\oplus \mathbb R^{N_s}, $$ and the block matrix $A$ acts on $\mathbb R^M$ by sending each summand $\mathbb R^{N_k}$ to the summand $\mathbb R^{N_{\alpha}}$ -- the action of the block matrix $A_{k,\alpha}$ -- and so the only way you can have a vector being sent to a multiple of itself is when it lies in $\mathbb R^{N_\alpha}$. Thus the eigenvalues of $A$ are precisely the eigenvalues of $A_{\alpha,\alpha}$, and in particular $\lambda_{\text{max}}(A) = \lambda_{\text{max}}(A_{\alpha,\alpha})$.