# Spectral radius of block matrices

Let $A = \begin{bmatrix}A^{11} & \mathbf{0} \\ \mathbf{0} & A^{22} \end{bmatrix}$ be a square block matrix where $A^{ii} \in R^{m_i \times m_i}$ ($i=1,2$) and $Q = \begin{bmatrix}Q^{11} & Q^{12} \\ Q^{21} & Q^{22} \end{bmatrix}$ be another square block matrix with the same shape as $A$ ($Q^{ii} \in R^{m_i \times m_i}$). We assume that $Q$ is symmetric and positive semi-definite.

Now, if we know that there exists a matrix $H$ (with the same shape as $A$ and $Q$) such that spectral radius of $A + HQ$ is less than 1 (which means maximum of absolute eigen-values of $A + HQ$ is less than 1), what can we say about the existence of a matrix $H^i$ (with the same shape as $A^{ii}$ and $Q^{ii}$) such that spectral radius of $A^{ii} + H^iQ^{ii}$ is less than 1? In other words, is there any matrix $H^i$ such that spectral radius of $A^{ii} + H^iQ^{ii}$ is less than 1?

Let $N$ be the null space of $Q$. Then $HQ$ can be any $(m_1+m_2)\times (m_1+m_2)$ matrix whose null space contains $N$. Namely, given such a matrix $B$, noting that the restriction $Q_{N^\perp}$ of $Q$ to the orthogonal complement $N^\perp$ of $N$ is invertible, we define $H v = 0$ for $v \in N$ and $H v = B Q_{N^\perp}^{-1} v$ for $v \in N^\perp$.
Thus the statement that there is $H$ such that the spectral radius of $A+HQ$ is less than $1$ means just that there is a matrix $B$ of spectral radius $<1$ such that $Bv = Av$ for $v \in N$. Of course this would not be true if $A$ has an eigenvector in $N$ with eigenvalue of absolute value $\ge 1$. Somewhat more generally, it would not be true if there exist $\epsilon > 0$ and a vector $v \in N$ such that for all positive integers $j$, $A^j v \in N$ and $\|A^j v\| > \epsilon$. I don't know if that is an "if and only if".
• I'm saying if such $\epsilon$ and $v$ exist, there is no $H$ such that the spectral radius of $A+HQ$ is less than $1$. So if such an $H$ does exist, there are no such $\epsilon$ and $v$. Dec 6, 2017 at 23:25