Lower spectral radius of matrices with an invariant subspace

Question

Let $\mathcal{A} = \{ A_1, \ldots, A_J, : A_j \in \mathbb{R}^{s\times s}\}$, and define the lower spectral radius (LSR) (aka joint subradius) by $LSR(\mathcal{A}) = \lim_{n\rightarrow\infty} \inf_{A_{j_i}\in\mathcal{A}} \rho(A_{j_n} \cdots A_{j_1})^{1/n}$,, where $\rho(\,\cdot\,)$ is the usual spectral radius of a matrix.

My question is about, what happens when the matrices in $\mathcal{A}$ have a common invariant (non-trivial) subspace.

It is well known that for the joint spectral radius (JSR) (where the $\inf$ is replaced by the $\sup$) that the $JSR$ then is the maximum of the $JSR$ of the sub-matrices restricted to the subspaces [Jungers 2009]. More precisely, if all $A_j$ have (under some basis $V$) a block triangular form $(A_j)_{[V]} = \begin{bmatrix}B_j & \ast \\ 0 & C_j\end{bmatrix}$, then $JSR(\mathcal{A}) = \max\{JSR(\mathcal{B}), JSR(\mathcal{C})\}$, where $\mathcal{B} = \{B_1,\ldots,B_J\}$, $\mathcal{C}=\{C_1,\ldots,C_J\}$.

I would guess for the $LSR$ the situation is similar, namely that $LSR(\mathcal{A}) = \max\{LSR(\mathcal{B}),LSR(\mathcal{C})\}$.

Is this already proven somewhere - Or is it maybe wrong?

Answer 1

This is false; the result is not perfectly analogous because the definition of the spectral radius contains a maximum. Here is a counterexample.

Take $$ A_1 = \begin{bmatrix}2 & 0\\ 0 & 1\end{bmatrix}, \quad A_2 = \begin{bmatrix}1 & 0\\ 0 & 2\end{bmatrix}. $$ Then, each product $A_{j_n}\dotsm A_{j_1}$ of length $n$ contains at least $n/2$ elements equal to $A_1$ or $n/2$ elements equal to $A_2$, and thus it has a diagonal element at least as large as $2^{n/2}$, and $\rho(A_{j_n}\dotsm A_{j_1}) \geq 2^{n/2}$. This proves that $LSR(\mathcal{A}) \geq 2^{1/2}$. On the other hand, $LSR(\mathcal{B}) = LSR(\mathcal{C}) = 1$, clearly.

If you replace $\rho$ with a function that computes the minimum modulus of an eigenvalue of a matrix, then the corresponding result should hold.