Suppose I have a contraction (or non-expansive) matrix $U \in \mathbb{R}^{n\times n}$, which satisfies $\left\lVert U \right\rVert_2 \leq 1$.
Given some matrix $A \in \mathbb{R}^{n \times n}$ can one say that the spectral radius of the product $AU$, denoted $\rho(AU)$, is less than the spectral radius of $A$? i.e., can you conclude $$\rho(AU) \leq \rho(A)$$
I know that if the matrices commute, you can say that $\rho(AU) \leq \rho(A)\rho(U) \leq \rho(A)$. But I'm interested in the case where these matrices don't necessarily commute.
I think that the spectral radius is the infimum over subordinate matrix norms (which I think are sub-multiplicative right?), (see, e.g., How to prove that the spectral radius of a linear operator is the infimum over all subordinate norms of the corresponding norm of the operator.). So along those lines $$ \rho(AU) = \inf_{\left\lVert \cdot \right\rVert} \left\lVert AU \right\rVert $$ Just suppose for a second that the infimum is achieved and let $\left\lVert U \right\rVert_{M}$ denote the norm. Using sub-multiplicativity of the subordinate norm, maybe you can pull out the $U$ and say something like $$ \left\lVert U \right\rVert_{M} \leq \left\lVert U \right\rVert_{2} \leq 1. $$ I know that all matrix norms are equivalent within a constant, but in this case, we would need to say that the constant is less than or equal to $1$. Also, what if you can't exactly pin down that norm $\left\lVert U \right\rVert_{M}$?
Does assuming that $U$ is a unitary matrix give you anything extra that you can leverage? For example, what if $U$ had the following block form $$ \begin{pmatrix} U_1 & 0 \\ 0 & U_2 \end{pmatrix} \in \mathbb{R}^{n \times n}, $$ where $U_1$ and $U_2$ are unitary matrices.