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.
Best Answer
The answer is no. As an example, consider $$ U = \epsilon \pmatrix{0 & 1\\1 & 0}, \qquad A = \pmatrix{0&1\\0&0} $$ where $\epsilon$ satisfies $0 < \epsilon \leq 1$. Then $U$ is contractive, but $\rho(AU) = \epsilon > 0 = \rho(A)$.