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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.

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  • $\begingroup$ Is suspect that you could probably say something like this in the case that $A$ is positive semidefinite. $\endgroup$
    – Ben Grossmann
    Jan 17 '20 at 10:04
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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 $\rho(AU) = \epsilon > 0 = \rho(A)$.

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  • $\begingroup$ Thanks! I don't want to read too much into your example, but to "fix things," if $A$ and $B$ have full rank and each have $\rho(A) < 1$ and $\rho(B)<1$, do you think you can say $\rho(AB) \leq \max\{\rho(A)^2, \rho(B)^2\}$? Or is there an obvious counter example here? Of course for symmetric matrices this holds, but what about non-symmetric matrices? $\endgroup$
    – Mido
    Jan 20 '20 at 2:30
  • $\begingroup$ Your fix is not quite enough. Try adding $\alpha I$ to $A$ for a sufficiently small $\alpha > 0$. $\endgroup$
    – Ben Grossmann
    Jan 20 '20 at 7:14
  • $\begingroup$ Ah yes. So for example, if you set $\alpha=1/2$ it works, but if you take $\alpha$ small enough, then $\rho(AU)$ is still pretty much equal to $\epsilon$ which is larger than $\epsilon^2$. $\endgroup$
    – Mido
    Jan 20 '20 at 18:21

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