# Spectral radius of a matrix multiplied by a contraction matrix

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.

• Is suspect that you could probably say something like this in the case that $A$ is positive semidefinite. Jan 17, 2020 at 10:04

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 $$\rho(AU) = \epsilon > 0 = \rho(A)$$.
• 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?
• Your fix is not quite enough. Try adding $\alpha I$ to $A$ for a sufficiently small $\alpha > 0$. Jan 20, 2020 at 7:14
• 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 $\epsilon^2$.