If $A, B \in \Bbb R^{n \times n}$ are nonsymmetric matrices, what can we do to bound $\lambda_{\max}(AB)$?

Question

I think my title illustrates my question pretty clear. I know that there is an obvious bound when A and B are normal matrices ($\lambda_\max(AB)=\rho(AB) \leq \rho(A) \rho(B)=\lambda_\max(A)\lambda_\max(B)$), but are there any bounds for nonsymmetric real matrices? Or, under what circumstances can on derive a bound? Thanks a lot!

Answer 1

For arbitrary matrices we always have a bound derived from the operator norm, namely

$$\rho(AB) \le \| AB \| \le \| A \| \| B \|.$$

If $A$ and $B$ aren't normal then this bound doesn't reduce to any kind of bound involving their eigenvalues; e.g. if $A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], B = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]$ then $A$ and $B$ are both nilpotent so have only zero eigenvalues, but

$$AB = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$$

satisfies $\rho(AB) = 1$. We do have $\| A \| = \| B \| = 1$ so the operator norm bound is tight in this case.