Let $\rho(\cdot)$ be the spectral radius operator (NOT the spectral norm). I'm aware that $\rho(AB)\leq\rho(A)\rho(B)$ holds if matrices $A$ and $B$ commute. But what is wrong with the following series of inequalities: let $z$ be an arbitrary vector then $||ABz||\leq \rho(A)||Bz||\leq \rho(A)\rho(B)||z||$, thus maximum absolute eigenvalue of $AB$ (i.e. $\rho(AB)$) must be less than $\rho(A)\rho(B)$.

It's simply not true in general that

$\| Ax \|_{2} \leq \rho(A) \| x \|_{2}$.

For example, consider

$A=\left[ \begin{array}{cc} 0 & 2 \\ 1/2 & 0 \\ \end{array} \right]$

The eigenvalues of $A$ are $\pm 1$, so $\rho(A)=1$. However, if we let

$x=\left[ \begin{array}{c} 0 \\ 1 \\ \end{array} \right]$

then

$Ax=\left[ \begin{array}{c} 2 \\ 0 \\ \end{array} \right]$

and

$\| Ax \|_{2}=2$

but $\rho(A)\| x \|_{2}=1$.

• Thanks a lot. I see. The spectral radius can only bound the matrix vector multiplication asymptotically, i.e. $\rho(A)=\lim_{n\rightarrow\infty}||A^n||^{1/n}$ – vulture Oct 31 '17 at 19:24