counterexample of $\rho(AB) \leq \rho(A)\rho(B)$

Question

Suppose $A$ and $B$ are normal complex $n \times n$ matrices. Then we know that $\rho(AB) \leq \rho(A)\rho(B)$ where $\rho(.)$ is the spectral radius of a matrix. Is there any counterexample if $A$ or $B$ is not normal?

Answer 1

Consider $A = \begin{bmatrix} 0 &1 \\ 0 &0 \end{bmatrix}$ and $B = A^T$. $\rho(A) = 0 = \rho(B)$ but $AB = \begin{bmatrix} 1 &0 \\ 0 &0 \end{bmatrix}$ then $\rho(AB)=1$.

Answer 2

This almost always occurs in the Cholesky factorisation $LL^T$ of a positive semidefinite matrix. When all entries in the strictly lower triangular part of $L$ are nonzero, we have $\rho(LL^T)>\rho(L)^2$.