Maximum eigenvalue of product of two matrices

Question

Let $A$ and $B$ be two Hermitian matrices. I wanted to know if there is any relation between the maximum eigenvalue of $AB$ and that of $A$ and $B$. Is the following relation true? $\lambda_{\text{max}}\left(AB\right)\le \lambda_{\text{max}}\left(A\right)\lambda_{\text{max}}\left(B\right)$

If it is true, then is this relation valid when $A$ and $B$ are real.

Answer 1

If $A$ and $B$ are $\color{blue}{\rm HPD}$, then $$ \begin{split} \color{red}{\lambda_{\max}(AB)} &= \lambda_{\max}(B^{1/2}AB^{1/2}) \\&= \max_{x\neq 0}\frac{x^*B^{1/2}AB^{1/2}x}{x^*x} \\&= \max_{x\neq 0}\frac{x^*Ax}{x^*B^{-1}x} \\&= \max_{x\neq 0}\left(\frac{x^*Ax}{x^*x}\right)\left(\frac{x^*x}{x^*B^{-1}x}\right) \\&\color{red}{\leq} \max_{x\neq 0}\left(\frac{x^*Ax}{x^*x}\right)\max_{y\neq 0}\left(\frac{y^*y}{y^*B^{-1}y}\right) \\&= \max_{x\neq 0}\left(\frac{x^*Ax}{x^*x}\right)\max_{y\neq 0}\left(\frac{y^*By}{y^*y}\right) \\&= \color{red}{\lambda_{\max}(A)\lambda_{\max}(B)}. \end{split} $$ Otherwise, as already indicated, the inequality does not need to be true if $A$ or $B$ is indefinite. By the continuity argument, it can also be extended for the case when $A$ and $B$ are only semi-definite.

Answer 2

Let $A=B=\operatorname{diag}(-2,1)$, then $AB = \operatorname{diag}(4,1)$, but $\lambda_\max (AB) = 4$, $\lambda_\max (A) = \lambda_\max (B)= 1$.

If the matrices are positive semi-definite, then $\|A\|=\lambda_\max(A)$ (since $A$ is unitarily diagonalisable) and the spectral norm is submultiplicatve, hence the result holds.