Spectral norm – trace inequality

Question

I am wondering whether the following is true under which assumptions on A and B?
$\operatorname{trace}(AB)\leqslant\|A\| \operatorname{trace}(B)$

The matrix norm is the spectral norm here. Maybe relevant

I tried to use the $\operatorname{trace}(AB)=\sum\limits_{i=1}^n\lambda_i,\lambda_i\in\sigma (AB)$, but then I need some relationship between eigs(AB) and eigs(A)*eigs(B), if I could have this, then because eigenvalues of A are less than its spectral norm, then the statement holds?

Answer 1

This certainly isn't true. Consider $A=B=\operatorname{diag}(1,-1)$ for instance. It is true, however, when $B$ is positive semidefinite. This follows directly from von Neumann's trace inequality $\operatorname{tr}(AB)\le\sum_i\sigma_i(A)\sigma_i(B)$: \begin{aligned} \operatorname{tr}(AB) \le\sum_i\sigma_i(A)\sigma_i(B) \le\sum_i\|A\|\sigma_i(B) =\sum_i\|A\|\lambda_i(B) =\|A\|\operatorname{tr}(B). \end{aligned}