# Spectral norm - trace inequality

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?

• What have you tried? See How to ask a good question. Commented Apr 21, 2020 at 10:23
• I saw this inequality used in a proof (for bounds of some matrix expressions) again and again but I just don't know why this is the case. Maybe this is relevant math.stackexchange.com/questions/3103958/… Commented Apr 21, 2020 at 10:28
• Yes I tried to use the trace of AB is the sum of eigenvalues of AB, but then I need some relationship between eigs(AB) and eigs(A)*eigs(B), if I could have this, then because eigs(A) is less than its spectral norm, then the statement holds? Commented Apr 21, 2020 at 11:09
• eigs means eigenvalues, sorry for the confusion. Commented Apr 21, 2020 at 11:32

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}
• @Stephanie We need $B$ to be symmetric (real positive semidefinite matrices are by definition symmetric), but $A$ can be any general real square matrix. Commented Apr 21, 2020 at 11:47