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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?

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  • $\begingroup$ What have you tried? See How to ask a good question. $\endgroup$ Commented Apr 21, 2020 at 10:23
  • $\begingroup$ 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/… $\endgroup$
    – Stephanie
    Commented Apr 21, 2020 at 10:28
  • $\begingroup$ 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? $\endgroup$
    – Stephanie
    Commented Apr 21, 2020 at 11:09
  • $\begingroup$ eigs means eigenvalues, sorry for the confusion. $\endgroup$
    – Stephanie
    Commented Apr 21, 2020 at 11:32

1 Answer 1

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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}

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  • $\begingroup$ Thanks! Do we also need real matrices A and B to be symmetric? $\endgroup$
    – Stephanie
    Commented Apr 21, 2020 at 11:42
  • $\begingroup$ @Stephanie We need $B$ to be symmetric (real positive semidefinite matrices are by definition symmetric), but $A$ can be any general real square matrix. $\endgroup$
    – user1551
    Commented Apr 21, 2020 at 11:47
  • $\begingroup$ I see thanks again! $\endgroup$
    – Stephanie
    Commented Apr 21, 2020 at 11:48

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