Inspired by this interesting problem (see Trace Norm properties), I have following question.

Let $X,Y\in\mathbb{R}^{n\times n}$ be real matrices. Is it true that $$|\mathrm{tr}(AB)|\leq\|A\|_{2}\|B\|_{2}$$ holds? Here, $\|\cdot\|_2$ is the operator $2$-norm for matrices.

Could you prove it or give a counterexample? I know the Von Neumann's trace inequality saying that $$|\mathrm{tr}(AB)|\leq\sum_{i=1}^n\sigma_i(A)\sigma_i(B)$$ where $\sigma_1(A)\geq\sigma_2(A)\geq\ldots\geq\sigma_n(A)$ and $\sigma_1(B)\geq\sigma_2(B)\geq\ldots\geq\sigma_n(B)$ are singular values of $A$ and $B$ respectively. Note that $\|A\|_{2}=\sigma_1(A)$ and $\|B\|_{2}=\sigma_1(B)$.

Any help is appreciated.


$A=B=I$ is a simple counterexample. If you take the 2 norm of the whole vector of singular values instead of just the largest one, then it holds.

  • $\begingroup$ If you take the 2-norm of the vector of singular values, then you are using the Frobenius norm, and the inequality is just the usual Cauchy-Schwarz inequality. $\endgroup$ – bartgol Nov 7 '17 at 5:53
  • $\begingroup$ Yeap. I should have thought about identity matrix. But thanks anyway. $\endgroup$ – Jay Nov 7 '17 at 6:04

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