# For non-negative definite symmetric matrices, $\mathrm{tr}(AB)\le \mathrm{tr}(A)\mathrm{tr}(B)$ [duplicate]

$$\DeclareMathOperator{\tr}{\mathrm{tr}}$$ Is the inequality in title true for non-negative definite matrices?? I could neither prove this result, nor provide a counter example.

Context

I was trying to prove that Frobenius norm is a matrix norm, i.e. $$||AB||_F\le||A||_F\cdot||B||_F\\ \iff\tr(B^\top A^\top AB)\le\tr(A^\top A)\cdot\tr(B^\top B)\\ \iff\tr(A^\top ABB^\top)\le\tr(A^\top A)\cdot\tr(B B^\top)$$ Now if that happens for all positive definite matrices $$A$$ and $$B$$, that means trace of product of two non-negative definite matrices is less than or equal to the product of their traces.

Please help, and provide a proof of this result which doesn't rely on the multiplicative inequality for Frobenius norm (since that result is proved to be equivalent to it, and I don't know the proof of that result).

• @DietrichBurde, no, it doesn't answer this question. It is using the inequality of Frobenius norms, which is equivalent to this result and I don't know proof of. – Martund May 27 '20 at 13:38
• I see. But it definitely answers your question "Is the inequality in title true for non-negative definite matrices?? I could neither prove this result, nor provide a counter example". Look up the proof for Frobenius norm inequality. – Dietrich Burde May 27 '20 at 13:40
• @DietrichBurde, and therefore, I need a proof of this result which doesn't rely on a result which I proved to be equivalent to it. – Martund May 27 '20 at 13:42
• Then start searching here. I found this post. Did you search already? – Dietrich Burde May 27 '20 at 13:44
• @DietrichBurde, Thank You very much, it answers the question. (I did search already, but used 'product of trace' and 'trace of product', didn't use the Frobenius norm terminology.) – Martund May 27 '20 at 13:53

Let $$(e_j)$$ be an orthonormal basis of eigenvectors of $$B$$ and let $$(\lambda_j)$$ denote the corresponding eigenvalues. Then $$\mathrm{tr}(AB)=\sum_j\langle AB e_j,e_j\rangle=\sum_j \lambda_j \langle A e_j, e_j\rangle\leq \max_j \lambda_j \mathrm{tr}(A)\leq \mathrm{tr}(A)\mathrm{tr}(B).$$