A trace inequality of the product of two matrices

Question

Given an arbitrary matrix $A$ and Hermitian positive definite matrix $B$, how can one prove this inequality: $$ |\text{Tr}(AB)| \le \|A\|_2 \text{Tr}(B) $$ where $\|A\|_2$ is the spectral norm of $A$. Under what condition(s) does the equality hold?

Answer 1

By a change of orthonormal basis you may assume that $B$ is a positive diagonal matrix. Then $$ |\operatorname{tr}(AB)| =\left|\sum_ia_{ii}b_{ii}\right| \le\sum_i|a_{ii}|b_{ii} \le\sum_i\left(\max_i|a_{ii}|\right)b_{ii} \le\|A\|_2\operatorname{tr}(B). $$ Tie occurs in the triangle (i.e., the first) inequality iff all $a_{ii}$s have the same phase; tie occurs in the second inequality iff all $a_{ii}$s have the same modulus. It follows that $|\operatorname{tr}(AB)|=\|A\|_2\operatorname{tr}(B)$ iff $A$ is a scalar multiple of the identity matrix.

By a continuity argument, we also have $|\operatorname{tr}(AB)|\le\|A\|_2\operatorname{tr}(B)$ when $B$ is positive semidefinite.

More generally, let $X$ and $Y$ be two arbitrary complex matrices. Let $Y=U|Y|$ be a polar decomposition, where $U$ is unitary and $|Y|=(Y^\ast Y)^{1/2}$. Let $A=XU$ and $B=|Y|$. Then $$ |\operatorname{tr}(XY)| =|\operatorname{tr}(AB)| \le\|A\|_2\operatorname{tr}(B) =\|X\|_2\operatorname{tr}|Y|. $$ When $Y$ is nonsingular, $B=|Y|$ is positive definite. In this case, from our previous discussion, we see that equality holds iff $A$ is a scalar multiple of $I$, i.e., iff $X=c|Y|Y^{-1}$ for some complex number $c$.