Inequality for Hermitian matrices

Question

Does the following equality hold for $A,B$ ( both Hermitian matrices): $$\vert \text{tr}(AB) \vert \leq \displaystyle\left(\text{tr}(\vert A\vert^k) \text{tr}(\vert B \vert ^k)\right)^{\frac{1}{k}}?$$

I remember seeing this for $k=2$, which can be treated roughly as Cauchy-Schwartz for Hermitian operators. But I don't recall the proof for it. If it helps, you can assume k to be even.

Answer 1

$\def\tr{\operatorname{Tr}}$You run into all kind of problems if you omit absolute values. For instance with $k=1$ and $$A=B=\begin{bmatrix}0&1\\1&0\end{bmatrix}$$ you have $$ \tr(AB)=1>\tr(A)\,\tr(B)=0. $$ In fact, the inequality fails even if $A,B$ are positive and commute. For instance with $A=B=I_2$, and any $k>2$, $$ \tr(AB)=2>2^{2/k}= \tr(A^k)^{1/k}\tr(B^k)^{1/k}. $$ The inequality that does hold is Holder's. That is, for $p,q≥1$ with $\frac1p+\frac1q=1$, $$ |\tr(AB)|≤\tr(|A|^p)^{1/p}\tr(|B|^q)^{1/q}. $$

Answer 2

One can show that for matrices of the same shape (not necessarily square), the following is an inner product. $$\langle A, B \rangle := \operatorname{tr}(A^\top B).$$ Then, Cauchy-Schwarz directly implies $$\operatorname{tr}(A^\top B) \le \sqrt{\operatorname{tr}(A^\top A) \operatorname{tr}(B^\top B)}.$$ In the special case where $A$ and $B$ are square and symmetric, this becomes $\operatorname{tr}(AB) \le \sqrt{\operatorname{tr}(A^2) \operatorname{tr}(B^2)}$. I don't know whether your claim holds for other $k$ though.