I came across an inequality for (finite dimensional) positive semi-definite operators and I tried to prove it, but I falied. Therefore, I hope some of you have a hint for me, what I missed.
Before I start, I give the definition of what is meant by $|A| := \sqrt{A^*A}$, where $A^*$ is transposing the matix $A$ and taking the element-wise coplex conjugate. As this question occurres in in connection with a physical problem I hope that it is ok to use Dirac's BraKet notation.
Suppose now $A,B \in \text{Pos(V)}$, where $V$ is some finite-dimensional vector space. I want to show that $| \text{Tr}[AB] | \leq \text{Tr}[AB] $ holds.
In my approach to prove this statement, I used that every positive operator is Hermitian and can be diagonalized, $A = \sum_i \lambda_i |v_i\rangle \langle v_i |$ and $B = \sum_j \mu_j | u_j \rangle \langle u_j |$, where $\{v_i\}$ and $\{u_j\}$ are bases, consisting of eigenvectors of $A$ and $B$, respectively. Thus, one can rewrite $|\text{Tr}[MN]| = | \sum_i \sum_j \lambda_i \mu_j |\langle v_i|u_j\rangle|^2|$.
Once I try to rewrite the right hand-side of the inequality in a similar way, it becomes a real mess, as I cannot assume that both $A$ and $B$ are diagonalized with respect to the same basis, hence I am left with lots of inner products between basis vectors from $\{v_i\}$ and $\{u_j\}$ (under the root, which occurrs due to the definition of the absolute value of a matrix - see above).
Is there any trick, how I can prove this statement directly, or is there anything that I missed when tying my "brute force diagonalizing" approach?
I am very grateful for any help!
Theorem. For any matrix $A$, one has that $$ |\text{tr}(A)|\leq \text{tr}(|A|). $$
Proof. Considering the inner product defined on $M_n(\mathbb C)$ by $$ \langle X, Y\rangle =\text{tr}(X^*Y), $$ one has by the Cauchy-Schwartz inequality that $$ |\text{tr}(X^*Y)|\leq \text{tr}(X^*X)^{1/2}\text{tr}(Y^*Y)^{1/2}. $$
Using the polar decomposition, write $A=U|A|$, where $U$ is a unitary matrix, and set $X=|A|^{1/2}U^*$, and $Y=|A|^{1/2}$. Then $A=X^*Y$, so
$$ |\text{tr}(A)|=|\text{tr}(X^*Y)| \leq \text{tr}(X^*X)^{1/2}\text{tr}(Y^*Y)^{1/2} = $$ $$ = \text{tr}(U|A|U^*)^{1/2}\text{tr}(|A|)^{1/2} = \text{tr}(|A|). $$
PS: The same proof can be used to show that $|\tau(A)|\leq \tau(|A|)$, for every positive trace $\tau$ on any $C^*$-algebra $\mathscr A$, that is, any linear functional $$ \tau:\mathscr A \to \mathbb C, $$ such that $\tau(A^*A)\geq 0$, and $\tau(BA)=\tau(AB)$, for every $A$ and $B$ in $\mathscr A$.