# Does $\operatorname{tr}(A^\star A \rho) \geq 0$ hold for $\rho$ self-adjoint, positive semidefinite and $\operatorname{tr} \rho = 1$?

Let $V$ be a finite dimensional complex inner product space. Let $\rho$ be self-adjoint, positive semidefinite and $\operatorname{tr}\rho = 1$. Let $A,B \in \mathrm{End} (V)$. I want to show that

$$\langle A,B \rangle_\rho := \operatorname{tr}(A^\star B \rho)$$

is a positive semidefinite Hermitian form on $\mathrm{End}(V)$. So firstly, I want to show that $\langle A,A \rangle_\rho \geq 0$ holds. Now if $A$ or $A^\star A$ were positive semidefinite, it would be easy since the product of positive semidefinite matrices is positive semidefinite as well. But I don't see why that should be the case and if not, how I would show this. Could anyone give me a hint?

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Write $\rho=R^*R$, where $R$ is a $d\times d$ matrix, $d$ the dimension of the subspace. Then $$\langle A,A\rangle_{\rho}=\operatorname{Tr}(A^*AR^*R)=\operatorname{Tr}(RA^*AR^*)=\operatorname{Tr}((AR^*)^*AR^*)\geq 0,$$ as $(AR^*)^*AR^*$ positive semi-definite.
Is $R^\star R$ supposed to be the Cholesky decomposition of $\rho$? Also, why is $(AR^\star)^\star AR^\star$ positive semidefinite? –  studeth Nov 1 '12 at 14:17
Yes (we have such a decomposition provided we have the spectral theorem). If $B$ is a matrix, then $B^*B$ is positive semi-definite. –  Davide Giraudo Nov 1 '12 at 14:23
@studeth, a nice way to summarize all the information that you need to know is the following: an operator $T \in \operatorname{End}(V)$ is positive semidefinite if and only if there exists an operator $S$ such that $T = S^* S$ –  Manny Reyes Nov 1 '12 at 15:15
Basically, I was asking whether/why $S^\star S$ is positive semidefinite (see original post). The other direction is clear to me. –  studeth Nov 1 '12 at 15:22
$x*S^*Sx=\lVert Sx\rVert^2\geq 0$. –  Davide Giraudo Nov 1 '12 at 15:23