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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.

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    $x*S^*Sx=\lVert Sx\rVert^2\geq 0$.2012-11-01