Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.