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This is a generalization of question Positivity of the anti-commutator of two positive operators .

note: by positive operator, I mean positive semidefinite (i.e. $\ge 0$, not necessary $>0$).

Let $A$ and $B$ two positive operators on a Hilbert space (I'm interested in the finite-dimensional case, but I think the question is interesting also in infinite dimension). The anti-commutator of $A$ and $B$ is defined as $\{A,B\} = AB + BA$.

If $A$ and $B$ commute, then it's easy to show that $\{A,B\} = 2 AB $ is a positive operator.

If $A$ and $B$ don't commute, we have a counterexample that shows that $\{A, B\}$ can be not positive, e.g. $A = \begin{pmatrix} 1 & 0 \\ 0 & 0\\ \end{pmatrix} $ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1\\ \end{pmatrix} $.

Question:

If $\{ A, B \}$ is positive, does it imply that $A$ and $B$ must commute? Or do exist non-commuting positive $A$ and $B$ such that $\{A,B\}$ is positive?

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Edit: (previous answer didn't consider $A\geq0$, $B\geq0$)

Let $$ A=\begin{bmatrix}2&0\\0&4\end{bmatrix}, \ \ B=\begin{bmatrix}2&1\\1&2\end{bmatrix}. $$ Then $$ AB+BA=\begin{bmatrix}4&2 \\4&8 \end{bmatrix} +\begin{bmatrix} 4&4\\2&8\end{bmatrix}=\begin{bmatrix} 8&6\\6&16\end{bmatrix}\geq0. $$

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    In your examples $A$ and $B$ are not Hermitian, and thus not positive, at least under one definition of positivity. Which is probably the most common. Where you thinking of the generalization of positivity for non hermitian matrices?2012-10-19
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    No, I just read your question and forgot about $A,B$ being positive several lines above.2012-10-19
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    Edited the answer.2012-10-19
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    Nice! I was pretty convinced that it was true.2012-10-19
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    Note that it is not that hard to look for examples, because every time you consider positivity with regards to a product of selfadjoint matrices, you can assume that one of them is diagonal. That is, by assuming one of them is diagonal you do not lose any generality.2012-10-19
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    Yes, indeed. The reason I was expecting them to commute is that if we call $P_B$ the ortogonal projector onto the kernel of $B$, then the fact that $\{ A, B \} \ge 0$ **do** imply that $A$ commutes with $P_B$. (in your example, this is trivial, but is not in general)2012-10-20