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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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    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