# Positivity of the anti-commutator of two positive operators implies commutativity?

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?

-

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 $$AB+BA=\begin{bmatrix}4&2 \\4&8 \end{bmatrix} +\begin{bmatrix} 4&4\\2&8\end{bmatrix}=\begin{bmatrix} 8&6\\6&16\end{bmatrix}\geq0.$$
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? – Angelo Lucia Oct 19 '12 at 15:14
No, I just read your question and forgot about $A,B$ being positive several lines above. – Martin Argerami Oct 19 '12 at 16:05