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} $.


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 $$ AB+BA=\begin{bmatrix}4&2 \\4&8 \end{bmatrix} +\begin{bmatrix} 4&4\\2&8\end{bmatrix}=\begin{bmatrix} 8&6\\6&16\end{bmatrix}\geq0. $$

  • $\begingroup$ 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? $\endgroup$ – Angelo Lucia Oct 19 '12 at 15:14
  • $\begingroup$ No, I just read your question and forgot about $A,B$ being positive several lines above. $\endgroup$ – Martin Argerami Oct 19 '12 at 16:05
  • $\begingroup$ Edited the answer. $\endgroup$ – Martin Argerami Oct 19 '12 at 16:37
  • $\begingroup$ Nice! I was pretty convinced that it was true. $\endgroup$ – Angelo Lucia Oct 19 '12 at 16:47
  • 2
    $\begingroup$ 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. $\endgroup$ – Martin Argerami Oct 19 '12 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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