# Symmetrized products of positive matrices are not positive

.. duh. This is in relation to the symmetrized matrix AM-GM inequality conjectured by Recht and Re. The most obvious/direct way of maybe attempting to prove this inequality is to show the semidefinite inequality
$\left(\frac{1}{n}\sum\nolimits_{i=1}^n\mat{A}_i\right)^n \succeq \frac{1}{n!} \sum_\sigma \mat{A}_{\sigma(1)}\mat{A}_{\sigma(2)}\times\cdots\times \mat{A}_{\sigma(n)}.$

Unfortunately, the right hand side isn’t even a positive matrix in general. The proof is pretty ridiculously simple, but I console myself with the fact that it wasn’t immediately obvious to some folks I’ve discussed this problem with that the symmetrized sum on the rhs could be nonpositive in general.

Consider the case of just two summands. Then $$\vec{x}^T (\mat{B}\mat{A} + \mat{A}\mat{B}) \vec{x} = 2\vec{x}^t \mat{B}\mat{A} \vec{x},$$ so $$\mat{A}\mat{B} + \mat{B}\mat{A}$$ is positive only if $$\mat{A}\mat{B}$$ is positive. This in turn implies $$\mat{A} \mat{B} = \mat{B}\mat{A},$$ so at least in this case, the matrices must commute for the symmetrized sum to be positive.

• If we take ‘n’ to be very large (tending to infinity), bringing n! to the LHS, it becomes (n!)/(n^n) [\sum_i A_i]^n >= RHS , which seems to be false since LHS term goes to a zero matrix (because of (n!)/(n^n) term). Right?

• swiftset

That could be true, maybe, but you’d have to find an example. In general it isn’t true: take $$\mat{A}_i$$ to be the identity, then the two sides are equal and always the identity.