Skip to main content
1 of 4

For positive definite $A,B$ why does $AB+BA$ tend to be positive definite?

Let $A$ and $B$ be two positive definite $n \times n$ matrices. It is, of course, not true that $AB+BA$ is positive definite.

Consider, though, the results of the following numerical experiment. I generate $A$ by letting its eigenvalues be random in $[0,1]$ and getting its eigenvectors by generating a random matrix of standard Gaussians and applying Gram-Schmidt to it.

I did this 1000 times and checked what percentage of times the resulting matrix has negative eigenvalues [1]. Here are the results as a function of the dimension $n$:

  • $n=2, ~~~~94.8 \%$
  • $n=3, ~~~~89.4 \%$.
  • $n=4, ~~~~78 \%$.
  • $n=5, ~~~~72.7 \%$.
  • $n=10, ~~~40.3 \%$.
  • $n=15, ~~~20.1 \%$.
  • $n=20, ~~~11.4 \%$.
  • $n=50, ~~~0.3\%$.
  • $n=100, ~~0 \%$.

This suggests that, as a function of $n$, examples with $AB+BA$ not psd tend to get rarer and rarer. Is it possible to give a proof of this?

It may be more natural to consider a different random model of randomly generated psd matrices; I only generated them in the way I described above because it seemed easiest.

[1] Actually, I checked if there is an eigenvalue less then $-1 \cdot 10^{-5}$ to account for rounding error.