Do the eigenvalues of the product of a positive diagonal matrix and a skew-symmetric matrix still have zero real part? Let $A$ be a $n\times n$ diagonal matrix with real, positive entries on the diagonal. Further, let $B$ be a $n\times n$ invertible, skew-symmetric and real matrix. Now let $n$ be even. Then it is known that the eigenvalues of $B$ all have zero real part. 

Is this also true of the product $AB$?

I tried doing some simulations in Matlab: The real parts were not zero, but always $\sim 10^{15}$ times smaller than the imaginary parts, so I suspect this is a rounding error on Matlabs part. But how to prove it?

Let $R$ be the (symmetric) root of $A$. Then, $AB$ and $RBR$ are similar and we have 
$$ (RBR)^T = RB^T R = -RBR. $$
That is, $RBR$ is skew-symmetric too. In particular the eigenvalues of $RBR$ are imaginary (including zero) and the same as $AB$. 
