Let $S$ and $A$ be two real square matrices such that $S=AA$. I know that, if $A$ is skew-symmetric, then $S$ is symmetric; moreover, its nonzero eigenvalues are negative and have even multiplicity. I also know that, if $A$ is itself symmetric, then so is $S$.

I am interested in understanding what happens if $A$ has a different form, namely, if it is "skew-symmetric up to a sign".

More precisely, for any $A=(a_{ij})_{i,j=1}^{n}$, let $A^{\mathrm{abs}}$ be the elementwise absolute value of $A$, i.e., the matrix $(\lvert a_{ij} \rvert)_{i,j=1}^{n}$ whose $(i,j)$-entry is the absolute value of $a_{ij}$.

**Question**. Suppose that $S= AA$ for some matrix $A$ *whose diagonal elements are all zero*. Suppose, further, that $A^{\mathrm{abs}}$ is symmetric. Can we say anything interesting about the diagonalizability of $S$ in this case? What about the multiplicity and sign of the eigenvalues?