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?
I suppose that $S$ is defined to be $A^2$ rather than $A^{\text{abs}}A^{\text{abs}}$ in your question (if it were $A^{\text{abs}}A^{\text{abs}}$, it is real symmetric and hence diagonalisable). In this case, $S$ is not always diagonalisable. Here is a random counterexample: $$ A=\pmatrix{0&-1&1&-1\\ -1&0&-1&-1\\ 1&1&0&-1\\ 1&1&-1&0}. $$ One may readily verify that $(A-I)^2x=0\ne(A-I)x$ where $x=(1,0,2,-1)^T$. Hence $1$ is a non-semi-simple eigenvalue of $A$. It follows that $1$ is also a non-semi-simple eigenvalue of $S=A^2$. Thus $S$ is not diagonalisable over $\mathbb R$ or $\mathbb C$.