Let $B$ be an invertible real matrix and let $Q=\{A \text{ real}\mid AB^{T} \text{ is symmetric}\}$. Is the subset $S=\{ A \in Q\mid A+A(B^{-1}A)^{2} \text{ is symmetric}\}$ of measure zero in $Q$? I need this for the final step for my proof that I've been working on for 4 years straight; I think that it might intuitively be true, since e.g. the condition for the product of two symmetric matrices to be nonsymmetric is quite strict (that the two matrices be noncommutative) and therefore the conditions for the product of a nonsymmetric matrix and a symmetric matrix to be symmetric should be equally strict as well.


EDIT: It suffices that this holds for all $B$ save a different set of measure zero.



Question on density of certain set of matrices
It suffices to check whether $B^{-1}S=:S'$ has measure zero in $B^{-1}Q=:Q'$. We have
$$Q'={\bf Sym}\_n(\mathbb R),\qquad S'=\{{\bf Sym}\_n(\mathbb R)|B(\Sigma+\Sigma^3)\in{\bf Sym}\_n(\mathbb R)\}.$$


The subset $S'$ is algebraic in $Q'$. Either it has measure zero, or $S'=Q'$.


Let us consider the latter case: then the equation $B(\Sigma+\Sigma^3)=(\Sigma+\Sigma^3)B^T$ is satisfied identically in ${\bf Sym}\_n(\mathbb R)$. By homogeneity, this amounts to saying that $B\Sigma\equiv\Sigma B^T$ in ${\bf Sym}\_n(\mathbb R)$. One checks easily that $B=\beta I\_n$ for some $\beta\in\mathbb R$.


In conclusion, if $B\in{\bf GL}\_n(\mathbb R)$ is not a homothety, then $S$ has measure zero in $Q$.


