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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.

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  • $\begingroup$ If you want to have {..} inside MathJax you can use backslash like this: $\{a,b,c\}$ $\{a,b,c\}$. Similarly, if you want to use text, you can use \text. For example: $Q=\{A \text{ real } | AB^{T} \text{ is symmetric}\}$ gives you $Q=\{A \text{ real } | AB^{T} \text{ is symmetric}\}$. $\endgroup$
    – Martin Sleziak
    Commented Mar 17, 2023 at 6:46
  • $\begingroup$ If $B=I$ is an identity matrix, your set $Q$ consists of all symmetric matrices and $S=Q$ $\endgroup$
    – Fedor Petrov
    Commented Mar 17, 2023 at 8:10

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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 $Q$.

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  • $\begingroup$ Hmmm.... any simplifications may I ask? I'm only a junior in maths $\endgroup$
    – Kanghun Kim
    Commented Mar 17, 2023 at 9:04
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    $\begingroup$ @WillieWong If $AB^T=\Sigma$ is symmetric, then $B^{-1}A=B^{-1}\Sigma B^{-T}$ is congruent to $\Sigma$, thus symmetric. And conversely. $\endgroup$
    – Denis Serre
    Commented Mar 20, 2023 at 6:18
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    $\begingroup$ @KanghunKim: quick explanations. (a) The use of $Q'$ and $S'$ is to reparametrize stuff to make them easier to analyze, using the fact that when $B$ is invertible, the mapping $M\mapsto B^{-1} M$ is a invertible linear mapping of the space of all matrices, so do not change measurability properties. $\endgroup$
    – Willie Wong
    Commented Mar 20, 2023 at 13:41
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    $\begingroup$ @KanghunKim: (b) your $Q$ is a vector space. Your requirement "$A + A (B^{-1}A)^2$ is symmetric" can be expressed as "a cubic polynomial vanishes". So your set $S$ is an algebraic subvariety (in fact affine) of $Q$. It is a general fact that $S$ therefore either has measure zero, or $S = Q$. (Here we use that if it doesn't have measure zero, then $S$ must contain an open subset of $Q$, but if a polynomial vanishes on an open set it must be identically zero.) $\endgroup$
    – Willie Wong
    Commented Mar 20, 2023 at 13:49
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    $\begingroup$ @KanghunKim: (c) When $S = Q$, the next task is to characterize those $B$ that leads to this. In this case, the "cubic polynomial" that is involved must be identically zero. This means all of its coefficients must be zero. In particular, the cubic term and the linear term must be separately equal to zero. Hence $B \Sigma = \Sigma B^T$ for every $\Sigma$. (d) Now let $v$ be an arbitrary vector, let $\Sigma$ be the symmetric matrix $v v^T$. Evaluating $B(v) (v^Tv) = B\Sigma v = \Sigma B^T v = v (v^T B^T v)$ we see that every $v$ is an eigenvalue of $B$, hence $B$ must be a multiple of $I_n$. $\endgroup$
    – Willie Wong
    Commented Mar 20, 2023 at 13:57

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