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A skew symmetric matrix $J$ is "special" if for any matrix $X$ with determinant equal to one, it satisfies $$ XJX^{-1}=XX^TJ.$$ For $2 \times 2$ matrices one can easily verify that any multiple of the matrix $$J=\begin{vmatrix} 0 & -1\\1 & 0 \end{vmatrix}$$ is special. Are there any nonzero $n \times n$ matrices satisfying this condition?

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  • $\begingroup$ Certainly the zero matrix does. $\endgroup$
    – joriki
    Apr 21, 2016 at 21:28
  • $\begingroup$ It is not true that $\det X=1$ implies $X^{-1}JX=XX^tJ$. It doesn't work for $X=\pmatrix{1&1\cr0&1}$, nor for $X=\pmatrix{2&0\cr0&1/2\cr}$, for example. $\endgroup$ Apr 22, 2016 at 3:06
  • $\begingroup$ @Gerry Sorry I meant $XJX^{-1}=XX^TJ$...thanks for catching that up $\endgroup$
    – Mike Cocos
    Apr 22, 2016 at 20:22
  • $\begingroup$ OK, then, since $X$ is invertible, you can cancel it to get to the simpler equation $JX^{-1}=X^tJ$. How hard can that be? [I don't know] $\endgroup$ Apr 23, 2016 at 4:33
  • $\begingroup$ It's easy to show that, for $n=2$, only scalar multiples of your $J$ work. $\endgroup$ Apr 23, 2016 at 23:53

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The given condition implies that $J=X^TJX$ for every $X$ with determinant $1$. In particular, if $X$ is a diagonal matrix, this would mean $J_{ij}=x_ix_jJ_{ij}$. So, if $x_ix_j\ne1$ for every $i$ and every $j$, then $J$ has to be zero. Now, when $n\ge3$, such an $X$ does exist: just pick any $x>1$ and let $X=\operatorname{diag}(x,x^2,x^3,\ldots,x^{n-1},x^{-\frac{n(n-1)}2})$.

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