1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Certainly the zero matrix does. $\endgroup$ – joriki Apr 21 '16 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. $\endgroup$ – Gerry Myerson Apr 22 '16 at 3:06
  • $\begingroup$ @Gerry Sorry I meant $XJX^{-1}=XX^TJ$...thanks for catching that up $\endgroup$ – Mike Cocos Apr 22 '16 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] $\endgroup$ – Gerry Myerson Apr 23 '16 at 4:33
  • $\begingroup$ It's easy to show that, for $n=2$, only scalar multiples of your $J$ work. $\endgroup$ – Gerry Myerson Apr 23 '16 at 23:53
2
$\begingroup$

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 $X=\operatorname{diag}(x,x^2,x^3,\ldots,x^{n-1},x^{-\frac{n(n-1)}2})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.