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I’m trying to prove that for $A=J_n(i)$, that is, the Jordan block matrix corresponding to the eigenvalue $i$ of size $n$, where $n$ is even, the matrix equation $AX+XA^{-T}=0$ has a nonsingular anti-symmetric solution $X$.

I have tried it on small values of $n (= 2,4,6)$ by brute force computations and proved that it is true.

Any ideas on how can I prove this for an arbitrary even $n$? Ideas or suggestions would suffice. Thank you so much!

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  • $\begingroup$ Is $i$ here a square root of $-1$? $\endgroup$
    – Fedor Petrov
    Commented Jul 9 at 9:39
  • $\begingroup$ @FedorPetrov yes $\endgroup$
    – White Cat
    Commented Jul 9 at 9:42

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If you denote $X_{pq}=(-i)^{p+q}Y_{pq}$, then the equations are $Y_{p+1,q+1}=Y_{p+1,q}+Y_{p,q+1}$ for $p, q=1,\dots,n$, here $Y_{kl} =0$ by definition if $\max(k, l) >n$. So you may put $Y_{pq}=0$ for $p+q>n+1$, $Y_{pq}=(-1)^p$ when $p+q=n+1$, $Y_{pp}=0$, and then define the elements on diagonals $p+q=n$, then $p+q=n-1$, etc.

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  • $\begingroup$ Okay, I will just take time to process this information. Thank you so much! $\endgroup$
    – White Cat
    Commented Jul 9 at 11:49
  • $\begingroup$ To the person who downvoted this answer, care to explain why? $\endgroup$
    – White Cat
    Commented Jul 9 at 12:56
  • $\begingroup$ @WhiteCat Why are you interested? In general, on SE one should be able to downvote without the need to explain anything. $\endgroup$
    – Federico Poloni
    Commented Aug 8 at 12:11
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    $\begingroup$ @FedericoPoloni of course there is no need to explain, but maybe downvote means that there is something wrong with the answer that should be fixed? $\endgroup$
    – Fedor Petrov
    Commented Aug 8 at 13:25

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