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For any real $n \times n$ skew-symmetric matrix $A,$ show there exists an orthogonal matrix $P$ such that: $PAP^T = \begin{pmatrix}{} \Lambda_1 & \\ & \Lambda_2 & \\ & & \ddots & \\ & & & \Lambda_\frac{n}{2} \end{pmatrix}$

Where $\Lambda_i = \begin{pmatrix}{} 0 & \lambda_i \\ -\lambda_i & 0 \\ \end{pmatrix}$, $\lambda_i \geq 0.$

I have found a decent proof of the above but unfortunately it uses notions of linear (skew) transformations. I am having difficulty locating a proof that uses bilinear forms instead and I would like to try and understand that approach.

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  • $\begingroup$ I think it very likely that the algorithm I asked about at link below can be adapted to skew symmetric matrices, with all matrix entires used rational, if the original is rational. math.stackexchange.com/questions/1388421/… I would not expect to finish such a computer program today. The one for symmetric (integer) matrices took me six months, on and off. $\endgroup$ – Will Jagy Apr 26 '18 at 18:15
  • $\begingroup$ Oh, you want orthogonal.....If you can prove that, I cannot do any better. My change of variable matrices are determinant $\pm 1,$ but definitely not orthogonal. The whole point of that method is to keep all numbers used in the smallest field containing the entries of the beginning matrix. $\endgroup$ – Will Jagy Apr 26 '18 at 18:18

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