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I have come across the fact that a $4\times 4$ skew-symmmetric matrix of full-rank is equivalent to \begin{pmatrix} 0 &\theta_1& 0 &0 \\ -\theta_1& 0 &0 &0 \\ 0& 0&0 & \theta_2 \\ 0& 0& -\theta_2 & 0 \end{pmatrix} And I'm not sure why. Thanks for the help.

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    $\begingroup$ What do you mean "equivalent"? What are "equivalent matrices" $\endgroup$ – Timbuc Jan 28 '15 at 16:19
  • $\begingroup$ I think similar matrices, like a change of basis. $\endgroup$ – user46348 Jan 28 '15 at 16:22
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HINT: The eigenvalues of a real skew-symmetric matrix are pure imaginary and come in pairs of $(\lambda_k, - \lambda_k)$.

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  • $\begingroup$ Thanks! So this means that the companion matrix of the corresponding block is $x^2 +\lambda^2$ ($\lambda \in \mathbb{R}$) which has corresponding matrix $\begin{pmatrix} 0 & -\lambda^2\\ 1&0\end{pmatrix}$ (using rational canonical form) which can be converted to $\begin{pmatrix} 0& \lambda\\ -\lambda&0\end{pmatrix}$ under a simple change of basis. And then do this for the other block. $\endgroup$ – user46348 Jan 28 '15 at 18:49
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For a proof of this see the article Normal forms for skew-symmetric matrices, Proposition $2.1$. In particular, every skew-symmetric matrix in $M_4(\mathbb{R})$ of rank $4$ is congruent to $$ J_4(4)=\begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}. $$ Here $A$ and $B$ are congruent, if $B=PAP^T$ for some $P\in GL_4$.

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  • $\begingroup$ Thanks. This is an interesting proof as well. $\endgroup$ – user46348 Jan 28 '15 at 18:51
  • $\begingroup$ @ Dietrich Burde , the result requested by the OP is well-known ; yet I don't understand your argument (I am not well awakened). You write that a vector space formed by skew-matrices with full rank (except $0$) is a line. How to deduct the requested result ? $\endgroup$ – loup blanc Jan 30 '15 at 9:02
  • $\begingroup$ @loupblanc I wanted to use the result of this discussion, but a direct proof seems much better. $\endgroup$ – Dietrich Burde Jan 30 '15 at 10:55
  • $\begingroup$ @ Dietrich Burde , I read the link to MO. Clearly the maximum dimension of a vector space formed by skew-matrices with full rank (except $0$) over $\mathbb{R}$ is not $n-3=1$ but is $2$ (that is the result of David Speyer). $\endgroup$ – loup blanc Jan 30 '15 at 10:56

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