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I am asking for some hints to solve this excercise. Given an invertible skew symmetric matrix $A$, then show that there are invertible matrices $ R, R^T$ such that $R^T A R = \begin{pmatrix} 0 & Id \\ -Id & 0 \end{pmatrix}$, meaning that this is a block matrix that has the identity matrix in two of the four blocks and the lower one with a negative sign.

I am completely stuck!

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The key fact is that every real and normal matrix is orthogonally similar to its *real Jordan form*. However, as real Jordan forms are seldom taught in universities, we may need to start from scratch here.

- As $A$ is skew symmetric and invertible, its eigenvalues are conjugate pairs of nonzero pure imaginary numbers $\pm b_1i,\,\pm b_2i,\,\ldots,\pm b_ni$ (where the size of $A$ is $2n\times2n$).
- Furthermore, as $A$ is skew symmetric, it is a normal matrix. Hence it possesses an orthonormal eigenbasis. (In some lecture notes, this fact is sometimes stated as “if $A$ is normal, $A=UDU^\ast$ for some unitary matrix $U$ and diagonal matrix $D$”.) Let $u_1+iv_1,\,u_2+iv_2,\,\ldots,\,u_n+iv_n$ be $n$
*orthogonal*eigenvectors corresponding the eigenvalues $b_1i,\,b_2i,\,\ldots,\,b_ni$. - Note that $\left\{\frac{u_1}{\|u_1\|},\frac{u_2}{\|u_2\|},\ldots,\frac{u_n}{\|u_n\|},\frac{v_1}{\|v_1\|},\frac{v_2}{\|v_2\|},\ldots,\frac{v_n}{\|v_n\|}\right\}$ is an orthonormal basis of $\mathbb{R}^{2n}$ (why?). So, these $2n$ vectors form a real orthogonal matrix $Q$ and $A=Q\pmatrix{0&B\\ -B&0}Q^T$ (again, why?), where $B=\operatorname{diag}(b_1,b_2,\ldots,b_n)$. Now the rest should be straightforward.

**Hint:** a skew-symmetric matrix commutes with its transpose, and so it is diagonalizable. Your block matrix on your right hand side is also skew-symmetric, and so it is also diagonalizable.

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