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Suppose I have a skew-symmetric, real $4\times4$ matrix $M$. This implies that all its eigenvalues are imaginary and come in pairs. Let us call them $\pm i \lambda_1$ and $\pm i \lambda_2$ with $\lambda_1, \lambda_2 \in \mathbb{R}$ and let us call the corresponding eigenvectors $v^\pm_1$ and $v^\pm_2$ such that $M v^-_2= -i\lambda_2 v^-_2$, and so on.

Using these four eigenvectors one can construct the matrix $S = (v^+_1, v^-_1, v^+_2, v^-_2)$ such that $J = S^{-1}MS = \text{diag}( i\lambda_1, -i\lambda_1, i\lambda_2, -i\lambda_2 )$ is the diagonalized matrix. So far so good.

I seem to recall that under some additional circumstances (i.e. $\text{rank}M = 4$?) one can find a special orthogonal matrix $R \in SO(4)$ such that

$D = R^T M R = \begin{pmatrix} 0 & i\lambda_1 & 0 & 0 \\ -i\lambda_1 & 0 & 0 & 0 \\ 0 & 0 & 0 & i\lambda_2 \\ 0 & 0 & -i\lambda_2 & 0 \end{pmatrix}$.

//edit: The matrix should be brought to the form

$D = R^T M R = \begin{pmatrix} 0 & \lambda_1 & 0 & 0 \\ \lambda_1 & 0 & 0 & 0 \\ 0 & 0 & 0 & \lambda_2 \\ 0 & 0 & \lambda_2 & 0 \end{pmatrix}$.

My question: how do I construct this matrix $R$ from the four eigenvectors $v^\pm_{1,2}$ that I already know? This basis is sometimes called a Darboux basis I think.

Jens
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1 Answers1

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Note that the eigenvalues of your $D$ are $\pm\lambda_1$ and $\pm\lambda_2$, so it cannot be orthogonally similar to $M$.

You have $$ \begin{bmatrix}1/\sqrt2&-1/\sqrt2\\ 1/\sqrt2&1/\sqrt2\end{bmatrix} \begin{bmatrix}1&0\\ 0&-1\end{bmatrix} \begin{bmatrix}1/\sqrt2&1/\sqrt2\\ -1/\sqrt2&1/\sqrt2\end{bmatrix} =\begin{bmatrix} 0&1\\ 1&0 \end{bmatrix}. $$ So, with $$ L=\begin{bmatrix}1/\sqrt2&1/\sqrt2&0&0\\ -1/\sqrt2&1/\sqrt2&0&0\\ 0&0&1/\sqrt2&1/\sqrt2\\0&0& -1/\sqrt2&1/\sqrt2 \end{bmatrix}, $$ you get $$ L^TJL=\begin{bmatrix} 0 & i\lambda_1 & 0 & 0 \\ i\lambda_1 & 0 & 0 & 0 \\ 0 & 0 & 0 & i\lambda_2 \\ 0 & 0 & i\lambda_2 & 0 \end{bmatrix}.$$ Thus $R=SL$ seems to be what you are looking for.

Martin Argerami
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  • I fixed my typo above, there are no imaginary units in the matrix in the Darboux frame. The SO(4) matrix you gave me is some kind of a double rotation by Pi/4, isn't it? – Jens Nov 08 '16 at 06:15
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    What you are describing is the $4\times 4$ special case of "symplectic diagonalization" of $2n\times 2n$ antisymmetric matrices. Traditional diagonalization is done with the Gram-Schmidt process, and there is a symplectic version of Gram-Schmidt. – Jess Riedel Feb 02 '23 at 15:19
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    The reason you have a double $\pi/4$ rotation is that you are taking each pair of complex conjugate eigenvalue pairs $i\lambda_i$ and $-i\lambda_i$ in the diagonalized form and rotating them in the 2-dimensional eigenspace to get a real-valued $2 \times 2$ matrix block, just as you can rotate the complex matrix ${{i,0},{0,-i}}$ to get the (real) symplectic form ${{0,1},{-1,0}}$. This way you get a real-valued matrix while preserving most of the diagonal-ness. – Jess Riedel Feb 02 '23 at 15:26