Computation of a matrix exponential for general dimensions

Question

Originally I wanted to prove something else then I hit on this question that I find quite interesting but I don't know how to prove it elegantly.

Let $$J=\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$$ with I the identity matrix in an appropriate dimension. Then the matrix-exponential is given by $$e^{\nu J t}=\begin{pmatrix} COS & SIN \\ -SIN & COS \end{pmatrix}$$ where $COS$ is the matrix with $\cos(\nu t)$ on its diagonals, i.e. $$COS=\begin{pmatrix} \cos(\nu t) & \dots & 0 \\ 0 & \ddots & 0 \\ 0 & \dots & \cos(\nu t) \end{pmatrix} $$ and $SIN$ analogue. $\nu$ is a scalar.

I noticed it in two and four dimensions and computed it for higher dimensions in Matlab, see at the picture.

Of course it could be that it works only for lower dimensions but I doubt that. Now to the proof. I can only imagine a quite extensive proof. I can show you my approach.

The matrix exponential can be computed by using the transformation $J=S N S^{-1}$ where N is the Jordan Normalform of J and S the matrix with the eigenvectors.

Let $J \in \mathbb{R^{2n}}$, then its JNF is given by the matrix with $n$ times $i$ and $n$ times $-i$ on the diagonals, i.e. all eigenvalues are algebraically simple. Now to the eigenvectors, they are always complex conjugated to another one. But I don't want to compute them if it is not necessary.

Another approach would be the definition of the exponential function, i.e. $$e^{\nu Jt}=\sum_{m=0}^{\infty} \frac{(\nu J t)^m}{m!}$$

Maybe someone has a more elegant proof.

Best regards, Marvin

Answer 1

Note that this just reduces to the case where $J = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} \right)$, nothing interesting is happening in the higher dimensions.