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The Rodrigues rotation formula gives the exponential of a skew-symmetric matrix in three dimensions, and the exponential of a skew-symmetric matrix in two dimensions is given by Euler's formula. Is there a general formula (or set of formulas) for the exponential of a skew-symmetric matrix in any dimension?

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The spectral decomposition of any skew-symmetric matrix $A$ is given by $A=U Q U^\dagger$ where $U$ is unitary and \begin{align*} Q=\begin{bmatrix} 0 & \lambda_1 & \\ -\lambda_1 & 0 & \\ & & 0 & \lambda_2\\ & & -\lambda_2 &0\\ & & & & \ddots\\ & & & & & 0 & \lambda_r\\ & & & & & -\lambda_r & 0\\ & & & & & & & 0\\ & & & & & & & &\ddots\\ & & & & & & & & & 0 \end{bmatrix} \end{align*} where I don't put all the $0$s for visibility's sake.

The exponential of a matrix is defined as the extension of the tailor expansion (up to convergence matter you will need to take care of), hence \begin{align*} e^A = \sum_{n=0}^\infty \frac{1}{n!} A^n \end{align*}

and since $U$ is unitary, $A^n = U Q U^\dagger \dots U Q U^\dagger=U Q^n U^\dagger$ so we aim to get an expression for $Q^n$, this is not trivial but the right way to go is to compute it for several $n$ and try to see a pattern and then prove the pattern is right. It is easy to prove by induction that for any $k\in\mathbb N$ \begin{align*} Q^{2k} &= \begin{bmatrix} (-1)^k\lambda_1^{2k} & & \\ & (-1)^k\lambda_1^{2k} & \\ & & (-1)^k\lambda_2^{2k} & \\ & & & (-1)^k\lambda_2^{2k}\\ & & & & \ddots\\ & & & & & (-1)^k\lambda_r^{2k}\\ & & & & & & (-1)^k\lambda_r^{2k}\\ & & & & & & & 0\\ & & & & & & & &\ddots\\ & & & & & & & & & 0 \end{bmatrix}\\ Q^{2k+1} &=\begin{bmatrix} 0 & (-1)^k\lambda_1^{2k+1} & \\ -(-1)^k\lambda_1^{2k+1} & 0 & \\ & & \ddots\\ & & & 0 & (-1)^k\lambda_r^{2k+1}\\ & & & -(-1)^k\lambda_r^{2k+1} & 0\\ & & & & & 0\\ & & & & & &\ddots\\ & & & & & & & 0 \end{bmatrix} \end{align*} For this proof the base case would be $Q^0=I$ and $Q$ then you do induction on $k$.

We had \begin{align*} e^A &= \sum_{n=0}^\infty \frac{1}{n!} A^n\\ &= \sum_{n=0}^\infty \frac{1}{n!} U Q^n U^\dagger\\ &= U \left( \sum_{n=0}^\infty \frac{1}{n!} Q^n \right) U^\dagger \end{align*} and by plugin in the $Q^n$ we had, we get that the diagonal terms will be of the form $\sum_{k=0}^\infty \frac{1}{(2k)!} (-1)^k \lambda_p^{2k}=\cos(\lambda_p)$ and the other elements are of the form $\pm \sum_{k=0}^\infty \frac{1}{(2k+1)!} (-1)^k \lambda_p^{2k+1}=\pm \sin(\lambda_p)$ hence \begin{align*} e^A = U e^Q U^\dagger \end{align*} with \begin{align*} e^Q = \begin{bmatrix} \cos(\lambda_1) & \sin(\lambda_1) & \\ -\sin(\lambda_1) & \cos(\lambda_1) & \\ & & \cos(\lambda_2) & \sin(\lambda_2)\\ & & -\sin(\lambda_2) &\cos(\lambda_2)\\ & & & & \ddots\\ & & & & & \cos(\lambda_r) & \sin(\lambda_r)\\ & & & & & -\sin(\lambda_r) & \cos(\lambda_r)\\ & & & & & & & 1\\ & & & & & & & &\ddots\\ & & & & & & & & & 1 \end{bmatrix} \end{align*}

So an algorithmic way to find the exponential of your matrix is finding the spectral decomposition and then applying the last formula.

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  • $\begingroup$ put ones in the southeast corner of $e^Q$. $\endgroup$ – loup blanc Jan 10 at 11:04
  • $\begingroup$ @loupblanc thanks, can you confirm that this comes from the fact that $Q^0=I$ ? $\endgroup$ – P. Quinton Jan 10 at 12:42
  • $\begingroup$ Indeed I confirm. $\endgroup$ – loup blanc Jan 10 at 12:44

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