The exponential of a skew-symmetric matrix in any dimension. 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?
 A: 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.
