Jtdylktuy

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?

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 $kinmathbb N$
begin{align*}
Q^{2k} &= begin{bmatrix}
(-1)^klambda_1^{2k} & & \
& (-1)^klambda_1^{2k} & \
& & (-1)^klambda_2^{2k} & \
& & & (-1)^klambda_2^{2k}\
& & & & ddots\
& & & & & (-1)^klambda_r^{2k}\
& & & & & & (-1)^klambda_r^{2k}\
& & & & & & & 0\
& & & & & & & &ddots\
& & & & & & & & & 0
end{bmatrix}\
Q^{2k+1} &=begin{bmatrix}
0 & (-1)^klambda_1^{2k+1} & \
-(-1)^klambda_1^{2k+1} & 0 & \
& & ddots\
& & & 0 & (-1)^klambda_r^{2k+1}\
& & & -(-1)^klambda_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.