I have the following skew-symmetric matrix
$$C = \begin{bmatrix} 0 & -a_3 & a_2 \\
a_3 & 0 & -a_1 \\
-a_2 & a_1 & 0 \end{bmatrix}$$
How do I compute $e^{C}$ without resorting to the series expansion of $e^{C}$? Shall I get a finite expression for it?
NB: Values of $a_i$ s can't be changed.
Best Answer
Let $x = \sqrt{a_1^2+a_2^2+a_3^2}$. You can verify that $C^3 = -(a_1^2+a_2^2+a_3^2)C = -x^2C$.
Hence, $C^{2m+1} = (-1)^mx^{2m}C$ and $C^{2m} = (-1)^{m-1}x^{2m-2}C^2$.
Therefore, $e^C = \displaystyle\sum_{n=0}^{\infty}\dfrac{1}{n!}C^n = I + \sum_{m=0}^{\infty}\dfrac{1}{(2m+1)!}C^{2m+1} + \sum_{m=1}^{\infty}\dfrac{1}{(2m)!}C^{2m}$
$= \displaystyle I + \sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m}}{(2m+1)!}C + \sum_{m=1}^{\infty}\dfrac{(-1)^{m-1}x^{2m-2}}{(2m)!}C^{2}$
$= \displaystyle I + \dfrac{1}{x}\sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m+1}}{(2m+1)!}C - \dfrac{1}{x^2}\sum_{m=1}^{\infty}\dfrac{(-1)^{m}x^{2m}}{(2m)!}C^{2}$
$= I + \dfrac{\sin x}{x}C + \dfrac{1-\cos x}{x^2}C^2$