Matrix exponential of a skew-symmetric matrix without series expansion 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.
 A: A geometric proof.
$C$ is the matrix of $u\in \mathbb{R}^3\rightarrow a\times u$ where $a=[a_1,a_2,a_3]^T$. $\exp(C){\exp(C)}^T=I$ and $\det(\exp(C))>0$ ; then $\exp(C)\in O^+(3)$, that is a rotation $\exp(C)=R=Rot(\Delta,\theta)$. Note that $Ca=0$ implies $Ra=a$ and $\Delta$ is the oriented line defined by $a$. 
After some calculations, $Trace(R)=1+2\cos(\theta)=1+2\cos(||a||)$. Moreover, let $b=[a_2,-a_1,0]^T$ ($b$ is orthogonal to $a$) ; we obtain $\dfrac{1}{{a_1}^2+{a_2}^2}||b\times Rb||=\sin(||a||)$. Conclusion $\theta=||a||$.
A: 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, $e^C = \displaystyle\sum_{n=0}^{\infty}\dfrac{1}{n!}C^n = I + \sum_{m=0}^{\infty}\dfrac{1}{(2m+1)!}C^{2m+1} + \sum_{m=1}^{\infty}\dfrac{1}{(2m)!}C^{2m}$ 
$= \displaystyle I + \sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m}}{(2m+1)!}C + \sum_{m=1}^{\infty}\dfrac{(-1)^{m-1}x^{2m-2}}{(2m)!}C^{2}$ 
$= \displaystyle I + \dfrac{1}{x}\sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m+1}}{(2m+1)!}C - \dfrac{1}{x^2}\sum_{m=1}^{\infty}\dfrac{(-1)^{m}x^{2m}}{(2m)!}C^{2}$
$= I + \dfrac{\sin x}{x}C + \dfrac{1-\cos x}{x^2}C^2$
A: $
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$The following method can be utilized even if you do not have access to a simple series definition of the function but are able to evaluate it for any value in its domain by some black-box method.
The first insight is that, as a consequence of the Cayley-Hamilton theorem, any analytic function of an $n\times n$ matrix is equal to a polynomial of degree $(n-1)$ with scalar coefficients.
For example, in the case of the matrix $C$ above
$$f(C) = \b_0 I + \b_1C + \b_2C^2$$
The problem boils down to calculating the $\b_k$ coefficients.
The next insight is that, if $\l_k$ are the eigenvalues of $C$,
then $f(\l_k)$ are the eigenvalues of $f(C)$. Therefore, you can write an eigenvalue version of the polynomial
$$f(\l_k) = \b_0 + \b_1\l_k + \b_2\l_k^2$$
Substituting the eigenvalues of $C$ yields a $3\times 3$ linear system, which can be solved for the $\b_k$ coefficients. The eigenvalues of
$C$ are known to be $\{\pm i\a,0\}$ where
$${
a = \m{a_1\\a_2\\a_3},\qquad
\a = \big\|a\big\|_2=\frac{1}{\sqrt 2}\big\|C\big\|_F
}$$
The linear system
$$\eqalign{
&\b_0 + i\a\b_1 - \a^2\b_2 &= f(i\a) &\doteq f_p
 \quad&\big(f\;{\rm plus}\big) \\
&\b_0 - i\a\b_1 - \a^2\b_2 &= f(-i\a) &\doteq f_m
 \quad&\big(f\;{\rm minus}\big) \\
&\b_0  &= f(0) &\doteq f_z
 \quad&\big(f\;{\rm zero}\big) \\
}$$
can be solved by Cramer's Rule to obtain
$$\eqalign{
\b_0 &= f_z \qquad
\b_1 &= \frac{f_p-f_m}{2i\a} \qquad
\b_2 &= \frac{2f_z-(f_p+f_m)}{2\a^2} \\
}$$
Again, these formulas are valid for any function of $C$:
$\;\,$log()$,\;$ erf()$,\;$ sinh()$,\, \ldots$
Let's evaluate the coefficients in the case of exp()
$$\eqalign{
\b_0 &= e^0 = \o \\
\b_1 &= \frac{e^{i\a}-e^{-i\a}}{2i\a}
  = \frac{2i\sin(\a)}{2i\a}
  = \frac{\sin(\a)}{\a} \\
\b_2 &= \frac{2-(e^{i\a}+e^{-i\a})}{2\a^2}
 = \frac{2-2\cos(\a)}{2\a^2}
 = \frac{\o-\cos(\a)}{\a^2} \\
}$$
Therefore
$$\eqalign{
\exp(C) = I + \frac{\sin(\a)}{\a}\;C + \frac{\o-\cos(\a)}{\a^2}\;C^2 \\
\\
}$$
This idea of eigenvalue interpolation is associated with the
names of$\,$ Sylvester $\,$and$\,$ Hermite.
