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.
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 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 $\theta=||a||$.
$ \def\tss{\mathop{\bigotimes}\limits} \def\dss{\mathop{\odot}\limits} \def\bbR#1{{\mathbb R}^{#1}} \def\a{\alpha}\def\b{\beta}\def\g{\gamma}\def\t{\theta} \def\l{\lambda}\def\s{\sigma}\def\e{\varepsilon} \def\n{\nabla}\def\o{{\tt1}}\def\p{\partial} \def\E{{\cal E}}\def\F{{\cal F}}\def\G{{\cal G}} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\vec#1{\operatorname{vec}\LR{#1}} \def\diag#1{\operatorname{diag}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\hess#1#2#3{\frac{\p^2 #1}{\p #2\,\p #3}} \def\c#1{\color{red}{#1}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\notimplies{\;\not\!\!\!\implies} $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.
I think that the simplest method of computing the matrix exponential (though not always the most efficient) is Putzer's method. It works from the Cayley-Hamilton theorem gives the matrix exponential as a polynomial in $\bf{C}.$ It requires knowledge of the eigenvalues of $\bf{C}$ though not the eigenvectors and it works in the case where $\bf{C}$ is not diagonalizable -- no need to compute the Jordan form.
The basic algorithm is this. First find the eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_d$. They are listed according to algebraic multiplicity so there are always $d$ of them. It doesn't matter if there are fewer linearly independent eigenvectors. The order does not matter but of course must be used consistently. Secondly define the following matrix polynomials: $$ {\bf M}_0 = {\bf I} \\ {\bf M}_1 = {{\bf C}-\lambda_1{\bf I}} \\ {\bf M}_2 = \left({{\bf C}-\lambda_2{\bf I}}\right)\left({{\bf C}-\lambda_1{\bf I}}\right) \\ {\bf M}_3 = \left({{\bf C}-\lambda_3{\bf I}}\right)\left({{\bf C}-\lambda_2{\bf I}}\right)\left({{\bf C}-\lambda_1{\bf I}}\right) \\ \vdots \\ {\bf M}_{d-1} = \left({{\bf C}-\lambda_{d-1}{\bf I}}\right)\left({{\bf C}-\lambda_{d-2}{\bf I}}\right)\ldots\left({{\bf C}-\lambda_2{\bf I}}\right)\left({{\bf C}-\lambda_1{\bf I}}\right) $$ Obviously Cayley-Hamilton tells you that ${\bf M}_d=\bf{0}$. Next solve the following sequence of first order differential equations $$ \frac{dr_0}{dt} = \lambda_1 r_0 \qquad r_0(0)=1 \\ \frac{dr_1}{dt} = \lambda_2 r_1 + r_0(t)\qquad r_1(0)=0 \\ \frac{dr_2}{dt} = \lambda_3 r_2 + r_1(t)\qquad r_2(0)=0 \\ \vdots\\ \frac{dr_k}{dt} = \lambda_{k+1} r_k + r_{k-1}(t)\qquad r_k(0)=0 $$ Each equation has the solution to the previous equation as a forcing term. They all satisfy zero initial condition except for the first. Then $$ \exp(t {\bf C}) = {\bf F}(t)=\sum_{i=0}^{d-1} r_i(t) {\bf M}_i $$
I won't post a complete proof but you basically just compute $\frac{d}{dt}{\bf{F}}(t)-{\bf C}{\bf F}(t)$ and the series telescopes, with the $k=0$ term being trivially zero and the $k=d$ term being zero by Cayley-Hamilton.
