Computing the exponential of the operator $ H(\textbf{x}) = \alpha \textbf{n} \times\textbf{x}$ Define the vector operator: $$ H(\textbf{x}) \equiv \alpha \textbf{n} \times\textbf{x}$$ For unit vector $\textbf{n}$ and some constant $\alpha$. We define further the operator: $$G \equiv I + H + \frac{H^2}{2!} + \frac{H^3}{3!} + ... $$ Where powers of $H$ represent iterations and $I$ is defined as the identity such that $I(\textbf{x}) = \textbf{x}$
We are to prove $$ G\textbf{x} = \textbf{x} +(\textbf{n}\times\textbf{x})\sin{\alpha} + \textbf{n}\times(\textbf{n}\times\textbf{x})(1-\cos{\alpha}) $$
My attempt at the proof involved noting: $$ G = e^{H} $$ And thus $$G\textbf{x} = e^H\textbf{x}$$
But I can't seem to see where to go from here, or even if the last step is a legitimate one. Many thanks in advance.
 A: $H x = \alpha \, n \times x$ gives the matrix:
$$
H = \alpha \left(
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right) = \alpha N
$$
Then for your proof
$$
G  = I + \sin(\alpha) N + (1-\cos\alpha) N^2
$$
needs to be shown. 
So we calculate $N^2$ and $N^3$:
\begin{align}
N^2 
&=
\left( 
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right)
\left( 
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right)
\\
&=
\left( 
\begin{matrix}
-(n_2^2+n_3^2) & n_1 n_2 & n_1 n_3 \\
n_1 n_2 & -(n_1^2 + n_3^2) & n_2 n_3 \\
n_1 n_3 & n_2 n_3 & -(n_1^2 + n_2^2)
\end{matrix}
\right)
\\
&= 
\left( 
\begin{matrix}
n_1^2-1 & n_1 n_2 & n_1 n_3 \\
n_1 n_2 & n_2^2 - 1 & n_2 n_3 \\
n_1 n_3 & n_2 n_3 & n_3^2 - 1 
\end{matrix}
\right)   
\end{align}
where we used $1 = n_1^2 + n_2^2 + n_3^2$ for the unit vector $n$ and then
\begin{align}
N^3 
&=
\left( 
\begin{matrix}
n_1^2-1 & n_1 n_2 & n_1 n_3 \\
n_1 n_2 & n_2^2 - 1 & n_2 n_3 \\
n_1 n_3 & n_2 n_3 & n_3^2 - 1 
\end{matrix}
\right)   
\left( 
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right)
\\
&=
\left( 
\begin{matrix}
0 & n_3 & -n_2 \\
-n_3 & 0 & n_1 \\
n_2 & -n_1 & 0
\end{matrix}
\right) = -N
\end{align}
This means all $N^k$ can be expressed with $I$, $N$ and $N^2$: 
$$
N^0 = I \quad N^{2k} = (-1)^k N^2 \quad N^{2k+1} = (-1)^k N
$$
for $k > 0$ and we can calculate
\begin{align}
G = e^H 
&= 
\sum_{k=0}^\infty \frac{1}{k!} H^k \\
&= 
\sum_{k=0}^\infty \frac{1}{(2k)!} H^{2k} +
\sum_{k=0}^\infty \frac{1}{(2k+1)!} H^{2k+1} \\
&= 
\sum_{k=0}^\infty \frac{1}{(2k)!} \alpha^{2k} N^{2k} +
\sum_{k=0}^\infty \frac{1}{(2k+1)!} \alpha^{2k+1} N^{2k+1} \\
&= 
I + \sum_{k=1}^\infty \frac{(-1)^k}{(2k)!} \alpha^{2k} N^2 +
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \alpha^{2k+1} N \\
&= 
I + (1-\cos \alpha) N^2 + \sin(\alpha) N
\end{align}
A: I don’t quite know whether the $e^H$ approach takes us anywhere useful, because I’m not entirely sure what the exponential of a vector operator means. Anywhere, here's one method that seems to work:
$\newcommand{\myvec}[1]{\mathbf{#1}}
\newcommand{\cross}{\times}
\newcommand{\nn}{\myvec{n}}
\newcommand{\xx}{\myvec{x}}$
First, a couple of useful identities. Lagrange's formula, or triple product expansion:
$$\myvec{a} \cross (\myvec{b} \cross \myvec{c})
= \myvec{b}\left(\myvec{a} \cdot \myvec{c}\right)
- \myvec{c}\left(\myvec{a} \cdot \myvec{b}\right)$$
and the power series for sine and cosine:
$$
\begin{align*}
\cos \theta &= 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots
&= \sum_{k=0}^\infty (-1)^k \frac{\theta^{2k}}{(2k)!} \\[6pt]
\sin\theta &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots
&=  \sum_{k=0}^\infty (-1)^k\frac{\theta^{2k+1}}{(2k+1)!}
\end{align*}
$$
So here's what I'd do:


*

*Expand $I\myvec{x}$, $H\myvec{x}$, $H^2\myvec{x}$, for a few small terms, to see if I could spot a pattern. Here I've done it up to $H^4\xx$:

$$\begin{align*}I\xx &= \xx \\H\xx &= \alpha\left(\nn \cross \xx\right) \\H^2\xx &= \alpha\left(\nn \cross \alpha\left(\nn \cross \xx\right)\right) \\&= \alpha^2 \left(\nn \cross \left(\nn \cross \xx\right)\right)  \\H^3\xx &= -\alpha^3 \left(\nn \cross \xx\right) \\H^4\xx &= -\alpha^4 \left(\nn \cross \left(\nn \cross \xx\right)\right)\end{align*}$$


*Identify a general formula for $H^m\xx$, and prove that it works.

I'm omitting the proof, but here's the general formula I came up with:$$H^m\xx =\begin{cases}\xx & \text{if $m=0$,} \\\alpha^m (-1)^{(m-1)/2}\left(\nn \cross \xx\right) & \text{if $m$ odd,} \\\alpha^m (-1)^{m/2-1}\left(\nn \cross \left(\nn \cross \xx\right)\right) & \text{if $m$ even.}\end{cases}$$You can prove this by induction.


*Now that we have a general formula for $H^m\xx$, we substitute this into the definition of $G$ and rearrange terms.
We can now write $G$ as (being careful with the signs)

$$G \equiv\xx+\underbrace{\left(\nn \cross \xx\right) \sum_{m=0}^\infty (-1)^{(m-1)/2}\frac{\alpha^{2m+1}}{(2m+1)!}}_{\text{odd terms}}-\underbrace{\left(\nn \cross \left(\nn \cross \xx\right)\right) \sum_{m=1}^\infty (-1)^{m/2}\frac{\alpha^{2m}}{(2m!)}}_{\text{even terms}}$$

Looking at the identities above, we can simplify this expression as

$$G \equiv \xx+\left(\nn \cross \xx\right) \sin\alpha - \left(\nn \cross \left(\nn \cross \xx\right)\right) (\cos\alpha-1)$$

which is the desired expression.
