= Solution
Setting $\mathbf b=\mathbf a$ in <11/c/i> removes the <cross product> term and gives $\sigma(\mathbf a)^2=(\mathbf a\cdot\mathbf a)I$. Repeated multiplication therefore yields, for nonnegative integers $m$,
$$
\boxed{\sigma(\mathbf a)^{2m}=(\mathbf a\cdot\mathbf a)^mI,\qquad
\sigma(\mathbf a)^{2m+1}=(\mathbf a\cdot\mathbf a)^m\sigma(\mathbf a).}
$$
The zeroth power is $I$. If negative powers are intended, they require $\mathbf a\cdot\mathbf a\ne0$, and the same parity formulas then follow from $\sigma(\mathbf a)^{-1}=\sigma(\mathbf a)/(\mathbf a\cdot\mathbf a)$. Every nonzero real vector meets that condition; a nonzero complex null vector need not. For example, $\mathbf a=(1,i,0)$ has $\sigma(\mathbf a)^2=0$ and no inverse.
For a real <unit vector> $\hat{\mathbf n}$, let $S=\sigma(\hat{\mathbf n})$. Then $S^2=I$, so even powers are $I$ and odd powers are $S$. Split the absolutely convergent <matrix exponential> into its even and odd series:
$$
\begin{aligned}
e^{i\theta S}
&=I\sum_{m=0}^\infty\frac{(-1)^m\theta^{2m}}{(2m)!}
+iS\sum_{m=0}^\infty\frac{(-1)^m\theta^{2m+1}}{(2m+1)!}\\
&=\boxed{\cos\theta\,I+i\sin\theta\,\sigma(\hat{\mathbf n})}.
\end{aligned}
$$
The result holds for every real $\theta$, not merely small angles. Its two scalar series are the ordinary cosine and sine series.
Back to article page
