Past exam of the mathematical methods I course of the University of Oxford / 2013 / 11 / c / ii / Solution

Setting $\mathbf{b}=\mathbf{a}$ in 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 $m$,The zeroth power is $I$. If negative powers are intended, they require $\mathbf{a}\cdot \mathbf{a}\ne 0$, 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:The result holds for every real $\theta$, not merely small angles. Its two scalar series are the ordinary cosine and sine series.