derive a general formula for the elements of $e^{\zeta S}$ as an infinite sum of powers of $\zeta$

Question

Let $S$ be a $2 ×2$ symmetric matrix

$S =\begin{bmatrix} 0& -1\\ -1& 0 \end{bmatrix} $

compute the first four terms in the Taylor expansion of the exponential $e^{\zeta S}$ around $\zeta = 0$ and derive a general formula for the elements of $e^{\zeta S}$ as an infinite sum of powers of $\zeta$

Answer 1

$$\begin{aligned} e^{\zeta S} &= \sum_{k=0}^\infty \frac{(-\zeta)^k}{k!} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^k \\ &= I - \zeta \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + \frac{\zeta^2}{2!} I -\frac{\zeta^3}{3!} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + \textrm{higher order terms}\\ & = \begin{pmatrix} 1+\frac{\zeta^2}{2!} +\frac{\zeta^4}{4!} + \cdots & -\zeta-\frac{\zeta^3}{3!}-\frac{\zeta^5}{5!} -\cdots \\ -\zeta-\frac{\zeta^3}{3!}-\frac{\zeta^5}{5!} -\cdots & 1+\frac{\zeta^2}{2!} +\frac{\zeta^4}{4!} + \cdots \end{pmatrix}\\ &= \begin{pmatrix} \cosh \zeta & -\sinh \zeta \\ -\sinh \zeta & \cosh \zeta \end{pmatrix} \end{aligned} $$

Answer 2

Let $$f(z)= \sum_{k\ge 0} c_k z^{2k},\qquad g(z)= \sum_{k\ge 0} d_k z^{2k+1}$$ What can you say on $$f(S),\qquad g(S)$$ Then look at $$f(z)=e^z+e^{-z},\qquad g(z)=e^z-e^{-z}$$