Let $A = \left[ \begin{matrix} I & X \\ 0 & -I \end{matrix} \right]$, $X\in\mathbb R^{m\times n}$ arbitrary. Need to show value of $e^A$.

Let $A = \left[ \begin{matrix} I & X \\ 0 & -I \end{matrix} \right]$, $X\in\mathbb R^{m\times n}$ arbitrary. Need to show value of $e^A = \left[ \begin{matrix} eI & \sinh 1X \\ 0 & \frac 1 e I\end{matrix} \right]$.

We know that $A^2 = I$ and $A^k = \begin{cases} A \quad &k\text{ odd} \\ I & k\text{ even}. \end{cases}$ for $k = 0, 1,\dots$. A plausible solution is so far $$ \begin {split} e^{A} &= \sum_{k=0}^\infty \frac 1{k!}A^k \\ &= \left[ \begin{matrix} \sum_{k=0}^\infty \frac 1{k!}I & \sum_{k\text{ odd}} \frac 1{k!}X \\ 0 & \sum_{k=0}^\infty \frac { (-1)^k}{k!}I \end{matrix} \right] \\ &= \left[ \begin{matrix} eI & \frac 12\left(\sum_{k=0}^\infty \frac 1{k!}X - \sum_{k=0}^\infty \frac {(-1)^k}{k!}X\right) \\ 0 & \frac 1eI \end{matrix} \right] \\ &\vdots \\ &= \left[ \begin{matrix} eI & \sinh 1X \\ 0 & \frac 1 e I\end{matrix} \right].\end{split} $$

Note: The '$\vdots$' part is easily follows recognizing the power series expansion of $e^x$. What I don't understand is the following string of arguments in the upper right block regarding $X$ matrix. That is how does this work: $$\sum_{k=0}^\infty \frac 1{k!}X = \sum_{k\text{ odd}} \frac 1{k!}X = \frac 12\left(\sum_{k=0}^\infty \frac 1{k!}X - \sum_{k=0}^\infty \frac {(-1)^k}{k!}X\right).$$

EDIT: I understand the last equality now considering the power series expansion of $\sinh$. Now I just need clarification on going from LHS to RHS: $$\sum_{k=0}^\infty \frac 1{k!}X = \sum_{k\text{ odd}} \frac 1{k!}X $$

Any help is appreciated thanks!

Solution 1:

$$\sum_{k=0}^\infty{1\over k!}A^k = \sum_{k\text{ even}}{1\over k!}I + \sum_{k\text{ odd}}{1\over k!}A$$

Solution 2:

$ e^{A}=\sum \limits_{\underset{}{k=0}}^\infty \frac{A^{k}}{k!}= \sum \limits_{\underset{k Odd}{k=0}}^\infty \frac{A}{k!} + \sum \limits_{\underset{k even}{k=0}}^\infty \frac{I}{k!}.$

$\frac{1}{2}(e^1-e^{-1})= \frac{1}{2}(\sum \limits_{\underset{}{k=0}}^\infty \frac{1}{k!}- \sum \limits_{\underset{}{k=0}}^\infty \frac{(-1)^k}{k!} )\\= \sinh(1) \\=\frac{1}{2}(2\sum \limits_{\underset{k odd}{k=0}}^\infty \frac{1}{k!})\\= \sum \limits_{\underset{k odd}{k=0}}^\infty \frac{1}{k!}.$