Let $\mathbf A$ be a matrix such that $\mathbf A^2=-\mathbf I$. Prove that $\exp(\varphi\mathbf A)=\mathbf I\cos{\varphi}+\mathbf A\sin{\varphi}$

Question

Let $\mathbf A$ be a matrix such that $\mathbf A^2=-\mathbf I$. Prove that $\exp(\varphi\mathbf A)=\mathbf I\cos{\varphi}+\mathbf A\sin{\varphi}$

This is my attempt: $$\mathbf A^2=-\mathbf I \implies \mathbf A=\begin{pmatrix}i&0 \\0&i\end{pmatrix}$$

Using the definition of the matrix exponential function: $\sum_{n=0}^{\infty}\frac{1}{n!}\mathbf A^n$, I get

$$\begin{aligned}&\exp{(\varphi \mathbf A)}=\sum_{n=0}^{\infty}\frac{1}{n!}\begin{pmatrix}i\varphi&0\\0&i\varphi\end{pmatrix}^n\\ & \iff \sum_{n=0}^{\infty}\frac{1}{(2n+1)!}\begin{pmatrix}i\varphi&0\\0&i\varphi\end{pmatrix}^{(2n+1)}+\sum_{n=0}^{\infty}\frac{1}{(2n)!}\begin{pmatrix}i\varphi&0\\0&i\varphi\end{pmatrix}^{(2n+1)}\\ & \iff \sum_{n=0}^{\infty}\frac{(i)^{2n+1}}{(2n+1)!}\varphi^{2n+1}\begin{pmatrix}1&0\\0&1\end{pmatrix}+\sum_{n=0}^{\infty}\frac{(i)^{2n}}{(2n)!}\varphi^{2n}\begin{pmatrix}1&0\\0&1\end{pmatrix}\end{aligned}$$

These sums look very close to the cosine and sine series but I don't know how to deal with the imaginary unit $i$ in the sums. Maybe someone can give me a hint.

Answer 1

$A^2=-I$ does not imply that $A=iI$. For example, one such $A$ is $$ \left(\begin{array}{rr} 0 & -1 \\ 1 & 0\end{array}\right). $$ Now, for its exponential, note first that $$ A^{2n}=(A^2)^n=(-I)^n=(-1)^nI $$ and $$ A^{2n+1}=(A^2)^nA=(-I)^nA=(-1)^nA. $$ Then $$ \exp(t A)=\sum_{n=0}^\infty\frac{t^n}{n!}A^n= \sum_{k=0}^\infty\frac{t^{2k}}{(2k)!}A^{2k}+ \sum_{k=0}^\infty\frac{t^{2k+1}}{(2k+1)!}A^{2k+1}=\\ \sum_{k=0}^\infty\frac{(-1)^kt^{2k}}{(2k)!}I+ \sum_{k=0}^\infty\frac{(-1)^{k}t^{2k+1}}{(2k+1)!}A=(\cos t)I+(\sin t)A. $$

Answer 2

An easy induction shows that $$ A^n=\begin{cases}\hphantom{-}I&\text{if}\enspace n\equiv 0\mod4\\ \hphantom{-}A&\text{if}\enspace n\equiv 1\mod4\\ -I&\text{if}\enspace n\equiv 2\mod4\\ -A&\text{if}\enspace n\equiv 3\mod4 \end{cases} $$ There results that \begin{align*} \exp (\varphi A)&=\sum_{k=0}^\infty \frac{\varphi^{2k}}{2k!} A^{2k}+\sum_{k=0}^\infty \frac{\varphi^{2k+1}}{(2k+1)!} A^{2k+1}\\ &=\biggl(\sum_{k=0}^\infty (-1)^k\frac{\varphi^{2k}}{(2k)!}\biggr)I+\biggl(\sum_{k=0}^\infty (-1)^k\frac{\varphi^{2k+1}}{(2k)!}\biggr)A\\ &=(\cos\varphi)\, I+(\sin\varphi)\, A. \end{align*}