How do I show that this square matrix has this relationship? Given a square matrix $M$, I have 
$$\exp M = 1 + \sum^{\infty}_{n=1}\frac{M^n}{n!}$$
for the matrix
$$ M = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & -1 & 0
\end{bmatrix}$$
and any real number $\theta$, I want to show a)
$$\exp\theta M = 1 + M\sin \theta + M^2 (1-\cos \theta)$$
and hence show that b)
$$\exp\theta_1 M \ \exp\theta_2 M = \exp(\theta_1 + \theta_2) M$$
and finally show that c)
$$(\exp\theta M ) ( \exp \theta M)^T = 1$$
Thoughts:


*

*I'm a bit puzzled here since there's an infinity summation. If i let $\theta$ to be in the equation, I think I can get both the sin and cos series approximation? But the thing is these approximations are signed (i.e. they have negative numbers) instead of purely just positive.

*Even so, I'm unable to remove the $M^n$ factor. I tried to do a matrix multiplciation of $M^2$ and couldn't find any discernable pattern.

*perhaps i need to use fourier series?
 A: Note that $M$ is diagonalizable, i.e.,
$$
M=P^{-1}\Lambda P,
$$
where
$$
P=\left(
\begin{array}{ccc}
0&-i&1\\
0&i&1\\
1&0&0
\end{array}
\right),\quad\Lambda=\left(
\begin{array}{ccc}
-i&0&0\\
0&i&0\\
0&0&0
\end{array}
\right).
$$
Thanks to this result, we have
$$
\exp\left(\theta M\right)=\exp\left[P^{-1}\left(\theta\Lambda\right)P\right]=P^{-1}\exp\left(\theta\Lambda\right)P,
$$
where
$$
\exp\left(\theta\Lambda\right)=\exp\left(
\begin{array}{ccc}
-i\theta&0&0\\
0&i\theta&0\\
0&0&0
\end{array}
\right)=\left(
\begin{array}{ccc}
e^{-i\theta}&0&0\\
0&e^{i\theta}&0\\
0&0&1
\end{array}
\right).
$$
A: $$
e^{\theta M}=I+\sum_{n=1}^{\infty}\frac{\theta^{n}M^{n}}{n!}
$$
but the characteristic polynomial for $M$ is $M^{3}+M=0$ then
$$
I+\sum_{n=1}^{\infty}\frac{\theta^{n}M^{n}}{n!}=I+\frac{\theta M}{1!}+\frac{\theta^{2}M^{2}}{2!}-\frac{\theta^{3}M}{3!}-\frac{\theta^{4}M^{2}}{4!}+\cdots=I+M\left(\frac{\theta}{1!}-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}+\cdots\right)+M^{2}\left(\frac{\theta^{2}}{2!}-\frac{\theta^{4}}{4!}+\frac{\theta^{6}}{6!}+\cdots\right)
$$
etc.
A: We start by calculating several $M^n$ matrices
$$ M^1 =M = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & -1 & 0
\end{bmatrix}
$$
$$
M^2  = \begin{bmatrix}
0 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & -1
\end{bmatrix}
$$
$$ M^3 =M^2M = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & -1\\
0 & 1 & 0
\end{bmatrix}=-M
$$
$$
M^4  = M^2M^2=\begin{bmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}=-M^2
$$
By continuing this, you get $M^5=M$, $M^6=M^2$, ...
Now 
$$\exp \theta M = 1 + \sum^{\infty}_{n=1}\frac{\theta^n M^n}{n!}$$
We split this into odd and even powers of $n$
$$\exp \theta M=1+\sum^{\infty}_{n=1}\frac{\theta^{2n} M^{2n}}{(2n)!}+\sum^{\infty}_{n=0}\frac{\theta^{2n+1} M^{2n+1}}{(2n+1)!}$$
We can write the odd powers term as $$\sum^{\infty}_{n=0}\frac{\theta^{2n+1} M^{2n+1}}{(2n+1)!}=M\sum^{\infty}_{n=0}\frac{\theta^{2n+1} (-1)^{n}}{(2n+1)!}=M\sin\theta$$
Similarly,
$$\sum^{\infty}_{n=1}\frac{\theta^{2n} M^{2n}}{(2n)!}=M^2\sum^{\infty}_{n=1}\frac{\theta^{2n} (-1)^{n+1}}{(2n)!}=M^2\left(\sum^{\infty}_{n=1}\frac{\theta^{2n} (-1)^{n}}{(2n)!}+1-1\right)=M^2(1-\cos\theta)$$
