How do I show that this square matrix has this relationship?

Question

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:

1. 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.

2. 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.

3. perhaps i need to use fourier series?

Answer 1

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 $$ \exp\left(\theta\Lambda\right)=\exp\left( \begin{array}{ccc} -i\theta&0&0\\ 0&i\theta&0\\ 0&0&0 \end{array} \right)=\left( \begin{array}{ccc} e^{-i\theta}&0&0\\ 0&e^{i\theta}&0\\ 0&0&1 \end{array} \right). $$

Answer 2

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, $$\sum^{\infty}_{n=1}\frac{\theta^{2n} M^{2n}}{(2n)!}=M^2\sum^{\infty}_{n=1}\frac{\theta^{2n} (-1)^{n+1}}{(2n)!}=M^2\left(\sum^{\infty}_{n=1}\frac{\theta^{2n} (-1)^{n}}{(2n)!}+1-1\right)=M^2(1-\cos\theta)$$

Answer 3

$$ e^{\theta M}=I+\sum_{n=1}^{\infty}\frac{\theta^{n}M^{n}}{n!} $$

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.