Why $B=\sum_{n\geq 0}\frac{\theta^n}{n!}J^n$ where $J:=\left( \matrix{0&-1\\ 1&0}\right)$?

Question

I'm trying to understand the answer to this post Exponential map and the special orthogonal group, and what I can't understand yet is why $B=\sum_{n\geq 0}\frac{\theta^n}{n!}J^n$ where $J:=\left( \matrix{0&-1\\ 1&0}\right)$. Could someone tell me why this is true? Thank you.

Answer 1

$\sum_\limits{n=0}^\infty \frac{x^n}{n!}$ is a Taylor series for $e^x$

$\sum_\limits{n=0}^\infty \frac{\theta^n}{n!}J^n = e^{\theta J}$

The set of $2\times 2$ matrices of the form $\begin{bmatrix} a& -b\\b& a \end{bmatrix}$ form a field that is isomorphic to $\mathbb C$

With this in mind, there is an isomorphism to

$B = \sum_\limits{n=0}^\infty \frac{(i\theta)^n}{n!}$

and

$B = e^{i\theta}$

Alternatively, you could say:

$J^2 = -I, J^3 = -J, J^4 = I, J^{4n+k}= J^k$

Break up the summation into even and odd values.

$B = \sum_\limits{n=0}^\infty \frac{(-1)^n(\theta)^2n}{2n!}I + \sum_\limits{n=0}^\infty \frac{(-1)^n(\theta)^{2n+1}}{(2n+1)!}J$

$B = I \cos\theta + J\sin\theta$

Answer 2

HINT:

Note that:

$\theta J \left(\matrix{\ 1 \\ \ i}\right)=\left( \matrix{0&-\theta\\ \theta&0}\right)\left(\matrix{\ 1 \\ \ i}\right)=-i\theta \left(\matrix{\ 1 \\ \ i}\right)$

$\theta J \left(\matrix{\ i \\ \ 1}\right)=\left( \matrix{0&-\theta\\ \theta&0}\right)\left(\matrix{\ i \\ \ 1}\right)=i\theta \left(\matrix{\ i \\ \ 1}\right)$

So $i\theta$, $-i\theta$ are eigenvalues of $\theta J$ ​​associated to $\left(\matrix{\ i \\ \ 1}\right)$ and $\left(\matrix{\ 1 \\ \ i}\right)$ respectively. Then $\theta J=C\theta DC^{-1}$ where $C=\left( \matrix{1&i\\ i&1}\right)$, $D=\left( \matrix{-i&0\\ 0&i}\right)$. Hence $e^{\theta J}=Ce^{\theta D}C^{-1}=\left( \matrix{\cos\theta &-\sin\theta\\ \sin\theta&\cos\theta}\right)=B$.