If Α a skew-symmetric matrix: \begin{equation*} A = \begin{bmatrix} 0 & -ω \\ ω & 0 \end{bmatrix} \end{equation*}
Show, using Taylor, that:
\begin{equation*} e^{AΘ} = \begin{bmatrix} cosωθ & -sinωθ \\ sinωθ & cosωθ \end{bmatrix} \end{equation*}
i tried that: $$e^{AΘ} = 1+\frac{(AΘ)}{1!}+\frac{({AΘ})^2}{2!}+...$$ but i'm not sure how can i continue in order to export the above matrix. For SO(3), the procedure is defined,but for SO(2)?
Hint:
Write $A = \begin{bmatrix} 0 & -\omega \\ \omega & 0 \end{bmatrix} \ \ $ as $A=\ \ \omega\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Observe pattern which have the consecutive powers of $A$.
$A^2=\ \ \omega^2\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$,
$A^3=\ \ \omega^3\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$,
$A^4=\ \ \omega^4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\omega^4I$,
$A^5=\ \ \omega^5 \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ ...
Probably now you see what is the form of the next powers - here we have only four types of $2 \times 2$ matrices multiplied by the $k$-th power of $\omega$ which can be appropriately grouped together ...