Exponential of the matrix I want to calculate the matrix exponential $e^{tA}$ of the matrix with the first row being $(0,1)$ and the second $ (-1,0)$. It would be sufficient if you would me the most important steps.

At first, you can easily check that
$$\begin{bmatrix}
0&1\\
-1&0\\
\end{bmatrix}^2=\begin{bmatrix}
-1&0\\0&-1
\end{bmatrix}.$$
So
$$
\begin{aligned}
e^{tA}=\sum_{n=0}^\infty \frac{A^n}{n!}t^n &= \sum_{n=0}^\infty \frac{A^{2n}}{(2n)!} t^{2n}+\sum_{n=0}^\infty \frac{A^{2n+1}}{(2n+1)!} t^{2n+1}\\
&=\sum_{n=0}^\infty \frac{(-1)^n I}{(2n)!}t^{2n} + \sum_{n=0}^\infty \frac{(-1)^n A}{(2n+1)!} t^{2n+1}
\end{aligned}$$
where $A=\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}, \,I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$ By Taylor series of cosine and sine function we get
$$e^{tA}=I\cos t + A\sin t = \begin{bmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{bmatrix}.$$
It is similar to derive a imaginary exponential function use a Taylor series of exponential and trigometric functions.
