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.
 A: Firstly, you should expand an exponent in Taylor series.
Then, you should understand, what happens with matrix, when it is exponentiated with power n.
The last step is to sum up all the matrices and realize, if there are Teylor series of some functions as an entries of the aggregate matrix.  
A: 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.
A: Another approach to calculate the exponent $e^{tA}$ bases on the fact that this exponent solves the Cauchy problem
$$
\frac{d}{dt} Y(t) = AY(t), \quad t\in\Bbb R,\\Y(0)=I.
$$
Obviously, this problem splits to solving two Cauchy problems
$$
\frac{d}{dt} \begin{pmatrix}x(t)\\y(t)\end{pmatrix}  = A \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}y(t)\\-x(t)\end{pmatrix}, \quad t\in\Bbb R,\\x(0)=1,\quad y(0)=0\tag{1}.
$$
and
$$
\frac{d}{dt} \begin{pmatrix}x(t)\\y(t)\end{pmatrix}  = A \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}y(t)\\-x(t)\end{pmatrix}, \quad t\in\Bbb R,\\x(0)=0,\quad y(0)=1\tag{2}.
$$
Whatever method you use to solve problems (1) and (2) - passing to higher order equations, observing that $x^2(t)+y^2(t)=1$, etc -  you will arrive to solutions
$$\begin{pmatrix}x_1(t)\\y_1(t)\end{pmatrix}=\begin{pmatrix}\cos t\\-\sin t\end{pmatrix},\quad \begin{pmatrix}x_2(t)\\y_2(t)\end{pmatrix}=\begin{pmatrix}\sin t\\ \cos t\end{pmatrix}.$$
This allows to conclude that $$Y(t)=e^{tA} =  \begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}.$$
A: You have $e^{tA}=\sum\limits_{k=0}^\infty \cfrac{A^k}{k!}$ so you need the powers of $A^k$.
Usually, the way you do this is that you find $P$ so that $PAP^{-1}=D+N$ so that $D$ is diagonal, $N$ is nilpotent and they commute. That's the Dunford decomposition. Then, $e^{tA}=P^{-1}e^{tD}e^{tN}P$ where $e^{tD}$ is easy to compute because it is diagonal so the exponential of the matrix is the matrix of the exponentials and $e^{tN}$ is easy to compute because there is a finite number of terms in the sum.
In your case, you will have $N=0$ since your matrix can be diagonalized.
A: This kind of problem is often best solved by diagonalising the matrix over the complex numbers, if you can (or triangularising otherwise). Here since $A^2=-I$ the matrix is diagonalisable with eigenvalues $\def\i{\mathbf i}\i,-\i$, and corresponding eigenvectors are
$v_1=\binom 1{-\i}$ and $v_2=\binom1\i$. Then $\exp(tA)\cdot v_1=e^{t\i}v_1$ and $\exp(tA)\cdot v_2=e^{-t\i}v_2$.
Now to convert back to the standard basis $e_1,e_2$ note that $e_1=\frac12(v_1+v_2)$ and $e_2=\frac\i2(v_1-v_2)$, so one gets
$$
 \begin{align}
 \exp(tA)\cdot e_1=\frac12(e^{t\i}v_1+e^{-t\i}v_2)
 &=\frac{e^{t\i}+e^{-t\i}}2e_1+\frac{-\i e^{t\i}+\i e^{-t\i}}2e_2
 \\&=\cos(t)e_1+\sin(t)e_2
\end{align}
$$
and
$$
 \begin{align}
 \exp(tA)\cdot e_2=\frac\i 2(e^{t\i}v_1-e^{-t\i}v_2)
 &=\frac{\i e^{t\i}-\i e^{-t\i}}2e_1+\frac{ e^{t\i}+ e^{-t\i}}2e_2
 \\&=-\sin(t)e_1+\cos(t)e_2.
\end{align}
$$
Therefore
$$ \exp(tA)=\begin{pmatrix}\cos(t)&-\sin(t)\\\sin(t)&\cos(t)\end{pmatrix}.
$$
