<s>[INST] How to find matrix exponential $e^A$ I have the matrix $$A =\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}$$
and I have to find $e^A$
I've found two complex-conjugate eigenvalues $\lambda_{1,2} = \pm i$
so substracting $\lambda_1 = i$ from the matrix's diagonal I got:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix}$$
and therefore. to find eigenvector I have to solve the system:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
so the first eigenvector is $h_1 = \begin{pmatrix}1 \\ i\end{pmatrix}$
 and the second one is $h_2 = \begin{pmatrix}1 \\ -i\end{pmatrix}$
so the general solution is $$x(t) = C_1e^{it}\begin{pmatrix} 1 \\i\end{pmatrix} + C_2e^{-it}\begin{pmatrix} 1 \\-i\end{pmatrix}$$
I know that now I have to solve two Cauchy's problems for the standard basis $\mathbb{R}^2$ with vectors $v_1 = (1, 0)$ and $v_2 = (0,1)$ But I do not know how to approach it for complex numbers
 [/INST] Answer
Let a,b $\in \mathbb{R}$ and
$$
\begin{align}
I &= 
    \begin{pmatrix}
    1 & 0 \\
    0 & 1 \\
    \end{pmatrix} \\
J &= 
    \begin{pmatrix}
    0  & 1 \\
    -1 & 0 \\
    \end{pmatrix}
\end{align}
$$
Then the following formula is true,

$$  \exp(aI + bJ) = \exp(a)[I \cos(b) + J \sin(b)]$$

In the special case where a = 0 and b = 1, 

$$  \exp(J) = 
    \begin{pmatrix}
    \cos(1)  & \sin(1) \\
    -\sin(1) & \cos(1) \\
    \end{pmatrix}
$$


Proof
We will need the following lemmas to go through the chain of equalities. The lemmas are not going to be proved or explained here. 
Let n $\in \mathbb{N}$, x $\in \mathbb{R}$, A and B two matrices


*

*The product of $aI$ and $bJ$ is commutative

*$\exp(A + B) = \exp(A)\exp(B)$ if $AB = BA$

*$J^{2n} = (-1)^n I$ 

*$J^{2n+1} = (-1)^n J$
Therefore,
$$
\begin{align}
\exp(aI + bJ)
&= \exp(aI)\exp(bJ) \\
&= \exp(a)[I + bJ + (bJ)^2/2! + (bJ)^3/3! + \cdots] \\
&= \exp(a)[I(1 - b^2/2! + b^4/4! - \cdots) + J(b - b^3/3! + b^5/5! - \cdots)]  \\
&= \exp(a)[I\cos(b) + J \sin(b)]
\end{align}
$$
</s>