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

Let$$T=\begin{pmatrix}1&1\\i&-i\end{pmatrix};$$its columns are the eigenvectors that you found. Then$$T^{-1}.A.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}.$$Therefore,$$T^{-1}.A^n.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}^n=\begin{pmatrix}i^n&0\\0&(-i)^n\end{pmatrix}$$and so$$T^{-1}.e^A.T=\sum_{n=0}^\infty\frac1{n!}\begin{pmatrix}i^n&0\\0&(-i)^n\end{pmatrix}=\begin{pmatrix}e^i&0\\0&e^{-i}\end{pmatrix}.$$Therefore,\begin{align}e^A&=T.\begin{pmatrix}e^i&0\\0&e^{-i}\end{pmatrix}.T^{-1}\\&=\begin{pmatrix}\frac{e^i+e^{-i}}2&\frac{e^i-e^{-i}}{2i}\\-\frac{e^i-e^{-i}}{2i}&\frac{e^i+e^{-i}}2\end{pmatrix}\\&=\begin{pmatrix}\cos 1&\sin 1\\-\sin1&\cos1\end{pmatrix}.\end{align}
