Suppose we have a system of ode with complex eigenvalues $a\pm ib$, and we set $X=TY$ where $T$ is a transformation matrix. Since $X$ has complex eigenvalues, its eigenvectors must be complex as well. From eigenvalue decomposition, we can have $X=V\Lambda V^{-1}$ where $V$ is the right eigenvector matrix. So, if I let $T=V$, I have a non-real solution and I don't understand why.
$X'=\begin{pmatrix}0&3\\-3&0\end{pmatrix}X$, eigenvalues are $\pm3i$ and eigenvectors are $[-i,1]^T$ and $[i,1]^T$. Let $T=\begin{pmatrix}-i&i\\1&1\end{pmatrix}$. If I apply transformation, I will $Y'=\Lambda Y$, and the general solution is $$Y=a\begin{pmatrix}\cos(3t)\\-\sin(3t)\end{pmatrix}+b\begin{pmatrix}\sin(3t)\\\cos(3t)\end{pmatrix}$$ From above, we have $X=TY$, so the general solution to $X$ will be $$X=\begin{pmatrix}-i&i\\1&1\end{pmatrix}\left(a\begin{pmatrix}\cos(3t)\\-\sin(3t)\end{pmatrix}+b\begin{pmatrix}\sin(3t)\\\cos(3t)\end{pmatrix}\right)$$
English isn't my first language, there may have some confusions.