Find the complex eigenvectors, knowing the eigenvalues If $$A= \begin{pmatrix}1 & -1 \\ h^2 & 1\end{pmatrix},$$ I know the complex eigenvalues are $1+ih$ and $1-ih$. How do we find the complex eigenvectors? Can someone please explicitly show me the working?
 A: i will show you how to find the eigenvector corresponding to the eigenvalue $1 + ih.$ it follows the same lines as when the eigenvalue is real. you need to solve the homogeneous system of equations represented by the matrix $$\pmatrix{-ih&-1\\h^2&-ih}$$ we already know this matrix has rank < $2,$ and in fact is of rank $1.$ therefore the second row must be a multiple of the first row(this works only for a $2\times 2$ matrix). in our case if you multiply the first row by $ih,$ you get the second row. we basically have one equation $$-ihz_1 - z_2 = 0 $$ one easy solution is $$ z_1 = i, z_2 = h$$ 
so we have found an eigenvector $$\pmatrix{i\\h}$$ corresponding to the eigenvalue $1+ih.$
i will let you practice finding the eigenvector corresponding to the eigenvalue $1-ih$ (you may see a pattern and guess what it should be.)
A: Hint: If $e$ is an eigenvalue, its eigenspace is 
$$\ker(A-e\mathbb{I}_n)$$
Edit: Let's find the eigenspace of the second eigenvalue. This leads to finding the kernel of the matrix
$$\left \lgroup\begin{matrix}
ih & -1\cr
h^2 & ih
\end{matrix} \right \rgroup$$
If $h\neq 0$, we can multiply the second line by $\frac{i}{h}$, which yields
$$\left \lgroup\begin{matrix}
ih & -1\cr
ih & -1
\end{matrix} \right \rgroup$$
The vectors of this matrix's kernel are of the form $(ih)x=y$, so the eigenspace is 
$$\left \langle \begin{pmatrix}1\\ih\end{pmatrix}\right \rangle$$
Can you apply this process to the first eigenvalue, now?
A: Use the exact same method as for real eigenvectors:
$$\begin{pmatrix}1&-1\\h^2&1\end{pmatrix}\begin{pmatrix}x_i\\y_i\end{pmatrix}=\lambda_i\begin{pmatrix}x_i\\y_i\end{pmatrix}$$
So, for $\lambda_1=1+ih$:
$$\begin{pmatrix}1&-1\\h^2&1\end{pmatrix}\begin{pmatrix}x_1\\y_1\end{pmatrix}=\begin{pmatrix}(1+ih)x_1\\(1+ih)y_1\end{pmatrix}$$
$$\therefore\begin{cases}x_1-y_1=(1+ih)x_1\\h^2x_1+y_1=(1+ih)y_1\end{cases}\Longrightarrow ihx_1=-y_1$$
And a general eigenvector for $\lambda_1$ is:
$$\vec{v_1}=t\begin{pmatrix}1\\-ih\end{pmatrix},\quad t\in\mathbb{R}$$
