Finding eigenvectors with complex eigenvalue

Question

I have trouble with how to find eigenvectors when you have a complex eigenvalue

For example the matrix $$ \begin{pmatrix} 0 & 1\\ -2 & -2 \end{pmatrix}$$

Here you get the eigenvalues $-1$ and $ \pm i$

Where do i go from here to find a eigenvector. The solution says it should be the 2x1 matrix: $$\begin{pmatrix} 1\pm i \\ -2 \end{pmatrix}$$

Answer 1

$\newcommand{\x}{\mathbf{x}}$You use the same procedure to find the eigenvectors as in the real case, just now you may have to use complex number arithmetic.

Let $A = \begin{pmatrix} 0 & 1\\ -2 & -2 \end{pmatrix}.$ I'll get you started on how to find the eigenvectors for the eigenvalue $-1+i$. As usual, these are found by solving the equation $(A - (-1+i)I)\x = \mathbf{0}$. Thus we solve the linear system

$$\left(\begin{array}{cc|c}0-(-1+i) & 1 & 0 \\ -2 & -2-(-1+i) & 0\end{array}\right),$$ that is, $$\left(\begin{array}{cc|c}1-i& 1 & 0 \\ -2 & -1-i& 0\end{array}\right).$$

Can you solve this (if not, the first step would be to practise complex number arithmetic and row reductions with complex numbers)?

Answer 2

You do it just as you would for real eigenvalues.

For instance, if you want an eigenvector corresponding to the eigen value $-1+i$, you solve the system$$\left\{\begin{array}{ll}y=(-1+i)x\\-2x-2y=(-1+i)y,\end{array}\right.$$you will get that the solutions are the vectors of the form $\bigl((1+i)x,-2x\bigr)$. In particular, $(1+i,-2)$ is an eigenvector corresponding to the eigenvalue $-1+i$.