# Solve for eigenvector (complex eigenvalue)

The complex eigenvalue is confusing me. $\lambda_{1,2} = \frac{1}{2}(5 \pm i\sqrt{3})$

for $$A=\begin{pmatrix}3&1\\-1&2\end{pmatrix}$$

so I get this for $\lambda_1 = \frac{1}{2}(5 - i\sqrt{3})$ (I multiplied both equations by $2$):

$$(1+i\sqrt{3})x+2y=0 \\ -2x - (1+i\sqrt{3})y = 0$$

How do I solve for the eigenvector now? Thanks.

Edit

I don't think I fully understand the process of calculating eigenvectors $$(1-i\sqrt{3})[(1+i\sqrt{3})x+2y=0]\\ -2x - (1+i\sqrt{3})y = 0$$

$$4x+(2-2i\sqrt{3})y=0\\ -2x - (1+i\sqrt{3})y = 0$$

This gives me $x=0$ and $y=0$. Does that mean there is no eigenvector?

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There is really no difference when solving for complex eigenvectors. You substitute your eigenvalue into $A-\lambda I$ and you row reduce as usual for the nullspace. – EuYu Dec 10 '12 at 2:13

Try multiplying the first equation by $1-i\sqrt{3}$. It should make things easier for you.
Ah! I see the problem. Your second equation should be $$-2x-(1-i\sqrt{3})y=0,$$ not $$-2x-(1+i\sqrt{3})y=0.$$ That should fix things for you. – Cameron Buie Dec 10 '12 at 2:58