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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 $2$):

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

How do I solve for the eigenvector now? Thanks.


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

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

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
up vote 2 down vote accepted

Try multiplying the first equation by $1-i\sqrt{3}$. It should make things easier for you.

Keep in mind that complex eigenvalues don't really change your procedure for finding eigenvectors at all. Of course, if you've never encountered them before, they can be a bit alarming.

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I did what you suggested and I get x=0, y=0. Please check my edited question. Thanks. – Jey Dec 10 '12 at 2:50
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
Thank you. It works perfectly. – Jey Dec 10 '12 at 4:49

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