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I want to compute the eigenvectors of the matrix $$\begin{pmatrix}6&2\\-10&-1\end{pmatrix}$$ and thus far I got the eigenvalues $\lambda_{1,2}=\frac{1}{2}(5\pm\sqrt{31}i)$. However solving for example the system $$\begin{pmatrix}6-\frac{1}{2}(5+\sqrt{31}i)&2\\-10&-1-\frac{1}{2}(5+\sqrt{31}i)\end{pmatrix}\cdot v=0$$ for the first eigenvector was pretty difficult and I couldn't come to a proper solution whatsoever. Do you have suggestions on how to simplify the system?

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  • $\begingroup$ The idea is the same regardless of what numbers are there, namely row reduce. Remember now though that you are working in $\mathbf{C}$ instead of the perhaps more familiar $\mathbf{R}$. $\endgroup$ – Future Jan 21 '16 at 23:42
  • $\begingroup$ @Prospect I do know the algorithm but I am struggling with the particular steps in this specific case. $\endgroup$ – Christian Ivicevic Jan 21 '16 at 23:48
  • $\begingroup$ Well there are several steps. Which step in particular are you struggling with? $\endgroup$ – Future Jan 21 '16 at 23:49
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We have:

$$\begin{pmatrix}6-\frac{1}{2}(5+\sqrt{31}i)&2\\-10&-1-\frac{1}{2}(5+\sqrt{31}i)\end{pmatrix}\cdot v=0$$

Lets simplify:

$$\begin{pmatrix}\frac{1}{2}(7-\sqrt{31}i)&2\\-10&-\frac{1}{2}(7+\sqrt{31}i)\end{pmatrix}\cdot v=0$$

Divide the first row by $\frac{1}{2}(7-\sqrt{31}i)$ and the second row by $-10$:

$$\begin{pmatrix}1 & \dfrac{4}{(7-\sqrt{31}i)}\\1&\dfrac{1}{20}(7+\sqrt{31}i)\end{pmatrix}\cdot v=0$$

Multiply $a_{12}$ by the conjugate:

$$\begin{pmatrix}1 & \dfrac{1}{20}(7+\sqrt{31}i)\\1&\dfrac{1}{20}(7+\sqrt{31}i)\end{pmatrix}\cdot v=0$$

$R_2 - R_1 \rightarrow R_2$:

$$\begin{pmatrix}1 & \dfrac{1}{20}(7+\sqrt{31}i)\\0&0\end{pmatrix}\cdot v=0$$

So:

$$v_1 = \left(-\dfrac{1}{20}(7+\sqrt{31}i), ~~1 \right)$$

Given that the eigenvalues are complex cojugates, so too are the eigenvectors, so you already know $v_2$.

However, you might want to give that one a go and make sure you can verify that.

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