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I am trying to find an eigen vector given $ \lambda = 1+i$:

Given:

$A=\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 2\end{bmatrix}$

$A- \lambda I = \begin{pmatrix} -i & -1 & 1 \\ 1 & -i & 1 \\ 0 & 0 & 1-i\end{pmatrix}$

I know that $-ix - y + z = 0$

However, how should I solve for or get rid of the 'i' in the equation? Do I need to use the complex conjugate for this?

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    $\begingroup$ Please format you post using MathJax. If you don't know how to use it, a quick reference can be found here. $\endgroup$ – memerson Nov 6 '18 at 0:24
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    $\begingroup$ To find an eigenvector, it suffices to "guess" the solution $x=i,y=1,z=0$ $\endgroup$ – Omnomnomnom Nov 6 '18 at 0:35
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    $\begingroup$ And the actual answer is solve the system of linear equations. E.g. use Gaussian elimination or something. $\endgroup$ – jgon Nov 6 '18 at 0:36
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    $\begingroup$ @APorter1031 It's not necessarily to format $-ix-y+z=0$ with MathJax. However, it looks way better, might attract more users and is helpful for other people, who want to recreate the problem later. $\endgroup$ – Doesbaddel Nov 6 '18 at 0:37
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    $\begingroup$ @APorter1031 well just from looking at it I could see that the left two columns were linearly dependent. If I multiplied the left column by $i$ it would equal minus the right column. Hence I knew what the eigenvector should be. It isn't an algorithm tho. The algorithm is gaussian elimination, or however you solve linear equations. $\endgroup$ – jgon Nov 6 '18 at 0:43
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You only stated the first equation for the eigenvector

$$ -i x - y + z = 0\\ $$

There are also the other 2

$$ x - i y + z = 0\\ 0 x + 0 y + (1-i) z = 0 $$

The last equation tells you $z=0$ by dividing by $1-i$. What's left?

$$ -i x - y =0\\ x - i y = 0 $$

Multiply the second by $-i$

$$ - i x - y = 0\\ $$

so you get nothing new from that. Just give a solution to the first. Suppose $x=1$ then $y = - ix=-i$ so together you get $(1,-i,0)$ as the eigenvector. Rescaling $x$ just rescales the entire vector.

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