# Solving for complex Eigen Values

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?

• Please format you post using MathJax. If you don't know how to use it, a quick reference can be found here. – memerson Nov 6 '18 at 0:24
• To find an eigenvector, it suffices to "guess" the solution $x=i,y=1,z=0$ – Omnomnomnom Nov 6 '18 at 0:35
• And the actual answer is solve the system of linear equations. E.g. use Gaussian elimination or something. – jgon Nov 6 '18 at 0:36
• @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. – Doesbaddel Nov 6 '18 at 0:37
• @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. – jgon Nov 6 '18 at 0:43

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.