Solving for complex Eigen Values

Question

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?

Answer 1

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.