Finding Eigenvectors [Confused] I would like to find the eigenvalues for the matrix 
$$\begin{pmatrix}
2 & 3-3i \\
3+3i & 5
\end{pmatrix}$$

I find that the eigenvalues are $8$ and $-1$.
For eigenvalue of $8$ I get
$$
\begin{pmatrix}
-2 & 1-1i \\
1+1i & -1
\end{pmatrix}=0
$$
and I get the equation
$$-2x + (1-1i)y = 0\\ 
(1+1i)x - y = 0
$$
My question is how do I solve this system? I've tried multiple times and have ended up with the wrong eigenvalue when comparing with the back of my textbook.
 A: First the eigenvalues
$\begin{align}\det(A-tI)&=\begin{vmatrix}2-t & 3-3i\\ 3+3i & 5-t\end{vmatrix}=(2-t)(5-t)-(3+3i)\overline{(3+3i)}=10-7t+t^2-18\\\\&=t^2-7t-8=(t+1)(t-8)\end{align}$
So you are correct, $\operatorname{Sp}(A)=\{-1,8\}$.
Now for the eigenvectors you want to solve two systems for $v=(x,y)^T$:


*

*$Av=-v$ 


$\begin{cases}2x+(3-3i)y=-x\\(3+3i)x+5y=-y\end{cases}\iff\begin{cases}(3-3i)y=-3x\\(3+3i)x=-6y\end{cases}\overset{\div 3}{\iff}\begin{cases}(1-i)y=-x\\(1+i)x=-2y\end{cases}$
Since $(1+i)(1-i)=2$ both equations are equivalent, meaning we can choose either $x$ or $y$ and the other is determined.
If you actually report one equation into the other you find 
either y=y or x=x, thus either y or x is a free variable. 

For instance (1+i)x=(1+i)(1-i)(-y)=-2y <=> -2y=-2y <=> y=y

This is expected since eigenvectors are only defined modulo
proportionality.

We can choose for instance $y=-1$ then $v_{-1}=\begin{pmatrix}1-i\\-1\end{pmatrix}$.


*

*$Av=8v$ 


$\begin{cases}2x+(3-3i)y=8x\\(3+3i)x+5y=8y\end{cases}\iff\begin{cases}(3-3i)y=6x\\(3+3i)x=3y\end{cases}\overset{\div 3}{\iff}\begin{cases}(1-i)y=2x\\(1+i)x=y\end{cases}$
Since $(1+i)(1-i)=2$ both equations are equivalent, meaning we can choose either $x$ or $y$ and the other is determined.
We can choose for instance $x=1$ then $v_{8}=\begin{pmatrix}1\\1+i\end{pmatrix}$.

The change of basis matrix is then $P=(v_8,v_{-1})=\begin{pmatrix}1 & 1-i\\ 1+i & -1\end{pmatrix}$
And now $P^{-1}AP=\begin{pmatrix}8 & 0\\ 0 & -1\end{pmatrix}$
