Eigenvectors of a complex matrix Given the following matrix 
$\begin{pmatrix}
0 & 1-i & 0\\
1+i & 0 &1-i\\
0& 1+i &0\\
\end{pmatrix}$
I have found the Eigenvalues $0, 2,-2$. But I have no idea how to calculate the corresponding Eigenvectors and I failed with Gaussian method. What could you recommend?
Thanks in advance!
 A: the eigenvalues $\lambda$ are given by 
$$det\, A = det \pmatrix{-\lambda&1-i&0\\1+i& -\lambda&1-i\\0&1+i&-\lambda} = 0$$ expanding by the first row, we have $$0=-\lambda\left(\lambda^2 -(1-i)(1+i)\right)+(1-i)\lambda(1+i)=-\lambda^3 + 4\lambda = -\lambda(\lambda - 2)(\lambda + 2)$$ the eigenvalues of $A$ are $0, 2, -2.$ 
the eigenvector corresponding to the eigenvalue $0.$ we need to solve $$   \pmatrix{0&1-i&0\\1+i& 0&1-i\\0&1+i&0}\pmatrix{x\\y\\z} = \pmatrix{0\\0\\0}.$$
we can take an eigenvector to be  $\pmatrix{1 - i\\0\\ -1 - i}.$  
i will leave you the task of finding the other eigenvectors.  
$\bf edit:$ now that egreg has done the eigenvalue $-2,$  i will finish it by finding the eigenvector corresponding to $2.$  if you look at the first equation $$-2x + (1-i)y = 0$$ so we will choose $$y = 2, x = 1-i. $$ now look at the last equation: that give $$(1+i)y -2z = 0 \to z = 1 + i $$ therefore an eigenvector corresponding to the eigenvalue $2$ is $$\pmatrix{1-i\\2\\1+i} $$
A: It's exactly the same as with real numbers, except that you have to do arithmetic with complex numbers.  Thus for the eigenvalue $0$, start by interchanging the first and second rows, 
then multiply the first row by $1/(1+i) = (1-i)/2$, ...
A: I'll find an eigenvector corresponding to $\lambda = 0$.  We want to find a vector $v = \left [ \begin{array}{ccc}
v_1 & v_2 & v_3 \\
\end{array} \right ]^T$ such that 
$$
\left [ \begin{array}{ccc}
0 & 1 -i & 0 \\
1 + i & 0 & 1 - i \\
0 & 1+i & 0 \\
\end{array} \right ] \left [ \begin{array}{c}
v_1\\
v_2\\
v_3\\
\end{array} \right ] \;\; =\;\; \textbf{0}_{3\times 3}.
$$
It should be clear from multiplication that $v_2 = 0$.  The only other constraint is $(1+i)v_1 = (1-i)v_3$ or rather 
$$
v_1 \;\; = \;\; \frac{1-i}{1+i} v_3 \;\; =\;\; \frac{1-i}{1+i}\cdot \frac{1-i}{1-i}v_3 \;\; =\;\; \frac{-2i}{2} v_3
$$
Or $v_1 = -iv_3$.  The technique used above can always be used to get rid of complex numbers in the denominator of a fraction.  Take $v_3 = 1$ and we have 
$$
v \;\; =\;\; \left [ \begin{array}{c}
-i \\
0 \\
1 \\
\end{array} \right]
$$
Is one of the eigenvectors.  The others can be found similarly, except you need to compute $(A - \lambda I)w=0$ where $A$ is the matrix given, $\lambda$ is an eigenvalue, and $w$ is a proposed eigenvector.
A: For the eigenvalue $-2$:
\begin{align}
\begin{bmatrix}
2 & 1-i & 0\\
1+i & 2 &1-i\\
0& 1+i &2\\
\end{bmatrix}
&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
1+i & 2 &1-i\\
0& 1+i &2\\
\end{bmatrix} && R_1\gets \tfrac{1}{2}R_1
\\[6px]&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
0 & 1 &1-i\\
0& 1+i &2\\
\end{bmatrix} && R_2\gets R_2-(1+i)R_1
\\[6px]&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
0 & 1 &1-i\\
0& 0 &0\\
\end{bmatrix} && R_3\gets R_3-(1+i)R_2
\\[6px]&\to
\begin{bmatrix}
1 & 0 & i\\
0 & 1 &1-i\\
0& 0 &0\\
\end{bmatrix} && R_1\gets R_1-\tfrac{1-i}{2}R_2
\end{align}
Thus the equations can be written
$$
\begin{cases}
x_1=-ix_3\\
x_2=(i-1)x_3
\end{cases}
$$
so an eigenvector is
$$
\begin{bmatrix}
-i\\
i-1\\
1
\end{bmatrix}
$$
