TITLE: Eigenvalues and Eigenvectors for matrix. Complex Eigenvalues
QUESTION [2 upvotes]: How can I find out the eigenvectors for this matrix: 
$$A=
\begin{pmatrix}
-3 &0&0\\
0&3&-2\\
0&1&1
\end{pmatrix}
$$
I found the eigenvalues: $\lambda_{1}=-3$, $\lambda_{2}=2-i$, $\lambda_{3}=2+i$. 
The eigenvector for $\lambda_{1}=-3$ is $$\begin{pmatrix}
1\\
0\\
0
\end{pmatrix}.$$
For $\lambda_{3}=i+2$ I write $$A(u+iv)=(i+2)(u+iw),$$ but I don't know how to do$\ldots$ Thanks!

REPLY [2 votes]: I do not recommend separating the eigenvector into real and imaginary parts. In my opinion it only makes it more troublesome, complex numbers are perfectly good scalars to work with.
Note the following 
$$\begin{align}
Av=\lambda _2v&\iff (A-\lambda _2I)v=0_{3\times 1}\\
&\iff \begin{pmatrix} -5+i & 0 & 0\\ 0 & 1+i & -2\\ 0 & 1 & -1+i\end{pmatrix}\begin{pmatrix} v_1\\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0\\0\\0 \end{pmatrix}\\
&\iff \begin{cases}(-5+i)v_1&=0\\ (1+i)v_2-2v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ 2v_2-2(1-i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ 2v_2+2(-1+i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ 2v_2-2(1-i)v_3&=0\\ v_2+(-1+i)v_3&=0 \end{cases}\\
&\iff \begin{cases}v_1&=0\\ v_2+(-1+i)v_3&=0\end{cases}\\
&\iff (v_1, v_2, v_3)=(0,(1-i)v_3, v_3).
\end{align}$$
You can thus take the eigenpair $\left(2-i, \begin{pmatrix} 0\\ 1-i\\ 1\end{pmatrix}\right)$.
It is very important to note that given a square real matrix $M$ and $(\lambda, u)$ one of its eigenpairs, the following holds:
$$Mu=\lambda u\iff \overline{Mu}=\overline{\lambda u}\iff M\overline u=\overline \lambda \overline u.$$
What does this tell you?
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