<s>[INST] How to find eigenvectors given complex eigenvalues I am given that: $\vec{x}' =  A\vec{x}$, where $A=\begin{pmatrix} -3 & 0 & 2 \\ 1 & -1 & 0 \\ -2 & -1 & 0  \end{pmatrix}$
I want to find the general solution of this in terms of real-valued functions.
Computing $\det(A-\lambda I)=0$, I found the characteristic equation to be $(\lambda +2)(\lambda^2 +2\lambda+3)=0.$ So $\lambda_1=-2,\lambda_{2,3} = -1\pm\sqrt{2}i$.
I'm having trouble dealing with $\lambda_{2,3}$. Can anyone carry me through the computations?
Thanks.
 [/INST] Let's write $\lambda_2=-1+i\sqrt{2}$, $\lambda_3=-1-i\sqrt{2}$.
We have 
$$A-\lambda_2 I=\begin{pmatrix} -2-i\sqrt{2} & 0 & 2 \\ 1 & -i\sqrt{2} & 0 \\ -2 & -1 & 1-i\sqrt{2}  \end{pmatrix} $$
Applying row operations, we obtain
\begin{align*}
A-\lambda_2 I&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 1 & -i\sqrt{2} & 0 \\ -2 & -1 & 1-i\sqrt{2}  \end{pmatrix}  \\
&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 0 & -i\sqrt{2} & \frac{2-i\sqrt{2}}{3} \\ 0 & -1 & \frac{-1-i\sqrt{2}}{3}  \end{pmatrix} \\
&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 0 & 1 & \frac{1+i\sqrt{2}}{3} \\ 0 & -1 & \frac{-1-i\sqrt{2}}{3}  \end{pmatrix} \\
&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 0 & 1 & \frac{1+i\sqrt{2}}{3} \\ 0 & 0 & 0  \end{pmatrix}.
\end{align*}
From this, we can see that a vector $x=(x_1,x_2,x_3)^T$ is mapped to $0$ under $A-\lambda_2 I$ if 
\begin{align*}
&x_1+\frac{-2+i\sqrt{2}}{3}x_3=0 \\
&x_2+\frac{1+i\sqrt{2}}{3}x_3=0,
\end{align*} 
or,
\begin{align*}
x_1&=\frac{2-i\sqrt{2}}{3}x_3,\\
x_2&=\frac{-1-i\sqrt{2}}{3}x_3.
\end{align*}
Therefore, the eigenspace of $A$ associated with eigenvalue $\lambda_2$ is spanned by
$$x=\begin{pmatrix} \frac{2-i\sqrt{2}}{3} \\ \frac{-1-i\sqrt{2}}{3} \\1\end{pmatrix},$$
i.e., $x$ is an eigenvector with eigenvalue $\lambda_2$. This gives your solution for $\lambda_2$.  To find the eigenvector associated with $\lambda_3$, essentially do the same as what I did.  If you require further explanation of what I did, please let me know and I will edit.
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