<s>[INST] Solving a linear system with complex eigenvalues I have the system:
\begin{equation}
x' = \begin{pmatrix}5&10\\-1&-1\end{pmatrix}x
\end{equation}
The corresponding characteristic equation is:
\begin{equation}
\lambda^2-4\lambda+5 \\
\implies \lambda_1 = 2+i \land \lambda_2 = 2-i
\end{equation}
I am having trouble solving for the eigenvector:
\begin{align}
(A-\lambda I)x_1 =& 0\\
\begin{pmatrix}3-i&10\\-1&-3-i\end{pmatrix}x_1 =&0
\end{align}
I really don't know what to do with the complex numbers. I thought about multiplying by the conjugate in the rows, but then I will have complex numbers in opposite columns again...


*

*How do I solve for the eigenvectors in the case of complex numbers?

 [/INST] If we set $x_1 = (a, b)^T$, we get
$$
\begin{pmatrix}3-i&10\\-1&-3-i\end{pmatrix}x_1 = \begin{pmatrix}3-i&10\\-1&-3-i\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}\\\\
= \begin{pmatrix}(3-i)a + 10b\\-a - (3 + i)b\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}
$$
which is just a set of two equations of two (complex) unknowns. Multiplying the lower equation by $-(3-i)$ (multiplying by conjugates is a good trick against both square roots and complex numbers, you're right bout that) shows that these equations are linearly dependant (which is good), and we get that $b = \left(-\frac{3}{10} + \frac{i}{10}\right)a$ solves them both:
$$
(3-i)a +10b = 0\\\\
(3-i)a = -10b\\\\
\frac{3-i}{-10}a = b \\\\
\left(-\frac{3}{10} + \frac{i}{10}\right)a = b
$$
And there you have your eigenspace for the first eigenvalue.
</s>