I have to solve the following equation by diagonalization.
$ X' = \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} X$
I was able to determine the complex eigenvalue roots:
$det(A-\lambda I)=0$
$det\begin{bmatrix}1-\lambda & 1\\1 & -1-\lambda\end{bmatrix} =0$
$(1-\lambda)(-1-\lambda)-1 =0$
$\lambda^2 - 2 = 0$
$\lambda_1 = \sqrt2 i$
$\lambda_2 = -\sqrt2 i$
Now I need to find the eigenvectors for diagonalization. However, I get stuck here: there does not seem to be a solution for the system.
$For \lambda_1 = \sqrt2 i : $
$(A-\lambda_1 I)K_1 = 0$
$ \begin{bmatrix}1-\sqrt2 i & 1\\1 & -1-\sqrt2 i\end{bmatrix}$ $\begin{bmatrix} k_1 \\ k_2 \end{bmatrix} =0$
$ \begin{bmatrix}(1-\sqrt2 i)k_1 + k_2\\k_1 + (-1-\sqrt2 i)k_2\end{bmatrix} = 0$
This is where I get stuck. When I attempt to assign values for $k_1$ and $k_2$ to satisfy one of the equations (an example: $k_1 = -(-1-\sqrt2 i)$ and $k_2 = 1$), I end up with $4=0$ for the other equation, which is clearly wrong. This is the first time I ever struggle with finding an eigenvector of $k_1$ and $k_2$ for a system, and I'm wondering if it even has a solution at all.
Is my assignment question wrong? Does this homogeneous linear system of differential equations have a solution? How can I solve for the eigenvectors in this problem?