The problem I'm trying to solve is this:

$X'(t) \in \mathbb{R}^3 \,, \, \omega = (\omega\_1,\omega\_2,\omega\_3) $ Find the general solution for

$$X'(t) = \omega \times X(t)$$

After doing the cross product and rearranging a bit I got to

$$\begin{pmatrix} x\_1' (t) \\ x\_2' (t) \\ x\_3' (t) \end{pmatrix} = \begin{pmatrix} 0 & -\omega\_3 & \omega\_2 \\ \omega\_3 & 0 &-\omega\_1 \\ -\omega\_2 & \omega\_1 & 0 \end{pmatrix} \begin{pmatrix} x\_1 (t) \\ x\_2 (t) \\ x\_3 (t) \end{pmatrix}$$

Then I looked for the eigenvalues of the matrix, which are 0, $\sqrt{-\omega\_1 ^2 -\omega\_2 ^2 -\omega\_3 ^2}$ and $-\sqrt{-\omega\_1 ^2 -\omega\_2 ^2 -\omega\_3 ^2}$.

Once I have the eigenvectors, I know how to proceed to find the general solution, but finding the eigenvectors of the second and third eigenvalues lead me to really weird looking stuff, which makes me think that there's another way of solving this.

I've been thinking about this all day, and I can't think of anything else, any help would be greatly appreciated.