$$\vec x'=A\vec x=\begin{bmatrix}5&-3&-2\\8&-5&-4\\-4&3&3\end{bmatrix}\vec x$$ Show that $r=1$ is a triple eigenvalue of $A$ and that there are only two linearly independent corresponding eigenvectors $a_1=(1 ,0 ,2)$ and $a_2=(0, 2, -3)$.
I'm wondering why we have only two linearly independent eigenvectors; could I add $(2,2,1)$ to the mix? I know how to show it if there's only one corresponding eigenvector, but not if there's two.
$(2,2,1)$ is just $2a_1+a_2$, so it's not linearly independent. The eigenspace could be 1, 2 or 3-dimensional. In this case, it's 2. I don't know what "it" is in your last sentence "show it." But you can get a generalized eigenvector in the usual way.