Generalized eigenvector question (ODE system)

Question

This is an example in Boyce-Diprima. I have the following system$$x'=\begin{bmatrix} 1 & -1 \\ 1 & 3 \end{bmatrix}x$$ I solve for the eigenvalues, which is just $\lambda=2$ in this case, and it has has eigenvector $\xi = (1,-1)$. I set up the equation to solve for the generalized eigenvector which gives me $ -\eta_1 + \eta_2=1$. Now, here is my question. If I set $\eta_1=k$ and the solve the equation for the generalized eigenvector, I get $$\eta= \begin{bmatrix} 0 \\ -1\end{bmatrix} + k\begin{bmatrix} 1 \\ -1\end{bmatrix}$$ But if I instead set $\eta_2=-k$ and solve the equation for the generalized eigenvector, I get

$$\eta= \begin{bmatrix} -1 \\ 0\end{bmatrix} + k\begin{bmatrix} 1 \\ -1\end{bmatrix}$$ They are not multiplicative constants of each other, which seems strange. Are both correct, and if so, why?

Thanks!

Answer 1

If $A=\begin{bmatrix} 1 & -1 \\ 1 & 3 \end{bmatrix}$, then $A-2I=\begin{bmatrix} -1& -1 \\ 1 & 1 \end{bmatrix}$.

Therefore $(A-2I)^2=\begin{bmatrix} -1& -1 \\ 1 & 1 \end{bmatrix}^2=\begin{bmatrix} 0&0 \\ 0&0\end{bmatrix}$.

We can therefore see that $(A-2I)^2$ will reduce every vector to zero.

Equivalently, we can say that $(A-2I)$ will map any vector to a multiple of the eigenvector $\begin{bmatrix} 1 \\ -1\end{bmatrix}$.