I have this exercise, which I am having some problems with. It says, let $\alpha,\beta$ be scalars in a field characteristic $\neq 2$, and let $A\in$ End(V) be the linear transformation of $V=\mathcal{F}^3$ which is represented by:
$$[A]=\begin{pmatrix} 0&\alpha&1\\ 0&1&0 \\ 1&\beta&0\\ \end{pmatrix}$$
with respect to the standard basis. Determine the generalized eigenspaces of A, for every $\alpha$ and $\beta$ from $\mathcal{F}$.
I've started by finding the eigenvalues for A, by solving finding the characteristic polynomial for $A-\lambda I$ to be $(1-\lambda)(\lambda^2-1)$, and then solving $(1-\lambda)(\lambda^2-1)=0$, and then I found $\lambda=\pm 1$. Where $\lambda=1$ has multiplicity 2, and $\lambda=-1$ has multiplicity 1.
To find the generalized eigenvectors I have to solve $(A-\lambda I)^k x=0$. Since $\lambda=1$ has multiplicity 2, I have $(A-\lambda I)^k x=0$ for k=2, so I determined:
$(A-\lambda I)^2= \begin{pmatrix} -1&\alpha&1\\ 0&0&0\\ 1&\beta&-1\\ \end{pmatrix}^2=\begin{pmatrix} 2&\alpha-\beta&-2\\ 0&0&0\\ -2&\alpha-\beta&2\\ \end{pmatrix}$
So to determine the generalized eigenvectors, I have to solve:
$$\begin{pmatrix} 2&\alpha-\beta&-2\\ 0&0&0\\ -2&\alpha-\beta&2\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix}=0$$
So, I have to solve:
$$2x_1+(\alpha-\beta)x_2-2x_3=0$$
$$-2x_1+(\alpha-\beta)x_2+2x_3=0$$
And that's what I don't really know how to do.
And I have the same problem for $\lambda = -1$, where I have:
$$A-\lambda I=\begin{pmatrix} -\lambda&\alpha&1\\ 0&1-\lambda&0\\ 1&\beta&-\lambda\\ \end{pmatrix}$$
So here, I have to solve:
$$\begin{pmatrix} 1&\alpha&1\\ 0&2&0\\ 1&\beta&1\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix}=0 $$
So I have to solve:
$$x_1+\alpha x_2+x_3=0$$ $$2x_2=0$$ $$x_1+\beta x_2+x_3=0$$
And that's what I don't get how to do, so I'd appreciate some help