Let $\alpha, \beta$ be scalars in a field $F$ of characteristic $\neq 2$, and let $A \in End(V)$ be the linear transformation of $V=F^3$ which is represented by
\begin{equation} [A] = \begin{pmatrix} 1 & 0 & 0\\ \alpha & 0 & 1 \\ \beta & 1 & 0 \end{pmatrix} \end{equation}
with respect to the standard basis. Determine the generalized eigenspaces of A, for every $\alpha$ and $\beta$ from F.
I have calculated the characteristic polynomium and got the eigenvalues $\lambda =\pm 1$ and I know that the generalized eigenspace is given as the set: $$M_{\lambda} = \lbrace x \in V : \text{there is some } k \in N: (A- \lambda I)^kx=0 \rbrace$$
However I am stuck now. How do I proceed from here? I want to solve the equation $(A- \lambda I)^kx=0$ but I am not quite sure how to do it. Is there an explicit method or trick to this?
Yes, the eigenvalue are $\pm1$, but it's more than that: the eigenvalue $1$ has multiplicity $2$, whereas the eigenvalue $-1$ has multiplicity $1$. So, the generalized eigenspace which corresponds to the eigenvalue $-1$ is $1$-dimensional. It turns out that it is equal to $F(0,-1,1)$.
On the other hand,$$(A-\operatorname{Id})^2=\begin{bmatrix}0 & 0 & 0 \\ \beta-\alpha & 2 & -2 \\ \alpha-\beta & -2 & 2\end{bmatrix}$$and it happens that$$\ker(A-\operatorname{Id})^2=\operatorname{span}\bigl(\{(2,0,-\alpha+\beta),(2,\alpha-\beta,0)\}\bigr).$$