I want to find a Jordan basis for the linear transformation $T(v) = Av$, where $A$ is as below:
$$A=\begin{bmatrix} 0&0&1&0\\1&1&0&-1\\ 0&0&1&0\\0&0&1&1 \end{bmatrix}$$
I started by calculating the Jordan Normal Form.
$c_T(x) = x(x-1)^3$ so our evals are $\lambda_1 = 0, \lambda_2 = 1$. Calculating the geometric multiplicities to be $g(0) = 1, g(1) = 1$. Hence giving JCF $= J_1(0) \oplus J_3(1)$.
Now to calculate the Jordan basis I need the find the nilpotent maps $S_i = T_{V_i} - \lambda_i I$ where $V_i = ker(T-\lambda_i)^{n_i}$. Then finding subspaces up to $S^r(V)$ where $S^{r+1}(V) = 0$. Then finding a jordan basis for each restriction, however I am not sure how to continue.
Solve $A X = 0$, you find that $e_1 = (1, -1, 0, 0)$ is a solution. Solve now $(A - I) X = 0$, you find that $e_2 = (0, 1, 0, 0)$ is a solution. Solve $(A - I) X = e_2$, a solution is $e_3 = (0, 0, 0, -1)$. Finally solve $(A - I)X = e_3$, a solution is $e_4 = (-1, 0, -1, -1)$. A Jordan basis is $(e_1, e_2, e_3, e_4)$.