Finding the Jordan basis of a linear map

Question

A linear map $A$ is given in the canonical basis with the matrix $$ \begin{bmatrix} -2&0&-2&-2\\ 1&0&1&1\\ -1&1&-1&-1\\ 3&-1&3&3\\ \end{bmatrix} $$

Determine the Jordan basis $(j)$ and write the map $A$ in it.

What's the general procedure in solving these problems? I have calculated $A^2$ and $A^3$. $A^3$ is $0$.

If $A^3$ wasn't $0$ I believe I would know how to solve it (the basis then should be $e_1$, $Ae_1$, $A^2e_1$, $A^3e_1$, right?)

But since I can't do that, I don't know what I can do. I assume the representation in the basis I'm looking for should look like$$ \begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{bmatrix} $$

Answer 1

here are jordan chain for the matrix $A$ in your question. let $ej$ stand for the basis vector with all zeros execpt the $j$ the component. chain of length 3 is $e3 \to Ae3 = -2e1 + e2 - e3 + e4 \to A^2 e3 = e3-e4$ and a chain of length 1 is $e1-e3.$ in this basis $\{e3, Ae3, A^2e3, e1-e3\}$ the jordan form looks like the one at bottom of your question.