Consider the matrix $A \in \mathbb M_{5 \times 5}(\mathbb C)$
$$ A= \begin{pmatrix} 1 & 0 & 0& 1& 0\\ 0 & 1 & 0 &0 & 0\\ 0& 0 & 1& -1 & -1\\ 0 & 0 & 0 & 1& 0 \\ 0 & 0 & 0& 0& 1\\ \end{pmatrix}$$ Find a Jordan base for $A$ and determine the corresponding jordan form.
The characteristic polynomial is given by $\det (x\mathrm{Id}-A) = (x-1)^5$ $\dim \ker(T- 1\mathrm{Id}) = A -1\mathrm{Id} = [T -1\mathrm{Id}]:$
$ A= \begin{pmatrix} 0 & 0 & 0& 1& 0\\ 0 & 0 & 0 &0 & 0\\ 0& 0 & 0& -1 & -1\\ 0 & 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0& 0\\ \end{pmatrix}$
The rank of the matrix is 2 so $\dim (\mathrm{Im}(T - \mathrm{Id})) = 2$ And by the ker and im theorem $\dim (\ker(T - \mathrm{Id})) = 3$ so there are 5-2 = 3 jordan blocks.
$\ker[(T - Id)^2] = \begin{pmatrix} 0 & 0 & 0& 1& 0\\ 0 & 0 & 0 &0 & 0\\ 0& 0 & 0& -1 & -1\\ 0 & 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0& 0\\ \end{pmatrix} * \begin{pmatrix} 0 & 0 & 0& 1& 0\\ 0 & 0 & 0 &0 & 0\\ 0& 0 & 0& -1 & -1\\ 0 & 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0& 0\\ \end{pmatrix} = 0$
The rank is $0$ so $\dim\ker[(T - \mathrm{Id})^2] = 5$ so $\dim \ker[(T - \mathrm{Id})^2] - \dim \ker[(T - \mathrm{Id})] = 5 - 3 = 2$ so we have 2 block of $2 \times 2$ therefore
$$J= \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\\ \end{pmatrix}$$
So far I think it's right.Right?
Now, the base I'm not knowing how to find, I did it like this (I cannot use $J=P^{-1}AP$):
$A - I = T - \mathrm{Id} = \begin{pmatrix} 0 & 0 & 0& 1& 0\\ 0 & 0 & 0 &0 & 0\\ 0& 0 & 0& -1 & -1\\ 0 & 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0& 0\\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z\\ w \\ u\\ \end{pmatrix} = 0 \to w=0, -w -u = 0 \to w=u=0 \to e_5 = (1,1,1,0,0)$
But this vector should be part of the $P$ matrix and I saw in wolfram that it doesn't, that is, it's wrong and I don't know how to proceed.
Thanks.
Update of what I understood from the comment and from what I read in the book. I already calculated $e_5$ which is the last column of the matrix. To calculate the other columns I need to choose a vector that will be $e_1$ and from it determine the others as follows
$e_2 = (A - I)(e_1) \to $ $A - I = T - \mathrm{Id} = \begin{pmatrix} 0 & 0 & 0& 1& 0\\ 0 & 0 & 0 &0 & 0\\ 0& 0 & 0& -1 & -1\\ 0 & 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0& 0\\ \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 0\\ 1 \\ 1\\ \end{pmatrix} = \begin{pmatrix} x \\ y \\ z\\ w \\ u\\ \end{pmatrix} \to x=1;z=-2 \to e_2 = (1,0,-2,0,0)$
As $y, w, u$ are independent I can choose any value for them then $e_2 = (1,0,-2,1,0)$
$e_3 = (A - I)(e_2) \to $ $A - I = T - \mathrm{Id} = \begin{pmatrix} 0 & 0 & 0& 1& 0\\ 0 & 0 & 0 &0 & 0\\ 0& 0 & 0& -1 & -1\\ 0 & 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0& 0\\ \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -2\\ 1 \\ 0\\ \end{pmatrix} = \begin{pmatrix} x \\ y \\ z\\ w \\ u\\ \end{pmatrix} \to x=1; z=-1 \to e_3 = (1,0,-1,0,1)$
$e_4 = (A - I)(e_3) \to $ $A - I = T - \mathrm{Id} = \begin{pmatrix} 0 & 0 & 0& 1& 0\\ 0 & 0 & 0 &0 & 0\\ 0& 0 & 0& -1 & -1\\ 0 & 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0& 0\\ \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -1\\ 0 \\ 1\\ \end{pmatrix} = \begin{pmatrix} x \\ y \\ z\\ w \\ u\\ \end{pmatrix} \to z=-1 \to e_4 = (0,0,-1,0,0)$
$P = \begin{pmatrix} 0 & 1 & 1 & 0 & 1\\ 0 & 0 & 0 & 0 & 1\\ 0 & -2 & -1 & -1 & 1\\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0\\ \end{pmatrix} $
Your idea is nice. But there is a flaw. Let us see what it it.
$A$ has 3 eigenvectors $e_1, e_2, e_3$ (they are a part of standard basis) with eigenvalue 1. Now as you have shown, $\text{rank} (A - I) = 2$.
Let consider a vector $v \in \mathbb{C}^5$. Then $(A - I)v = v_4 e_1 + (-v_4 - v_5) e_3$.
Then for $v_1 = \left( \begin{matrix} 0\\0\\0\\1\\-1 \end{matrix}\right)$, we get $(A - I)v_1 = e_1$ and for $v_2 = \left( \begin{matrix} 0\\0\\0\\0\\-1 \end{matrix}\right)$, we get $(A-I)v_2 = e_3$.
Then if we take change of basis matrix $P$ such that $e_1 \mapsto e_2$, $e_2 \mapsto v_1$, $e_3 \mapsto e_1$, $e_4 \mapsto v_2$, and $e_5 \mapsto e_3$, we will get the Jordan canonical form for $A$ and which is $$\left( \begin{matrix} 1 &0 &0 &0 &0\\ 0 &1 &0 &0 &0\\ 0 &1 &1 &0 &0\\ 0 &0 &0 &1 &0\\ 0 &0 &0 &1 &1 \end{matrix}\right)$$