Define $A = \begin{pmatrix} -7 & -32 & -32 & -35 \\ 1 & 5 & 4 & 4 \\ 1 & 4 & 5 & 5 \\ 0 & 0 & 0 & 1 \end{pmatrix} \in \mathbb{C^{4x4}}$
I computed the Jordan normal form of this matrix:
$J=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}$.
But how to find a Jordan basis for this matrix?
Since $ker(A-I)^3=0$ I can't choose a $v \in ker(A-I)^3$ \ $ker(A-I)^2$
Also, I tried to use $v \mapsto Av$
$Ab_1=b_1$,
$Ab_2=b_2$,
$Ab_3=b_2+b_3$,
$Ab_4=b_3+b_4$
But I don't know how to continue.
If you write down $(A-I)^2$ you see that any vector with the final entry nonzero is outside the kernel, so you can pick $$ p_4 = \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right) $$ followed by $p_3 = (A-I) p_4 \;, \;$ $p_2 = (A-I) p_3 \;, \;$ which is automatically an eigenvector because $(A-I)p_2 = (A-I)^3 p_4 = 0.$ Then let the first column $p_1$ be a different eigenvector from $p_2.$ Notice that, with two eigenvectors available, I deliberately chose a linear combination so the entires of the first column would be small. I hoped it would reduce the determinant of $P,$ and this worked out nicely, the determinant came out to $2.$ You can see this is the factor of $\frac{1}{2}$ required to write all the matrices with integers.
$$ \frac{1}{2} \left( \begin{array}{cccc} 1&5&3&0 \\ -9&-35&-35&0 \\ 2&8&8&0 \\ 0&0&0&2 \end{array} \right) \left( \begin{array}{cccc} 0&-8&-35&0 \\ 1&1&4&0 \\ -1&1&5&0 \\ 0&0&0&1 \end{array} \right) = \left( \begin{array}{cccc} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array} \right) $$
$$ \frac{1}{2} \left( \begin{array}{cccc} 1&5&3&0 \\ -9&-35&-35&0 \\ 2&8&8&0 \\ 0&0&0&2 \end{array} \right) \left( \begin{array}{cccc} -7&-32&-32&-35 \\ 1&5&4&4 \\ 1&4&5&5 \\ 0&0&0&1 \end{array} \right) \left( \begin{array}{cccc} 0&-8&-35&0 \\ 1&1&4&0 \\ -1&1&5&0 \\ 0&0&0&1 \end{array} \right) = \left( \begin{array}{cccc} 1&0&0&0 \\ 0&1&1&0 \\ 0&0&1&1 \\ 0&0&0&1 \end{array} \right) $$ and $$ \frac{1}{2} \left( \begin{array}{cccc} 0&-8&-35&0 \\ 1&1&4&0 \\ -1&1&5&0 \\ 0&0&0&1 \end{array} \right) \left( \begin{array}{cccc} 1&0&0&0 \\ 0&1&1&0 \\ 0&0&1&1 \\ 0&0&0&1 \end{array} \right) \left( \begin{array}{cccc} 1&5&3&0 \\ -9&-35&-35&0 \\ 2&8&8&0 \\ 0&0&0&2 \end{array} \right) = \left( \begin{array}{cccc} -7&-32&-32&-35 \\ 1&5&4&4 \\ 1&4&5&5 \\ 0&0&0&1 \end{array} \right) $$