some time ago I found a 'trick' for getting a basis-transformation-matrix for jordan. I'd like to understand it, but at a certain point I stuck. Maybe you can help me?
Given is a matrix A:
$\begin{pmatrix} 1 & -2 & 0 &-1&2 \\ 1&-3&-1&0&3 \\ 0&2&1&-1&-3 \\ 1 &0&0&-1&-2 \\0&-1&0&0&2 \end{pmatrix}$
As we can see, we have $P_A(t) = t^5$ and our only eigenvalue is $0$.
Now we do forming with our raws, simultaneously with the identity matrix:
$\begin{pmatrix} 1&0&0&0&0|1&-2&0&-1&2 \\ 0&1&0&0&0|1&-3&-1&0&3 \\ 0&0&1&0&0|0&2&1&-1&-3 \\ 0&0&0&1&0|1&0&0&-1&-2 \\ 0&0&0&0&1|0&-1&0&0&2 \end{pmatrix}$
On the left is the identity matrix, on the right we have $(A - 0\cdot E_5)$
After some steps and changing the columns we get
$\begin{pmatrix} 0&0&0&1&0|1&0&0&-1&-2 \\ 0&0&0&0&-1|0&1&0&0&-2 \\0&0&1&0&2|0&0&1&-1&1 \\ 0&1&1&-1&-1|0&0&0&0&0 \\ 1&0&0&-1&-2|0&0&0&0&0 \end{pmatrix}$
On the left we have the matrix T, on the right A'. It always ends with having a matrix on the right side with some zero lines.
Take a look at the two last columns/vectors of A', and change those vectors (by changing prefix and adding 1/0 respecively 0/1) to:
$v_1 = (1, 0, 1, 1, 0)$ and $v_2 = (2, 2, -1, 0, 1)$. These are two of the five vectors that we need
Now do the following:
$T\cdot v_1 = v_1 ^{(2)} = (1, 0, 1, 0, 0)$
$T\cdot v_1^{(2)} = v_1^{(3)} = (0, 0, 1, 1, 1)$
...
$T \cdot v_2 = v_2^{(2)} = (0, -1, 1, 0, 0)$
$T \cdot v_2^{(2)} = v_2^{(3)} = (0, 0, 1, 0, 0)$
...
Our basis-transformation matrix S contains the following vectors in this order: $S := (v_1, v_1^{(2)}, v_2, v_2^{(2)}, v_2^{(3)})$
So WHY are those the vectors I need? Why isn't for example $v_1^{(3)}$ a vector that we need? WHY isn't $S = (v_1, v_1^{(2)}, v_1^{(3)}, v_1^{(4)}, v_1^{(5)})$, for example?