Given this matrix in size $n \times n$
$\begin{pmatrix} 1 &1 &1&\cdots & 1 & 1 & \\ 0 &1 & & & \\ . & &.& & & \\ . & & &.& & \\ . & & & &. & \\ 0& & & & &1 \end{pmatrix}$
EDIT: the matrix has all 1's above the diagonal.
I need to find Jordan form and base for this matrix.
I'd love your help understanding if I'm on the right track.
the characteristic polynomial is $f_{A}=(x-1)^{n}$ and the minimal one is $M_{A}=(x-1)$ (EDIT: it's clearly not right). So I can know that the dimension of the biggest Jordan block is 1, thus there are n Jordan blocks of size $1 \times 1$, the $\ker$ of eigenvalue 1 is all the dimension, So I simply assuemd that Jordan base is: $\left(\begin{smallmatrix} 1\\ 0\\ .\\ .\\ 0\\ \end{smallmatrix}\right)$, $\left(\begin{smallmatrix} 0\\ 1\\ .\\ .\\ 0\\ \end{smallmatrix}\right),\dots, \left(\begin{smallmatrix} 0\\ 0\\ .\\ .\\ 1\\ \end{smallmatrix}\right)$ .(If it's correct I will be happy to hear an explanation for that).
EDIT: For some stupid reason I first call this matrix the "Identity", Thanks for the comment below I changed it.
Trying to correct myself: So now I realize that I should sleep more and even much more significant fact that this matrix is nilpotent of rank n, so the minimal polynomial is $(x-1)^{n}$, and so finding the base will be a little more complicated. actually I'll need to find all the $\ker$s of $(x-1)^{i}$ for i=1,...n.
And After a while I think that Jordan base Will be $\left(\begin{smallmatrix} 1\\ 0\\ .\\ .\\ 0\\ \end{smallmatrix}\right)$, $(A-I)$ $\left(\begin{smallmatrix} 0\\ 1\\ .\\ .\\ 0\\ \end{smallmatrix}\right),\dots,(A-I)^{n-1}$ $\bigl(\begin{smallmatrix} 0\\ 0\\ .\\ .\\ 1\\ \end{smallmatrix}\bigr)$
Thank you, Have a great week.