Let $A=$$ \begin{bmatrix} 1&1&1&1&1&1 \\ 0 & 1 & 0 & 0&0&1 \\ 0 & 0 & 1 & 0&0&1 \\ 0 & 0 & 0 & 1&0&1 \\ 0 & 0 & 0 & 0&1&1 \\ 0 & 0 & 0 & 0&0&1 \\ \end{bmatrix} $
I want to calculate the characteristic polynomial $p_A$, minimal polynomial $ m_A$ and the Jordan normal form J.
As a hint is given that almost no calculations have to be done!
A is upper triangular thus we can see $p_A=(t-1)^6$.
EDIT Correct solution according to comments and the accepted answer
$J=$$
\begin{bmatrix}
1&1&0&0&0&0 \\
0 & 1 & 1 & 0&0&0 \\
0 & 0 & 1 & 0&0&0 \\
0 & 0 & 0 & 1&0&0 \\
0 & 0 & 0 & 0&1&0 \\
0 & 0 & 0 & 0&0&1 \\
\end{bmatrix}
$
This jordan normal form has one block of size 3x3 and three blocks of size 1x1. Since there is only one eigenvalue and the biggest block has size 3 we get $m_A=(t-1)^3$.
Denote the $k\times k$ Jordan block for an eigenvalue $\lambda$ by $J_k(\lambda)$. Clearly, $A-I$ is nilpotent and it has rank $2$. Therefore the Jordan normal form of $A-I$ is either $J_2(0)\oplus J_2(0)\oplus0_{2\times2}$ or $J_3(0)\oplus0_{3\times3}$. Now you may rule out one of the these two possibilities by considering the rank of $(A-I)^2$.