Compute Jordan Form of the following matrix

How to find the Jordan form of the following matrix ( It's not necessary to find the jordanizing basis )?

$A = \left(\begin{array}{ccccc} n & n-1 & \cdots & 2 & 1 \\ 0 & n & & 3 & 2 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & & n & n-1 \\ 0 & 0 & \cdots & 0 & n\end{array}\right) $

The characteristic polynomial of this matrix is $ \triangle_A(x) = ( x-n)^n $.

However, I don't see how to make any progress. I see that $ A = n\cdot I_n + B $ where $ B = \left(\begin{array}{ccccc} 0 & n-1 & \cdots & 2 & 1 \\ 0 & 0 & & 3 & 2 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & & 0 & n-1 \\ 0 & 0 & \cdots & 0 & 0 \end{array}\right) $ is a nilpotent matrix of index $ n $, but how can I proceed?

Solution 1:

Show by induction that if $(a_1, \cdots, a_n)^t$ is an eigenvector of $B$, then $a_n=a_{n-1}=\cdots=a_2=0$. Hence the geometric multiplicity of $0$ is $1$, which implies that $B$ has a single Jordan block.

However, in general the algebraic and geometric multiplicities are not sufficient to determine the Jordan canonical form: A $4\times 4$ matrix with a single eigevalue $\lambda$ of geometric multiplicity $2$ will have two Jordan blocks, but their sizes could be $1+3$ or $2+2$.