Jordan form of the "multiplicative table" matrix I have to find the Jordan form of the $(10\times10)-$matrix $A$ with the $n$th row formed by $n(1,2,3,4,5,6,7,8,9,10), \ \ 1 \leq n \leq 10$ 
I have calculated the determinant of $(A-xI)$ using Gaussian elimination, and I have that the result is $x^9(x-1)$, so there are two eigenvalues $x=0$ with algebraic multiplicity $9$ and $x=1$ with algebraic multiplicity $1$ (and so also the geometric one is $1$). Now I have to understand the geometric multiplicity of $0$, and I think that it could be $9$ (again using Gaussian elimination), is this right? 

I want to thank @Hagen von Eitzen to make me see where was the mistake.
Given $A-xI$ the matrix defined upon, with Gaussian elimination of the type: $n$-th column $-$ $n$ times the first column, we end up with a matrix of the form:
$\left(\begin{matrix}1-x & 2x & 3x &... \\ 2 & -x & 0 &... \\ 3 & 0 & -x &... \\ . & ... \\ . \\ . \end{matrix}\right)$
It's easy to prove with induction that the determinant of this matrix is $ \\x^9(x-(1+4+9+16+...+100))$
There are two eigenvalues:
$x=0$ which has algebraic multiplicity $9$ and geometric multiplicity $9$, this last because, given that $e_i:= (0,0,...,1,0,...,0)$ where $1$ is in the $i$th place, the vectors $e_2,...,e_{10}$ are a base for the eigenspace. 
$x=\sum_{k=1}^{10} k^2 $ which has algebraic and geometric multiplicity $1$
So the matrix in Jordan form will be:
$\left(\begin{matrix}0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \sum_{k=1}^{10} k^2=385 \\ \end{matrix}\right)$
