Jordan form of a Matrix with Ones over a Finite Field Question: Find the Jordan Form of $n\times n$ matrix whose elements are all one, over the field $\Bbb Z_p$.
I have found out that this matrix has a characteristic polynomial $x^{(n-1)}(x-n)$ and minimal polynomial $x(x-n)$, for every $n$ and $p$.
Here I have two cases:
If $n$ is not divisible by $p$, means $n\neq0 \pmod p$, Then the minimal polynomial is separable and thus the form is diagonal: $\operatorname{diag}(0,...,0,n)$.
I am stuck in the second case, where we have $p\mid n$, thus $n=0 \pmod p$. I know that there is a block of order $2$ with $0$, but is there only one as such - and why? I do not see how to prove this using the polynomials only.
Thanks
 A: Your matrix always has rank$~1$ (if $n>0$). This means (rank-nullity) that the eigenspace for $\lambda=0$ has dimension $n-1$; this is the geometric multiplicity of that eigenvalue, and the algebraic multiplicity of $\lambda=0$ is either $n-1$ or $n$. The trace$~n$ of the matrix is the remaining eigenvalue (as the sum of all eignevalues with their algebraic multiplicities equals the trace); if it differs from $0$ in the field (that is $p\nmid n$) then your matrix is diagonalisable, in agreement with what you found from the minimal polynomial.
If on the other hand $p\mid n$, then the matrix is not diagonalisable: the characteristic polynomial is $x^n$, but the eigenspace for $\lambda=0$ has dimension only $n-1$. But that's close enough to determine the Jordan type completely: every Jordan block (for $\lambda=0$) of size $d$ contributes $d-1$ to the rank of the matrix (the dimension of the image of the linear map), which rank is only$~1$. So the Jordan type must be $(2,1,1,\ldots,1)$.
The same by the way applies to any matrix of rank$~1$, except that the trace does not have to be$~n$ now, so certain instances of$~n$, those which originate from the trace, should be replaced by the trace of the matrix. See also this answer.
A: EEEEEEEEEEEEDDDDDDDDDDDDIIIIIIIIIIIIITTTTTTTTTTTTTTT
this is better, any prime $p$ and $n$ by $n$ matrix with $n \equiv 0 \pmod p$
$$
\color{blue}{\left(
\begin{array}{rrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & 1 \\
0 & 1 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 
\end{array}
\right).}
$$
Columns $c_2,c_3,\ldots, c_n$ are eigenvectors with eigenvalue $0,$ and the image of $c_1$ under the linear transformation is $c_n.$
Isn't that something, the inverse (over the rational field, say) is
$$ \color{magenta}{
\left(
\begin{array}{rrrrrrc}
1 & 1 & 1 & 1 & 1 & 1 & 1-n \\
0 & 1 & 0 & 0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 & 0 & 0 & -1 \\
0 & 0 & 0 & 1 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 
\end{array}
\right),}
$$
which you can confirm by just multiplying the two matrices. Then, since $n \equiv 0 \pmod p,$ the upper right corner is equivalent to $1.$ So, the usual calculation $P^{-1}A P$ can be done in full detail. Go figure.
