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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

Marc van Leeuwen
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Mike
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    do 2 by 2 with prime 2, then 3 by 3 with prime 3. by hand. – Will Jagy Sep 13 '14 at 19:34
  • @WillJagy I did. There is even simpler solution in the case of 2x2 or 3x3 matrix, since in such matrices the minimal and characteristic determine the Jordan Form. There is no proof that the case of n=2,3 is general, I want a proof for every n. – Mike Sep 13 '14 at 19:37
  • I want a proof for every n, and do not have א0 time to check all n and p there are. – Mike Sep 13 '14 at 20:01
  • It's a shame we are only mortals @ Will Jagy :) – Oria Gruber Sep 13 '14 at 20:04

2 Answers2

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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.

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Marc van Leeuwen
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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.

Will Jagy
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