matrix rank over finite field

Question

Let $M$ be a square matrix over finite field $\Bbb F_p$. Let $N$ be the matrix over $\Bbb F_p$ obtained replacing every non-zero entry of $M$ by $1$.

Is $Rank(N)\leq Rank(M)^{f_p}$? for some constant $f_p$?

Answer 1

In response to the original question:

Consider the following matrix in $\Bbb F_p$ with $p>2$: $$ M = \pmatrix{ 1&-1&0\\ 0&1&-1\\ -1&0&1}, N = \pmatrix{ 1&1&0\\ 0&1&1\\ 1&0&1} $$ Note that $\det(N) = 2$, but $M$ has a non-zero kernel. So, $N$ has rank $3$, but the $M$ has rank $2$.

So, for the new question: we know that if an $f_p$ exists, we have $f_p>1$ when $p>2$.

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For an arbitrary $n$: let $I_n$ denote the identity matrix, and let $J_n$ denote the permutation matrix $$ J_n = \pmatrix{ &1\\ &&1\\ &&&\ddots\\ &&&&1\\ 1 } $$ We note that $I_n - J_n$ is always singular, and that $I_n + J_n$ has eigenvalues $1 + \omega$, where $\omega$ is any solution to $\omega^n = 1$.

Note that $I_j + J_n$ is singular when $n$ is even.

It may be useful to consider the case $n = kp$, where $k$ is odd (in these cases, the eigenvalues are easy to determine).