I am trying to solve the following problem:

Given is a $3\times 3$ matrix $M$ over $\mathbb{F}_{7}$, such that for every vectors $v,w\in \mathbb{F}_{7}^3\setminus \{0\}$ there exists an integer $n$ with $M^{n}v=w$. Find this $M$.

Well, i think i have to look for matrices with full ranks, but i have no idea how and where to start... I've been thinking of idempotent matrices, but the problem is what to do with the vectors $v,w$. I can't just choose them to be (1,1,1).

Does anyone have an idea how can this probem be solved? I will be glad to read your comments and remarks. Thank you in advance!

  • $\begingroup$ I guess yes, because in this case $v=w$, which is very trivial. $M$ should not be the identity matrix. $\endgroup$ – Lullaby Nov 10 '13 at 15:44
  • $\begingroup$ I edited your question to show that the nonzero vectors are in $\mathbb{F}_7^3$. $\endgroup$ – Sammy Black Nov 10 '13 at 16:00

Let $g$ be a primitive element of the field $E=\Bbb{F}_{7^3}$, and let $M$ represent multiplication by $g$ on $E$ viewed as a 3-dimensional vector space over $\Bbb{F}_7$. Show that this $M$ works.

  • $\begingroup$ There $7^3-1=342$ non-zero vectors in $E$. The element $g$ has $342$ distinct powers, all of them invertible, because $g^{342}=I$ and no lower power will do that. $\endgroup$ – Jyrki Lahtonen Nov 11 '13 at 21:48

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