In a vector space over a finite field, can the orbit of a point under matrix multiplication have dependent subsets? Let $\mathbb{F}_q^n$ be the vector space of dimension n over the finite field of order q, $\vec{v}$ a vector in the space, and $M$ an invertible $n \times n$ matrix over $\mathbb{F}_q$.
We know that $M^p$ = $M^0$ = $Id$. Consider the orbit of $\vec{v}$ under repeated left multiplication by $M$. (Equivalently, under the action of the cyclic group of matrices generated by $M$ under left muliplication.) Suppose that the elements of the orbit are all unique (mod scaling), and they span $\mathbb{F}_q^n$. Is it possible that some n elements of the orbit are linearly dependent?
What if q is prime?
edit: M is invertible.
 A: This is possible.
Let us view $L=\Bbb{F}_{64}$ as a 6-dimensional vector space $V$ over $K=\Bbb{F}_2$. The field $L$ contains a primitive ninth root of unity $\alpha$, because $9\mid(64-1)$. Let $M:V\to V$ be multiplication by $\alpha$, so it generates a cyclic group of order nine. Let $\vec{v}=1$. The orbit consists of the powers $\alpha^i,i=0,1,\ldots,8$. These are all distinct and because $1$ is the only non-zero constant of $K$ they are also distinct mod scaling (if I understood that term correctly). They also span all of $V$ because the span of powers of $\alpha$ is a subfield, and no proper subfield of $L$ contains a primitive ninth root of unity.
But the minimal polynomial of $\alpha$ is $x^6+x^3+1$, so the triplets
$$\{1,\alpha^3,\alpha^6\},\quad\{\alpha,\alpha^4,\alpha^7\},\quad\{\alpha^2,\alpha^5,\alpha^8\}$$
are all linearly dependent over $K$. Here $3<\dim_K V=6$.
A: In general, every projective geometry over a finite field (in which the "points" are the one-dimensional subspaces) admits what is called a Singer cycle, which is a cyclic permutation that acts sharply transitively on the points.  In this case any selection of $n$ linearly dependent points lie in the same orbit.
