Sum of all vectors in a vector space over a finite field

Question

Let $F_n$ be a finite field of $n$ elements, where $n$ is a power of a prime.

Now, let $F_n^k$ be the set of all row vectors of length $k$ over $F_n$.

Is it true that the sum of all of the none zero vectors is equal to the zero vector?

I have experimented with small cases and this seems to be true, but I can’t prove the general result?

Answer 1

It is true. Suppose $n$ is not a power of 2. There are $n$ possible first entries for a vector. Note, the sum of all $n$ elements in the finite field is $0$ when $n$ isn't a power of $2$. Further, for each choice of a first entry, except $0$, there are the same number of non-zero vectors with this first entry. Thus, the sum of all the non-zero vectors will have first entry equal to $0$. The same logic shows the other entries will also sum to $0$.

If $n$ is a power of $2$ and $k>1$, there will be $n^{k-1}$ vectors with a given non-zero first entry. This is because there are $n$ choices for every other entry. The result follows once one multiplies the first entry by $n^{k-1}$.

If $k=1$ and $n$ is a power of $2$, then the result doesn't hold.

Answer 2

The sum of all vectors $s = \sum_{x \in F_n^k} x$ is $\operatorname{GL}_k(F_n)$-invariant, i.e. for every $A \in \operatorname{GL}_k(F_n)$ we have that $As = s$. For any two non-zero vectors $x, y \in F_n^k$ there exists some $A \in \operatorname{GL}_k(F_n)$ with $Ax = y$, so if $F_n^k$ contains at least three elements then it follows that $s = 0$. If $F_n^k$ has at most two elements than either $k = 0$, in which case $s = 0$, or $F_n = F_2$ and $k = 1$, in which case $s = 1$ (where we identitfy $F_2^1$ with $F_2$).