TITLE: Choosing a linear map $(\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ which is nonzero on half of a sequence of vectors
QUESTION [10 upvotes]: Let $v_1,\ldots,v_m \in (\mathbb{Z}/2\mathbb{Z})^n$ be nonzero vectors.  Is it always possible to choose a linear map $f : (\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ such that $f$ is nonzero on at least half of the $v_i$, i.e. such that
$$|\{\text{$i$ $|$ $1 \leq i \leq m$ and $f(v_i) \neq 0$}\}| \geq \frac{1}{2}m?$$
My guess is that the answer is yes; at the very least, it is true for $m=1$ and when the $v_i$ enumerate all of the nonzero vectors.

REPLY [3 votes]: My coauthors and I needed this result recently (to prove something about pseudo-Anosov dilatations), and I was googling to see if it was known.  I'm amazed that it was asked here so recently!
I'm answering now to record a short alternative proof that we found.  I'll prove the same more general result that Landscape proved, namely:
Claim : Let $\vec{v}_1,\ldots,\vec{v}_m \in \mathbb{F}_q^n$ be nonzero vectors (not necessarily distinct).  Then there
exists a linear map $f : \mathbb{F}_q^n \rightarrow \mathbb{F}_q$ such that $f(\vec{v}_i) = 1$ for at least $\frac{q-1}{q}$
of the $\vec{v}_i$, i.e. such that $\{\text{$i$ $|$ $1 \leq i \leq m$, $f(\vec{v}_i) \neq 0$}\}$ has cardinality at least $\frac{q-1}{q}m$.
Proof : Let $\Omega$ be the probability space consisting of all linear maps from $\mathbb{F}_q^n \rightarrow \mathbb{F}_q$, each given
equal probability.  Let $\mathcal{X} : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to the
cardinality of the set $\{\text{$i$ $|$ $1 \leq i \leq m$, $f(\vec{v}_i) \neq 0$}\}$.  We will prove that the expected value $E(\mathcal{X})$ of $\mathcal{X}$
is $\frac{q-1}{q} m$, which clearly implies that there exists some element $f \in \Omega$ such that $\mathcal{X}(f) \geq \frac{q-1}{q} m$.
To prove the desired claim, for $1 \leq i \leq m$ let $\mathcal{X}_i : \Omega \rightarrow \mathbb{R}$ be the random variable that
takes $f \in \Omega$ to $1$ if $f(\vec{v}_i) \neq 0$ and to $0$ if $f(\vec{v}_i) = 0$.  Viewing $\vec{v}_i$ as an element of the double dual
$(\mathbb{F}_q^n)^{\ast \ast}$, the kernel of $\vec{v}_i$ consists of exactly $\frac{1}{q}$ of the elements of $(\mathbb{F}_q^n)^{\ast}$.  This
implies $E(\mathcal{X}_i) = \frac{q-1}{q}$.  Using linearity of expectation (which recall does not require that the random
variables be independent), we get that
$$E(\mathcal{X}) = E(\sum_{i=1}^m \mathcal{X}_i) = \sum_{i=1}^m E(\mathcal{X}_i) = \sum_{i=1}^m \frac{q-1}{q} = \frac{q-1}{q} m,$$
as desired.