Let $V=\mathbb{F}^n$, for a finite field $\mathbb{F}$. Prove the equivalence of the following statements:
- There is a linear subspace $C$ of $V$ with the property that every vector $v$ of $C\setminus\{0\}$ has at least $t$ nonzero entries, and $\dim(C)=n-k$.
- There are $k$ vectors $\phi_1, \ldots, \phi_k$ of $V^*$ such that, for any nonzero vector $v \in V$ with fewer than $t$ nonzero entries, $v \not \in \ker(\phi_i)$ for some $1 \leq i \leq k$.
My (first attempt) solution:
$i \implies ii$: Suppose that there is a linear subspace $C$ of $V$ with the property that every vector $v$ of $C\setminus\{0\}$ has at least $t$ nonzero entries, and $\dim(C)=n-k$. Let $\{\phi_1,\dots,\phi_k\}$ be a basis for $Ann(C)$. (The annihilator has dimension k, since this is the dimension of $C^\perp$). Suppose that $v\not\in C$ is a vector with fewer than t non-zero entries. Since $v\not\in C$ we must have $v\not\in \ker(\phi_i)$ for some $i$. Thus, there are $k$ vectors $\phi_1, \ldots, \phi_k$ of $V^*$ such that, for any nonzero vector $v \in V$ with fewer than $t$ nonzero entries, $v \not \in \ker(\phi_i)$ for some $1 \leq i \leq k$.
$ii \implies i$: Suppose that there are $k$ vectors $\phi_1, \ldots, \phi_k$ of $V^*$ such that, for any nonzero vector $v \in V$ with fewer than $t$ nonzero entries, $v \not \in \ker(\phi_i)$ for some $1 \leq i \leq k$. Let $C=\{v\ |\ v\in\ker(\phi_i)\ \forall\ i\}$. Then for all $v\in C$, $v$ has at least $t$ nonzero entries. Further, $\{\phi_1,\dots,\phi_k\}=Ann^*(C)$, so $dim(C)=dim(V)-dim(Ann^*(C))=n-k$.