Counting zero-sum subspaces for the multiplicative inverse function

Let $q=2^n$ with $n\ge 6$, and let ${\mathbb{F}}_q$ be the finite field with $q$ elements. For a subspace $E$ of ${\mathbb{F}}_q$, define $$S(E)=\sum _{x\in E\setminus \{0\}}{x}^{-1}\mathrm{.}$$ Let $N_{n,k}$ denote the number of $k$-dimensional ${\mathbb{F}}_2$-subspaces $E$ for which $S(E)=0$. We prove that, if $k\ge 3$ and $n\ge 2k+1$, then $$\mid \frac{2^n{N}_{n,k}}{{\left[\genfrac{}{}{0px}{}{n}{k}\right]}_2}-1\mid <2^{r^2+r}{2}^{n(1-r\mathrm{/}2)},$$ where $r=\lfloor \frac{k-1}{2}\rfloor$, and ${\left[\genfrac{}{}{0px}{}{n}{k}\right]}_2$ is the Gaussian binomial coefficient. In particular, $$N_{n,k}\sim 2^{-n}{\left[\genfrac{}{}{0px}{}{n}{k}\right]}_2(n\to \mathrm{\infty })$$ uniformly for $n\mathrm{/}3<k\le (n-1)\mathrm{/}2$. Combined with the known low-dimensional cases and the symmetry between dimensions $k$ and $n-k$, and the elementary middle-dimensional subfield construction, this proves a conjecture of Carlet: for every $3\le k\le n-3$, there exists a $k$-dimensional ${\mathbb{F}}_2$-subspace $E$ such that $S(E)=0$.