Can every hyperplane be described as $\{(x_1,\dots,x_n)\in\Bbb F^n\mid a_1x_1+\dots+a_nx_n=0\}$ for some $a_1,...,a_n$ with $a_i\neq0$ for some $i$? I know from this answer that if $H = \{ (x_1, …, x_n) \in \mathbb{F}^n \mid a_1x_1 + … + a_n x_n = 0 \}$ for some $a_1, …, a_n$ with $a_i \neq 0$ for some $i$, then $H$ is a hyperplane. My question is about the converse: 

Can every hyperplane $H$ be described as $H = \{ (x_1, …, x_n) \in \mathbb{F}^n \mid a_1x_1 + … + a_n x_n = 0 \}$ for some $a_1, …, a_n$ with $a_i \neq 0$ for some $i$?

(My definition of hyperplane is a subspace whose dimension is 1 less than that of the ambient space.)
Here is my attempt:
Let $H$ be a hyperplane. Write $\mathbb{F}^n = H \oplus H^\perp$, and note that $H^\perp$ has dimension 1, so it is spanned by some nonzero vector $a$, i.e. $H^\perp = \text{span}(a)$. Take the orthogonal complement of both sides to get $H = (\text{span}(a))^\perp$. So we have $H = (\text{span}(a))^\perp = \{ a \}^\perp = \{ (x_1, …, x_n) \in \mathbb{F}^n \mid a_1x_1 + … + a_nx_n = 0 \}$. Since $a$ was nonzero, some $a_i$ is nonzero, as required.
 A: Let $v_2,...,v_n$ span $H$. Choose $v_1 \notin H$, then the $v_k$ span the ambient space.
Define a linear functional by defining it on the basis: $f(v_1) = 1$ and $f(v_k) = 0$ for $k >1$.
Then $H = \{ x | f(x) = 0 \}$. Note that $f(v_1) = 1$ so $f \neq 0$.
We have
$f(x) = f(\sum_k x_k e_k) = \sum_k x_k f(e_k) $ and note that $a_k=f(e_k) \neq 0$ for at least one $k$ (otherwise this would contradict $f \neq 0$).
Then
$H= \{ x | \sum_k a_k x_k = 0 \}$ as required.
A: An important observation is,$\mathbb{F}^n$ is equipped with the natural basis vectors $e_{k}$ where $e_{k}$ is the vector which is $1$ at the $k^{th}$ place and $0$ elsewhere.Now you have the collection of linear functionals $e_{k}^{*}$ where $e_{k}^{*}(e_{m})=\delta_{km}$,which form a basis for the dual of $\mathbb{F}^{n}$.Now notice that the hyperplane $H$ is the kernel of a linear functional $f$,which can be expressed uniquely in terms of these dual basis vectors as $f=\sum_{k=1}^{n} a_{k}e_{k}^{*}$.Being the kernel of $f$,$H$ is the set of all $(x_{1},x_{2},..,x_{n})$ such that $f(x_{1},x_{2},..,x_{n})=0$,which,when $f$ is expressed as above and $(x_{1},x_{2},..,x_{n})$ is expressed as a linear combination of the natural basis vectors in $\mathbb{F}^{n}$,gives what you are looking for.
