I was asked to prove that a subset
$H = \{(x_1,...,x_n) \in\Bbb{F}^n|a_1x_1 + a_2x_2...+a_nx_n = 0\}$
of $F^n$ is a hyperplane if and only if $a_i \neq 0$ for some $\mathit{i}$
When trying to prove this I understand that at least one $a_i$ has to be nonzero since if every $a_i$ were zero then any vector in the space would satisfy, thus making H not a hyperplane, but I don't really understand the intuition besides that going either way. Please help.
Geometrically you can think of $H$ as consisting of the vectors $\vec{x}=(x_i)$ that are perpendicular to some given non-zero vector $\vec a = (a_i)$ i.e. such that $\vec x . \vec a = 0$.
Algebraically, to show that $H$ is a hyperplane, you have to show two things:
i) $H$ is a subspace of $\mathbb{F}^n$ - this is clear because if $\vec x . \vec a =0$ and $\vec y . \vec a = 0$ then $(\vec x + \vec y).\vec a = 0$.
ii) The dimension of $H$ is $n-1$. One way to show this is to pick one of the $a_i$ that is non-zero (we know there is at least one) - let's say this is $a_n$. Then if we set $x_1, x_2, \dots, x_{n-1}$ to any values we like, there is exactly one value of $x_n$ that will satisfy $\vec x . \vec a = 0$ - that value is
$x_n = \displaystyle -\frac { \sum_{j=1}^{n-1} a_jx_j }{a_n}$
So we have $n-1$ degrees of freedom when choosing an element of $H$, and so $H$ has dimension $n-1$.
An alternative but equivalent approach is to show that the $n-1$ vectors
$b_1 = (1,0,0,\dots,0,-\frac{a_1}{a_n}) \\ b_2 = (0,1,0,\dots,0,-\frac{a_2}{a_n}) \\ \dots \\ b_{n-1} = (0,0,0,\dots,1,-\frac{a_{n-1}}{a_n})$
form a basis for $H$.