Hyperplanes of a Vector Space

Definition: Let

$E$

be a vector space. A Hyperplane of

$E$

is a maximal and proper subspace

$H$

$E$

Equivalently, a hyperplane of $$E$$ is a subspace $$H$$ of $$E$$ such that $\mathrm{codim}(H) = 1$ .

Proposition 1: Let

$E$

be a vector space. Then

$H$

is a hyperplane of

$E$

if and only if there exists a nonzero linear form

$f$

$E$

such that

f^{-1}(0) = H

Proof: $\Rightarrow$ Let $$H$$ be a hyperplane of $$E$$ and let $a \in E \setminus H$ . For each $x \in E$ , write $x = h + \lambda_x a$ where $h \in H$ and $\lambda_x \in \mathbf{F}$ . Let $f(x) := \lambda_x$ . Then $$f$$ is a linear functional on $$E$$ , and is such that $f^{-1}(0) = H$ .

$\Leftarrow$ Suppose that $$f$$ is a nonzero linear form on $$E$$ . Then clearly $f^{-1}(0)$ is a proper subspace of $$E$$ . All that remains to be shown is that it is maximal.

Let $a \in E$ be such that $f(a) \neq 0$ , and without loss of generality, assume that $$f(a) = 1$$ (for if $f(a) \neq 1$ , take $b := \frac{a}{f(a)}$ instead). Then observe that every $x \in E$ can be written in the form:

(1)

\begin{align} \quad x = \left ( x - \frac{f(x)}{f(a)} a \right ) + \frac{f(x)}{f(a)}a \end{align}

But $x - \frac{f(x)}{f(a)}a \in f^{-1}(0)$ , and $\frac{f(x)}{f(a)}a \in \mathbf{F} a$ . So $f^{-1}(0) + \mathbf{F} a = E$ . Clearly, $f^{-1}(0) \cap \mathbf{F}a = \{ 0 \}$ , since $f(\lambda a) = 0$ if and only if $\lambda = 0$ in which case $\lambda a = 0$ . Thus $f^{-1}(0) \cap \mathbf{F} a = E$ . But since $\mathrm{dim}(\mathbf{F}a) = 1$ , we see that $f^{-1}(0)$ is maximal.

Hyperplanes of a Topological Vector Space

Proposition 2: Let

$E$

be a topological vector space. If

$H$

is a hyperplane of

$E$

, then either

$H$

is closed or

$H$

is dense in

$E$

Proof: Let $$H$$ be a hyperplane of $$E$$ . Then by definition, $$H$$ is a subspace of $$E$$ . But since $$E$$ is a topological vector space and by the proposition on the Closures of Subspaces of a Topological Vector Space page, we have that $\overline{H}$ is subspace of $$E$$ such that $H \subseteq \overline{H} \subseteq E$ .

By definition, $$H$$ is maximal. So either $H = \overline{H}$ , in which case $$H$$ is closed, OR, $\overline{H} = E$ , in which case, $$H$$ is dense in $$E$$ . $\blacksquare$