Hyperplanes of a Vector Space

# Hyperplanes of a Vector Space

## Hyperplanes of a Vector Space

 Definition: Let $E$ be a vector space. A Hyperplane of $E$ is a maximal and proper subspace $H$ of $E$.

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 $f$ on $E$ 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 $E$. $\blacksquare$