Characterization of Hyperplanes

== Theorem ==
Let $\Bbb F$ be a field.
Let $X$ be a vector space over $\Bbb F$.
Let $U$ be a subspace of $X$. 
:$(1): \quad$ $U$ is a hyperplane
:$(2): \quad$ $U \ne X$, and for any $x \in X \setminus U$ we have $\map \span {U \cup \set x} = X$
:$(3): \quad$ there exists a non-zero linear functional $\phi : X \to \Bbb F$ such that $U = \map \ker \phi$.
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose that: 
:$U$ is a hyperplane.
Let $x \in X \setminus U$. 
Then from Linear Span is Linear Subspace, we have: 
:$\map \span {U \cup \set x}$ is a subspace of $X$.
We have that: 
:$U \subseteq \map \span {U \cup \set x}$
So either: 
:$U = \map \span {U \cup \set x}$
or: 
:$X = \map \span {U \cup \set x}$ 
Since $x \in \map \span {U \cup \set x}$ but $x \not \in U$, we must have: 
:$X = \map \span {U \cup \set x}$ 
=== $(2)$ implies $(1)$ ===
Suppose that $U \ne X$ and:
:for any $x \in X \setminus U$ we have $\map \span {U \cup \set x} = X$.
Now, let $Z$ be a subspace of $X$ such that: 
:$U \subseteq Z \subseteq X$
Then either: 
:$U = Z$
in which case we are done, or:
:there exists $x \in X \setminus U$
Suppose that:
:there exists $x \in X \setminus U$
then we have: 
:$\map \span {U \cup \set x} = X$
We can also write any $z \in \map \span {U \cup \set x}$ in the form: 
:$z = u + \lambda x$
for $u \in U$ and $\lambda \in \Bbb F$ from the definition of linear span.
Since $Z$ is a linear subspace and $U \subseteq Z$, we have: 
:$u \in Z$
and:
:$\lambda x \in Z$
so:
:$u + \lambda x \in Z$ 
We can therefore see that: 
:$z \in Z$
so that: 
:$\map \span {U \cup \set x} \subseteq Z$
So we have: 
:$X \subseteq Z$
as well as:
:$Z \subseteq X$
giving:
:$X = Z$
Since $Z$ was arbitrary, we have that $U$ is a hyperplane.
=== $(2)$ implies $(3)$ ===
Suppose that: 
:$U \ne X$, and for any $x \in X \setminus U$ we have $\map \span {U \cup \set x} = X$
then:
:$U$ is a hyperplane. 
Let $x \in X \setminus U$ and consider: 
:$\map \span {U \cup \set x}$ 
From hypothesis, we have: 
:$\map \span {U \cup \set x} = X$ 
We prove that any: 
:$v \in \map \span {U \cup \set x} = X$
can be uniquely written in the form $v = u + \lambda x$ for $u \in U$ and $\lambda \in \Bbb F$. 
Suppose that: 
:$v = u + \lambda x = u' + \lambda' x$
Then: 
:$\paren {u - u'} + \paren {\lambda - \lambda'} x = 0$
Suppose that $\lambda \ne \lambda'$, we can write: 
:$\ds x = \frac 1 {\lambda' - \lambda} \paren {u - u'}$
So we must have $\lambda = \lambda'$, and hence $u = u'$. 
Since $U$ is a linear subspace, we have $x \in U$, a contradiction. 
So, we can define a map $\phi : X \to \Bbb F$ by: 
:$\map \phi {u + \lambda x} = \lambda$
Clearly $\phi$ is non-zero, since: 
:$\map \phi x = 1$ 
We show that $\phi$ is linear and $U = \map \ker \phi$. 
Let $v_1, v_2 \in \map \span {U \cup \set x}$ and $\alpha, \beta \in \Bbb F$ and write: 
:$v_1 = u_1 + \lambda_1 x$
and:
:$v_2 = u_2 + \lambda_2 x$
Then: 
so $\phi$ is linear.
We then have: 
:$\map \phi x = 0$
 the unique representation of $x$ in the form: 
:$x = u + \lambda x$
for $u \in U$ and $\lambda \in \Bbb F$ has $\lambda = 0$.
That is: 
:$\map \phi x = 0$
:
:$x \in U$
So from the definition of kernel, we have: 
:$U = \map \ker \phi$
=== $(3)$ implies $(2)$ ===
Suppose that there exists a non-zero linear functional $\phi : U \to \Bbb F$ such that: 
:$U = \map \ker \phi$ 
Since $\phi$ is non-zero, we have: 
:$X \ne \map \ker \phi$ 
Let $x \not \in U$.
We aim to show that: 
:$\map \span {U \cup \set x} = X$ 
From the definition of kernel, we have: 
:$\map \phi x \ne 0$ 
Now, let: 
:$z \in X \setminus U$
and let: 
:$\ds y = z - \frac {\map \phi z} {\map \phi x} x$
We then have: 
so:
:$y \in U$
Writing: 
:$\ds z = y + \paren {\frac {\map \phi z} {\map \phi x} } x$ 
we see that: 
:$z \in \map \span {U \cup \set x}$
So, we have: 
:$X \setminus U \subseteq \map \span {U \cup \set x}$
So:
:$\map \span {U \cup \set x} = X$
== Sources ==
* $21.3$: Linear Functionals and Hyperplanes
Category:Hyperplanes