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 $U = \map \ker \phi$.

$(1)$ 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:


 * $\map \span {U \cup \set x} = X$