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 $X$.

The following statements are equivalent:

$(1): \quad$ $U$ 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:

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

$\Box$

$(2)$ 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.

$\Box$

$(2)$ 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:

\(\ds \map \phi {\alpha v_1 + \beta v_2}\)

\(=\)

\(\ds \map \phi {\paren {\alpha u_1 + \beta u_2} + \paren {\alpha \lambda_1 + \beta \lambda_2} x}\)

\(\ds \)

\(=\)

\(\ds \alpha \lambda_1 + \beta \lambda_2\)

\(\ds \)

\(=\)

\(\ds \alpha \map \phi {u_1 + \lambda_1 x} + \beta \map \phi {u_2 + \lambda_2 x}\)

\(\ds \)

\(=\)

\(\ds \alpha \map \phi {v_1} + \beta \map \phi {v_2}\)

so $\phi$ is linear.

We then have:

$\map \phi x = 0$

if and only if 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:

$\map \phi x = 0$

if and only if:

$x \in U$

So from the definition of kernel, we have:

$U = \map \ker \phi$

$\Box$

$(3)$ 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:

\(\ds \map \phi y\)

\(=\)

\(\ds \map \phi z - \map \phi {\frac {\map \phi z} {\map \phi x} x}\)

Definition of Linear Functional

\(\ds \)

\(=\)

\(\ds \map \phi z - \frac {\map \phi z} {\map \phi x} \map \phi x\)

Definition of Linear Functional

\(\ds \)

\(=\)

\(\ds \map \phi z - \map \phi z\)

\(\ds \)

\(=\)

\(\ds 0\)

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:

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

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $21.3$: Linear Functionals and Hyperplanes