Dimension of hyperplanes

L. El Ghaoui

Dimension of hyperplanes

Theorem

A set ${\bf H}$ in $\mathbb{R}^n$ of the form

${\bf H}=\{x: a^Tx = b\},$

where $a \in \mathbb{R}^n$ , $a \neq 0$ , and $b \in \mathbb{R}$ are given, is an affine set of dimension $n-1$ .

Conversely, any affine set of dimension $n-1$ can be represented by a single affine equation of the form $a^Tx=b$ , as in the above.

Proof:

Consider a set ${\bf H}$ described by a single affine equation:

$a_1 x_1 + \cdots + a_n x_n = b,$

with $a \neq 0$ . Let us assume for example that $a_1 \neq 0$ . We can express x_1 as follows:

$x_1 = b - \frac{a_2}{a_1} x_2 - \cdots - \frac{a_n}{a_1}x_n.$

This shows that the set is of the form $z_0 + {\bf span}(z_1, \cdots, z_{n-1})$ , where

$z_0=\left(\begin{array}{c} b \\ 0 \\ 0 \\ \vdots \\ 0 \end{array}\right), z_1=\left(\begin{array}{c} -\frac{a_2}{a_1} \\ 1 \\ 0 \\ \vdots \\ 0 \end{array}\right), \ldots, z_{n-1}=\left(\begin{array}{c} -\frac{a_n}{a_1} \\ 0 \\ \vdots \\ 0 \\ 1 \end{array}\right).$

Since the vectors $z_1, \cdots, z_{n-1}$ are independent, the dimension of ${\bf H}$ is $n-1$ . This proves that ${\bf H}$ is indeed an affine set of dimension $n-1$ .

The converse is also true. Any subspace ${\bf L}$ of dimension $n-1$ can be represented via an equation $a^Tx=0$ for some $a \neq 0$ . A sketch of the proof is as follows. We use the fact that we can form a basis $(z_1, \cdots, z_{n-1})$ for the subspace ${\bf L}$ . We can then construct a vector $a$ that is orthogonal to all of these basis vectors. By definition, ${\bf L}$ is the set of vectors that are orthogonal to $a$ .

Dimension of hyperplanes

License

Share This Book