Dimension of hyperplanes

Alicia Y. Tsai; Authors: Laurent EI Ghaoui; Contributors: Phan Hoang, Tony Tin, Ha To Tam, Nguyen Van Kep, Le Dinh Nam, Juliette Decugis, Nguyen Quoc Cuong.; Giuseppe C. Calafiore

Dimension of hyperplanes

Theorem:

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

$\mathbf{H} = \left\{ x \, : \, a^T x = b \right\},$

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 $a^T x = b$ , as in the above.

Proof:

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

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

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

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

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

$z_0 = \begin{pmatrix} b \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad z_1 = \begin{pmatrix} -\dfrac{a_2}{a_1} \\[0.8em] 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad \ldots, \quad z_{n-1} = \begin{pmatrix} -\dfrac{a_n}{a_1} \\[0.8em] 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}.$

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

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

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