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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