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Theorem:
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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 \(a\).
