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My textbook says the following:

A hyperplane is a set of the form

$$\{ x \mid a^T x = b \},$$

where $a \in \mathbb{R}^n$, $a \not= 0$, and $b \in \mathbb{R}$. Analytically it is the solution set of a nontrivial linear equation among the components of $x$ (and hence an affine set). Geometrically, the hyperplane $\{ x \mid a^T x = b \}$ can be interpreted as the set of points with a constant inner product to a given vector $a$, or as a hyperplane with normal vector $a$; the constant $b \in \mathbb{R}$ determines the offset of the hyperplane from the origin. This geoemtric interpretation can be understood by expressing the hyperplane in the form

$$\{ x \mid a^T(x - x_0) = 0 \},$$

where $x_0$ is any point in the hyperplane (i.e., any point that satisfies $a^T x_0 = b$).

I find the coherency of the hyperplane definition $\{ x \mid a^T(x - x_0) = 0 \}$ to be questionable. We are looking to define the hyperplane in the first place, which the author defines as $\{ x \mid a^T(x - x_0) = 0 \}$; but this definition has imbedded in it that $x_0$ is any point in the hyperplane. So in the process of constructing a full definition of "hyperplane", the author explicitly imbeds in the as-of-yet incomplete definition of "hyperplane" a term that is defined in terms of the as-of-yet incomplete definition, which therefore presumes that we already have a full definition of the hyperplane. In other words, the "definition" presumes that we have already defined what a hyperplane is. It seems to me that this is incoherent, since, it seems to me, a mathematical definition (indeed, any definition) cannot be self-referential in this way?

I would appreciate it if people could please take the time to clarify this.

EDIT:

The way I see it, there are two possibilities: (1) I am misunderstanding $\{ x \mid a^T(x - x_0) = 0 \}$, and it is actually a special case (a specific type of) $\{ x \mid a^T x = b \}$, or (2) both $\{ x \mid a^T x = b \}$ and $\{ x \mid a^T(x - x_0) = 0 \}$ are equivalent, which means that they must both be definitions of hyperplane, which means that the second is an incoherent "definition" because it is self-referential and cannot stand on its own. I'm honestly not sure and would appreciate help.

EDIT2:

I just did some sketching of hyperplanes in the ambient space $\mathbb{R}^2$ to get a better idea of the role of $x_0$ in hyperplanes. We first begin by selecting any (random) points $a$ and $x_0$ in the ambient space $\mathbb{R}^2$. After this step, it seems that the geometry of the hyperplane is "locked-in"; in other words, the selection of $a$ and $x_0$ determines the hyperplane. We then, as the definition $\{ x \mid a^T(x - x_0) = 0 \}$ suggests, find the point $x$ such that $x - x_0$ is orthogonal to $a$. The set of all $x$ such that $x - x_0$ is orthogonal to $a$ is a line, and this line is the hyperplane, which can also be expressed equivalently as $\{ x \mid a^T x = b \}$.

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  • $\begingroup$ Well the hyperplane is defined in the first equation, and the second equation follows from the definition, i.e. from the first equation we can write the hyperplane in the form of the second equation. $\endgroup$ – Dave Apr 5 at 13:27
  • $\begingroup$ The definition of the hyperplane is in the first equation. The second equation isn't its definition, but another form in which to express it given that it already is defined by the first form. $\endgroup$ – MPW Apr 5 at 13:28
  • $\begingroup$ @Dave But the point is that definitions are supposed to be able to stand on their own, no? Isn't this a necessity of mathematical logic? So we should be able to erase the first part and the second part should still make sense. $\endgroup$ – The Pointer Apr 5 at 13:29
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    $\begingroup$ You should understand this as follows: (1) A hyperplane, is a set $H(a,b)$, given two initial data $a\in R^n$ and $b\in \mathbb{R}$, defined as $H(a,b)=\{x\in \mathbb{R}^n\mid a^Tx=b\}$. (2) A hyperplane is a set $H(a,x_0)$, given two initial data $a,x_0\in \mathbb{R}^n$, defined as $H(a,x_0)=\{x\in \mathbb{R}^n\mid a^T(x-x_0)=0\}$. Consequence of the second definition is $x_0\in H(a,x_0)$. The two definitions (1) and (2) are equivalent. $\endgroup$ – Hamed Apr 5 at 13:45
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    $\begingroup$ @ThePointer A hyperplane is a set of the form $\{x~|~a^T(x-x_0)=0\}$ where $a,x_0\in R^n$, $a\neq 0$. $\endgroup$ – Michal Adamaszek Apr 5 at 13:47
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The two definitions of a hyperplane can be written in the following format:

Definition 1: A hyperplane in $\mathbb R^n$ is a subset which can be written in the form $\{x \mid a^T x = b\}$, for some $a \in \mathbb R^n$ such that $a \ne 0$, and for some $b \in \mathbb R$.

Definition 2: A hyperplane in $\mathbb R^n$ is a set which can be written in the form $\{x \mid a^T(x-x_0) = 0\}$ for some $a \in \mathbb R^n$ such that $a \ne 0$, and some $x_0 \in \mathbb R^n$.

These definitions are equivalent.

To see why definition 1 implies definition 2, if a set $H$ satisfies definition 1 as witnessed by $a,b$, then $H$ is not empty, as one can prove by linear algebra. Pick any $x_0 \in H$. It follows that $H$ also satisfies definition 2 as witnessed by $a,x_0$.

Conversely, if $H$ satisfies definition 2 as witnessed by $a,x_0$, it also satisfies definition 1 as witnessed by $a,b$ where $b = a^T x_0$.

To summarize, the "coherence" issue that you bring up is a valid issue, but is settled by demonstrating that any $H$ that satisfies Definition 1 is not empty.

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In both cases you need two things to specify a hyperplane: a normal vector, and a point on the hyperplane. In the first case, the special point on the hyperplane is not made explicit, but is hidden in the affine translation $b$. Thinking by analogy to a line in a plane, recall that a line can be described in slope-intercept form, i.e. by an equation of the form $$ y = ax + b, $$ where $a$ is the slope and $b$ is the $y$-intercept. In particular, this line passes through the point $(0,b)$ in the plane. This point is not made explict, but is part of this definition of a line. Rearranging things a bit, this becomes $$ \langle -a, 1 \rangle \cdot \langle x, y \rangle = b, $$ which is of the form $\mathbf{a}^T \mathbf{x} = b$.

Lines may also be described by a point-slope equation, i.e. $$ y - y_0 = a(x-x_0), $$ where $(x_0, y_0)$ is a point that the line passes through, and $a$ is the slope of the line. In a more vector-y notation, this can be written as $$ a(x-x_0) + (-1)(y-y_0) = \langle a, -1 \rangle \cdot \langle x-x_0, y-y_0 \rangle = 0, $$ which is of the form $\mathbf{a}(\mathbf{x} - \mathbf{x}_0) = 0$. In this description, a point on the line is made explicit.

Note that any line in point-slope form can be expressed in slope-intercept form, and vice versa: $$ y - y_0 = a(x-x_0) \implies y = ax + \underbrace{(y_0 - ax_0)}_{=b}, $$ and $$ y = ax + b \implies y - b = a(x-0). \tag{$(x_0,y_0) = (0,b)$}$$ We could therefore define a line via either description. They are completely equivalent. The definition of a hyperplane can be seen as a generalization of this same idea.

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