I am reading the Convex Optimization book from Stephen Boyd and I went back to hyperplane. I don't understand it's definition:

A hyperplane is a set of the form $\{x ~|~a^Tx=b \},~a\in I\!R^n,~x \in I\!R^n,~ b \in I\!R$

Ok, why not. How can you get a plane from this ?

If I refer to another definition of the hyperplane :

Let $a_1,...,a_n$ be scalars not all equal to 0. Then the set S consisting of all vectors $\begin{align} X &= \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{bmatrix} \end{align}$ in $I\!R^n$ such that $a_1x_1+...+a_nx_n = c$, for $c$ a constant, is a hyperplane. (I suppose that this is in fact a scalar product, or I don't get how you can have a constant).

Comparing both definitions, I suppose that $a^Tx = b$ in Boyd is equivalent to $a_1x_1+...+a_nx_n = c$ in the other definition. But Boyd is also saying that $a$ is the normal vector of the hyperplane. How can $a$ be a normal vector if $a \in I\!R^n$ ? For example, if $a \in I\!R^2$, you will have $a = (x, y)$, where $x, y \in I\!R$, not $I\!R^2$. So how can you say that $a$ is a normal vector to the hyperplane if $a$ is only one point of $I\!R^n$ ?

I am pretty sure I have misunderstand something, so if someone could explain it to me with a simple numercial example, it will be great. I will be able to answer to my questions after that. I am not able to find anything clear on the net about hyperplanes.

Thanks a lot.

  • $\begingroup$ Do you know what a linear transformation is? A subspace? $\endgroup$ – D_S Apr 25 '17 at 3:38
  • $\begingroup$ So, Define normal vector. In your words, or in a strict definition that you feel is reasonable for this case. $\endgroup$ – Sentinel135 Apr 25 '17 at 3:52
  • $\begingroup$ @Sentinel135: a normal vector is a vector perpendicular to an object (plane, point, another vector ...) D_S: Sadly no, I didn't follow courses on these points, that is probably why I don't understand. $\endgroup$ – Ecterion Apr 25 '17 at 4:01
  • $\begingroup$ ok now the interesting thing is for $a := <x,y>$ s.t. $x,y \in I\mathbb R$, $a$ is not just a point but a vector. The reason it can be normal to $\vec x$ is that $a^T \vec x = b$ where $b\in \mathbb R$. So what happens if $b=0$? $\endgroup$ – Sentinel135 Apr 25 '17 at 4:32
  • $\begingroup$ @Sentinel135 $a$ and $x$ are orthogonal if $b = 0$ right ? So $x$ is the null vector as $a$ cannot be null everywhere. $\endgroup$ – Ecterion Apr 25 '17 at 4:43

Suppose $a \neq 0$. Let $f(x) = a^Tx = \sum_k a_k x_k$, note that $f$ is a linear functional on $\mathbb{R}^n$ and that $\ker f$ is a linear space of dimension $n-1$. Also note that ${\cal R} f = \mathbb{R}$.

In particular, $\ker f$ is a plane in $\mathbb{R}^n$ that passes through the origin.

Hence for any $x_0$ the set $\{ x | f(x) = f(x_0) \}$ is a plane parallel to $\ker f$ that passes through the point $x_0$. Since $f$ is surjective,it follows that $H=\{ x | f(x) = b \}$ is also a plane parallel to $\ker f$ that passes through some (any) point $x_b$ such that $f(x_b) = b$.

Regarding normals, suppose $x_0$ lies on the hyperplane $H$, that is, $f(x_0) = b$. Now pick any other point $x_1 \in H$ and note that $f(x_1) = b$ and so $f(x_1-x_0) = 0$ or $(x_1-x_0) \bot a$. That is, $a$ is perpendicular to any of the directions $x_1-x_0$ with $x_1 \in H$. That is, for any two points $x_0,x_1 \in H$ we have $(x_1-x_0) \bot a$. This is what we mean when we say $a$ is a normal to $H$.

  • $\begingroup$ You kind of beat me to the punch here. $\endgroup$ – Sentinel135 Apr 25 '17 at 4:34
  • $\begingroup$ No one was hurt in the making of this answer. $\endgroup$ – copper.hat Apr 25 '17 at 4:49
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    $\begingroup$ Haha that's cute $\endgroup$ – Sentinel135 Apr 25 '17 at 4:52

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