3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $H$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Haha that's cute $\endgroup$ – Sentinel135 Apr 25 '17 at 4:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.