Geometric interpretation of adding two hyperplanes

Question

maybe a bit of a novice question, but I would like to get a geometric interpretation of the addition of two hyperplanes. You can interpret a hyperplane described by:

$ax -b = 0$ where $a,x \in R^{d}$ and $b \in R$

as

$(x-p)a = 0$ s.t. $pw = b$, where $p \in R^{d}$.

If I understand correctly, this intuitively means you translate the hyperplane $ax=0$ by a vector $p$. I would like to know how you can describe this vector $p$ in the case of adding two hyperplanes:

$(a_1 x - b_1) + (a_2x - b_2) = 0 = (a_1 + a_2)x - p_1a_1 -p_2a_2 $

So, I would like to describe this hyperplane in the same way as the previous, namely the translation of the hyperplane $(a_1+a_2)x=0$ by some vector $p^`$, but I cant figure out what this $p^`$ would be.. Is there some nice way of describing $-(a_1p_1+a_2p_2)$ as a factor of $(a_1+a_2)$ such that we can translate the hyperplane by this vector?

Answer 1

You are making a mistake in thinking that there is some particular $p$ that must be identified for $ax - b = 0$ to be changed to $a(x - p) = 0$ ($a$ is a matrix and $x$ is a vector, so $ax \ne xa$, in fact, $xa$ is not even defined unless $a$ is $1\times d$). But in fact, every point on the hyperplane defined by $ax - b = 0$ can be used as $p$.

If $x$ and $p$ are any two points on the hyperplane, then $$\begin{matrix} &ax - b = 0\\-&ap - b = 0\\\hline&ax - ap = 0\\&a(x - p) = 0\end{matrix}$$

Conversely, if $a(x - p) = 0$ we can distribute: $ax - ap = 0$ and set $b = ap$ to get $ax - b = 0$. Since $ap = b$, we have $ap - b = 0$, which says that $p$ must be a point on the hyperplane as well.

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Now as to your question, First note that this "sum" of hyperplanes does not generally contain either of the original hyperplanes, but only their intersection. If $a_1x + b_1 = 0$ then for $a_1x + b_1 + a_2x + b_2 = 0$ to be true, we must have $a_2x + b_2 = 0$. That is, for points of $a_1x + b_1 + a_2x + b_2 = 0$ to be one either of the two original hyperplanes, they have to be on both.

Since any point on the combined hyperplane will work for its $p$, and the combined hyperplane includes the intersection of the originals, we could just choose one of the intersection points, assuming any exist: Solve the system of equations $$a_1p = b_1\\a_2p = b_2$$ If $p$ is a solution, then $(a_1 + a_2)(x - p) = 0$ is your combined plane.

Of course, it could easily be that there is no intersection. In this case, the combined plane does not intersection either of the original planes. Before examining that case, I think some examples might be helpful:

Let's look at what happens in $\Bbb R^2$, and set $k = 1$. If $a_1 = \begin{bmatrix} 1 & 0\end{bmatrix}$ and $a_2 = \begin{bmatrix} 0 & 1\end{bmatrix}$, and $b_1 = b_2 = 0$, the first hyperplane $L_1$ is vertical axis $x_1 = 0$ and the second $L_2$ is the horizontal axis $x_2 = 0$. However, your "sum" $L_3$ is the equation $x_1 + x_2 = 0$. In more familiar form $y = -x$, the line at angle $-45^\circ$ through $0$.

Suppose we now set the $a_2$ line to be $L_3$, while leaving $a_1$ as the vertical axis. The next line $L_4$ has equation $2x + y = 0$. Setting $a_2$ to the $L_4$ line gives $L_5$ with equation $3x + y = 0$, and so on:

As you can see, the behavior of the combined lines is not some readily recognizable geometric combination.

What happens if the two planes are parallel: $a_1 = a_2 = a$ but $b_1 \ne b_2$? The combined equation is $2ax - (b_1 + b_2) = 0$ or $$ax - \frac{b_1 + b_2}2 = 0$$ I.e., you get the midplane between them.

What happens if they are skew? Here, you can use gaussian elimination to solve the system of equations $$(a_1 + a_2)p = b_1 + b_2$$. If there were some slick general trick for determining it, that trick would be an easier method of solving such systems, so we can be fairly sure there isn't one. Examining this as skew lines in $\Bbb R^3$ doesn't reveal any evident geometric relationship either (there almost certainly is one, but it will be something esoteric, not obvious).