Proving linear independence about unit outward normals of a pyramid

Question

In proving that every pyramid (4 faces) admits a unique inscribed ball, I encountered the following problem:

Given any pyramid whose unit outward normal vectors are $n_{1,2,3,4}$, prove that the set of vectors $\{n_1-n_{2,3,4}\}$ are linearly independent.

This turns out surprisingly complicated. I feel we will have to do some inner product calculation, since if their lengths were flexible, the claim would be clearly false. So I suppose $$\sum_{j=2}^4 \alpha_j (n_1 - n_j) = 0$$ and end up with $$n_1 =\frac{\sum_j \alpha_j n_j}{\sum_j\alpha_j}\tag{*}$$ But after performing inner product (with themselves) on both sides, I get little information.

One fact I'm aware of: any three of $\{n_{1,2,3,4}\}$ are linearly independent, since three faces intersect at a single point iff their outward normals are independent.

Could anybody give any insight? Thanks!

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I attempted another approach to simplifying this problem. Since $n_{2,3,4}$ form a basis in $\Bbb R^3$, there exist unique $\lambda_{2,3,4}$ such that $$n_1 = \sum_{j=2}^4\lambda_j n_j.$$ But from $(*)$ we also must have that $\sum_{j=2}^4\lambda_j = 1$. Therefore, it suffices to show the following system of linear equations has no solution $$\begin{bmatrix} n_2 & n_3 & n_4 \\ 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{bmatrix} = \begin{bmatrix} n_1 \\ 1 \end{bmatrix}$$ which again amounts to showing that $$A:=\begin{bmatrix} n_1 & n_2 & n_3 & n_4\\ 1 & 1 & 1 & 1 \end{bmatrix}$$ has full rank, i.e. rank 4 which is greater than the coefficient matrix's rank 3. Givien that each $n_{1,2,3,4}$ is unit length, and any three of them are independent, is there any easy way to show $A$ is full rank? Thanks!

Answer 1

Finally, I figured out this relatively easy approach. From $(*)$ we know that if $\{n_1 - n_{2,3,4}\}$ are dependent then we must have $\lambda_{j}\in\Bbb R,\,j=2,3,4$ such that $\sum_j\lambda_j=1$ and $n_1 = \sum_j\lambda_j n_j$.

We put the starting points of all $n_i$s to the origin $O$. Then the endpoints of $n_{2,3,4}$ must form a plane located away from the origin, since they form a basis. Now since $\sum_j\lambda_j=1$, the endpoint of $n_1$ lies on this plane too. Let $n$ be the unit normal of this plane that points away from the origin, then if we denote the distance of this plane to the origin by $d$, we have $$0<d = n\cdot n_i,\quad i=1,2,3,4.$$

However, as a result of Gauss-Ostrogradski theorem (aka divergence theorem), we must have $$\sum_{i=1}^4 S_i n_i = 0 \tag{**}$$ where $S_i>0$ denote areas of the faces corresponding to $n_i$ respectively. Now dot multiply $n$ on both sides of $(**)$ to get a contradiction.

Answer 2

I will write out my take on this problem as a series of hints. Currently is it incomplete. This proof is admittedly a bit long, but it is does not take any hand-wavy approaches to steps. There may be a much better way to approach this problem. You can stop reading at any time if you feel you know how to proceed.

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Hint 1: Use the Inner Product

As you stated, we wish to show that

$$a_1(n_1-n_2)+a_2(n_1-n_3)+a_3(n_1-n_4)=0$$

has no nontrivial solution. We can rearrange this to

$$(a_1+a_2+a_3)n_1=a_1n_2+a_2n_3+a_3n_4$$

Assume we have a nontrivial solution to this. That is, not all $a_1,a_2,a_3=0$. As you suggested, the inner product may be able to help us here. We will assume these normal vectors have unit length for simplicity. Note that

$$n_i\cdot n_j=||n_i||\cdot||n_j||\cos\theta=\cos\theta\leq1$$

How could we use this to simplify the previous equation?

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Hint 2: Simplifying the Equation

The inner products tell us

$$(a_1+a_2+a_3)(n_1\cdot n_1)=a_1(n_2\cdot n_1)+a_2(n_3\cdot n_1)+a_3(n_4\cdot n_1)$$ $$a_1+a_2+a_3=a_1\cos\theta_2+a_2\cos\theta_3+a_3\cos\theta_4$$ $$a_1(1-\cos\theta_2)+a_2(1-\cos\theta_3)+a_3(1-\cos\theta_4)=0$$

Why does this imply linear independence? At this point, you could wave your hands and say this equation can't be true in general. That would result in a much shorter proof. However, I will prove that below.

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Hint 3: Restriction on the Inner Product

Say we orient the pyramid so it has a "bottom" flush with the $x-y$ plane, and let that face have normal vector $n_1$. Logically, it will be pointing straight down, and at least one other $n_{2,3,4}$ will be pointing somewhat up. Call it $n_2$. Thus, we then have

$$n_1\cdot n_2=\cos\theta_2<0$$

Since $\theta_2>\pi/2$. We must now split into cases:

Case A: One $n_{3,4}$ is pointing up, and the other down.

Case B: All $n_{2,3,4}$ are pointing up.

In case A, we have, for simplicity, $\theta_3>\pi/2$ and $\theta_4<\pi/2$

In case B, we have all $\theta\leq\pi/2$.

What does this tell us about our simplified equation?

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Hint 4: Multiple Cases

Note that, if $\cos\theta_j\leq0$,

$$|a_i(1-\cos\theta_j)|\geq|a_i|$$

If $\cos\theta_j>0$

$$|a_i(1-\cos\theta_j)|<|a_i|$$

Note that the equation from hint 2 is equivalent to

$$|a_1(1-\cos\theta_2)+a_2(1-\cos\theta_3)+a_2(1-\cos\theta_4)|=0$$

There are two possible cases here:

1. One or two of these $a_i<0$.
2. All or none of these $a_i<0$.

You should see why I grouped these conditions like this. Can either of these cases be true?

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Hint 5: Neither Case is Correct

For case B2, we break up the above equation to see

$$0=|a_1(1-\cos\theta_2)|+|a_2(1-\cos\theta_3)|+|a_3(1-\cos\theta_4)|\geq|a_1|+|a_2|+|a_3|$$

The right side is a positive number, but we have it less than or equal to $0$. So case B2 can't hold. I do not currently have time to handle the other three cases, hopefully you can work them out.

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Wrapping Up

We've now exhausted all possible routes. Everything we can do has ended in a contradiction. Thus, the only possibility is that our initial non-trivial assumption was false. That is, we must have $a_1=a_2=a_3=0$, showing that $\{n_1-n_{2,3,4}\}$ are LI.