Functional Analysis: Cannot find a linear combination that involves large scalars but represents a small vector

Question

Theorem:

Let $\{x_1,\ldots,x_n\}$ be a linearly independent set of vectors in a normed space $X$ (of any dimension). Then there is a number $c>0$ such that for every choice of scalars $a_1,\ldots,a_n$:

$$\left\lVert a_1x_1+a_2x_2+\cdots+a_nx_n \right\rVert\ge c \left( \lvert a_1 \rvert + \cdots + \lvert a_n \rvert \right)$$

My question is: why is this theorem relevant? It seems that I could pick an extremely small $c$ value close to zero that would satisfy the final equation. I realize that this is likely wrong but I hoping to get some intuition on what the theorem seems to be getting at. The author gets at it a bit with the following statement:

Very roughly speaking it states that in the case of linear independence of vectors we cannot find a linear combination that involves large scalars but represents a small vector.

But the fact that a $c$ value is needed seems to suggest that the $c$ value is needed to reduce the total value of the the scalars (if $c<1$) and so it would seem that we can find cases where we have small vectors with large scalars but we simply do away with that by imposing a constant $c$ to make the scalars "less big" (the RHS of the inequality).

This lemma is from Kreyszig's Introductory Functional Analysis book.

Answer 1

The fact that you are saying "... a $c$ value is needed", and "Imposing a constant $c$ ..." makes me think that your are looking at this the wrong way round. The theorem says (rewriting it a little bit) that if someone gives us a linear independent set $\{x_1, ..., x_n\}$ then the ratio $$\left\lVert a_1x_1 + a_2x_2 + ... + a_nx_n \right\rVert /(\lvert a_1 \rvert+ ... + \lvert a_n \rvert)$$ stays bounded away from zero by some definite amount (where we skip the case of all $a_i = 0$).

Simple linear independence says that the ratio never equals zero. The theorem gives a stronger result. If we assume that we take the $c$ in your statement to be as large as possible, i.e. set it equal to the $\min$ of the ratio, then we can see $c$ as a measure of how close to linear dependence our set of vectors is.

To me it looks to be an analog of the "angle between two vectors" that doesn't use an inner product. Let's look at the simplest possible example: two unit vectors in $\mathbb{R}^2$. In this case the ratio can be shown to equal $$\sqrt{\dfrac{a_1^2 + a_2^2 + 2a_1a_2 \cos(\theta)}{a_1^2 + a_2^2}},$$ where $cos( \theta) = \langle x_1, x_2\rangle$. This has a minimum of $\sqrt{1+\cos \theta}$, the value of $c$ in your theorem. Consider what happens to vectors similar to those used by SmileyCraft: $(1,0)$ and $(\cos(\alpha), \sin(\alpha))$ for $\alpha$ going from $\pi/2$ to $\pi$. $\cos(\theta)$ goes from $0$ to $-1$ and your value of $c$ goes from $1$ down to $0$ - as the vectors get closer to being linearly dependent the value of c gets closer to $0$.

Answer 2

It is interesting to note that the converse holds as well.

Assume $\sum_{i=1}^n a_ix_i = 0$ for some scalars $a_i$. If such constant $c > 0$ exists, then for we would have $$0 = \|a_1x_1 + \cdots+ a_nx_n\| \ge c(|a_1| + \cdots + |a_n|) $$ which implies $|a_1| + \cdots + |a_n| = 0$, or $a_1=\cdots=a_n = 0$. Hence $\{x_1, \ldots, x_n\}$ is linearly independent.

Therefore the property of not being able to take linear combinations with large scalars and yield small vectors actually characterizes finite linearly independent sets. Consider what happens with the standard basis $e_1, \ldots, e_n$ of $\mathbb{R}^n$:

$$\|a_1e_1 + \cdots + a_ne_n\|_2 = \|(a_1, \ldots, a_n)\|_2 = \sqrt{|a_1|^2+\cdots + |a_n|^2} \ge \frac1{\sqrt{n}}(|a_1|+\cdots+|a_n|)$$