# Linearly independent set of vectors in a normed space

Lemma:

Let $$\left \{ v_{1},\cdot \cdot \cdot ,v_{n} \right \}$$ be a linearly independent set of vectors in a normed space $$V,\left \| \cdot \right \|$$. Then, there exists a constant $$c>0$$ such that for any scalars $$\alpha_{1},\cdot \cdot \cdot ,\alpha_{n}$$:

$$\left \| \alpha^{1}v_{1}+\cdot \cdot \cdot +\alpha^{n}v_{n} \right \|\leq c\left ( \left | \alpha^{1} \right |+\cdot \cdot \cdot + \left |\alpha^{n} \right | \right )$$.

If $$s=0$$, the proof is trivial.

So suppose $$s\neq 0$$ and define $$\beta^{i}:= \frac{\alpha^{i}}{s}$$.

With some manipulation we get $$\left \| \sum_{i}^{n} \beta^{i}v_{i}\right \| \leq c$$

Suppose this statement is false. Then there exists a sequence $$\left ( \beta_{m} \right )=\left ( b_{m}^{1}v_{1}+\cdot \cdot \cdot +\beta_{m}^{n}v_{n} \right )$$ in $$V$$ with $$\sum_{i=1}^{n}\left | \beta_{m}^{i} \right |=1$$ such that $$\lim_{m\rightarrow \infty}\left \| \beta_{m} \right \| = 0$$

I'll like to know why the limit of $$\beta_{m}$$ tends to $$0$$.

Any help is appreciated.

• We speak of "a proof" and "to prove", not "a prove" or "to proof". Nov 2, 2016 at 6:15
• Yes that is correct @Masacroso Nov 2, 2016 at 6:24
• @EwanDelanoy that was what I meant. Note that the superscripts are not exponents but rather are indices. Nov 2, 2016 at 6:25
• The index notation is standard in this context but it can cause some confusion. To solve this the common solution is to write $x^{(n)}$ instead of $x^n$. Nov 2, 2016 at 6:27
• I agree with you. Unfortunately, my notes are rather inconsistent in the use of notation. I'd much prefer it if $\left ( x_{n} \right )$ were used as a notation for sequences. In fact, the limit notation in the OP was properly notated by me. The note uses $\left \| \beta_{m} \right \|\rightarrow 0$ Nov 2, 2016 at 6:29

$$||\alpha^1v_1+...+\alpha^nv_n||\le ||\alpha^1v_1||+...+||\alpha^nv_n||=|\alpha^1| ||v_1||+...+|\alpha^n| ||v_n||\le max(||v_i||)(|\alpha^1|+...|\alpha^n|)$$
Any $c\ge max(||v_i||)$ is ok