# Linearly independent vectors on a normed space

I'm studying functional analysis and I have a problem I don't know how to solve or even how to start. Can anyone help me? The problem is the following:

Let $$X$$ be a normed space and $$x_1,...,x_n$$ a set of linearly independent vectors. Show that there exists $$\delta>0$$ so that if $$y_1,...,y_n \in X$$ satisfy $$\|y_i\|<\delta$$ for all $$i$$ then $$x_1+y_1,...,x_n+y_n$$ is a set of linearly independent vectors.

Thanks in advance.

• This has been answered here many times. May 18 at 8:36
• and this time the OP did not show any work. May 18 at 8:59

## 1 Answer

Suppose no such $$\delta$$ exists, then for each $$k\in \mathbb N$$, we can find a $$n$$-tuple $$(y_1,..,y_n)$$ such that $$||y_i||<\frac{1}{k}$$ for each $$i$$ and $$x_1+y_1,...,x_n+y_n$$ is linearly dependent. Thus for some scalars $$c_1,c_2,..,c_n$$ associated to $$k$$ (with atleast one of them non-zero), we have $$\sum_{i=1}^nc_ix_i=-\sum_{i=1}^nc_iy_i$$ Taking norm on the both sides gives $$||\sum_{i=1}^nc_ix_i||=||\sum_{i=1}^nc_iy_i||\leq \sum_{i=1}^n|c_i|||y_i||<\frac{1}{k}\sum_{i=1}^n|c_i|$$ If we set $$a_i=\frac{c_i}{\sum_{i=1}^n|c_i|}$$ and adjust the signs of $$y_i$$'s and $$x_i$$'s to make all $$c_i$$'s non-negative, then we see from above that for each $$k\in \mathbb N$$, we can find an associated element $$X_k=\sum_{i=1}^na_ix_i$$ in the symmetric convex hull of $$\{ x_1,..,x_n\}$$ with norm less than $$\frac{1}{k}$$. Thus $$\{X_k\}\to 0$$ and hence $$0$$ must be in the symmetric convex hull of $$x_1,..,x_n$$ which is contradiction as these are linearly independent elements.

Just to clarify that, by symmetric convex hull of $$\{𝑥_1,𝑥_2,..,𝑥_𝑛\}$$, I mean the union of the hulls of $$\{𝑠(1)𝑥_1,𝑠(2)𝑥_2,...,𝑠(𝑛)𝑥_𝑛\}$$ where $$𝑠:\{1,2,..,𝑛\}→\{+1,−1\}$$ are sign functions. This can easily seen to be closed and not containing 0