In this topic, we show that in a normed space, any finite linearly independent system is stable under small perturbations. To be exact, here is the statement.

Suppose is a normed space and is a set of linearly independent elements in . Then is stable under a small perturbation in the sense that there exists some small number such that for any with , the all elements of are also linearly independent.

We prove this result by way of contradiction. Indeed, for any , there exist elements with such that all elements of are linearly dependent, that is, there exist real numbers with such that

with

Clearly,

Observe that

Therefore, if we denote , then

Hence, we have just shown that for any , there exist numbers with

Set

Since is closed and bounded, it is compact. We also set

Clearly, is continuous. We now consecutively choose with . Then there exists a sequence such that . Hence

By the Weierstrass extreme value theorem, there exists some such that . However, this is impossible because is the set of linearly independent elements. The proof is complete.

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Define by with . is continous. Since is linearly independent then for any . Hence . By triangle inequality, we have for any

Choose , then for any with which implies is linearly independent. 🙂

Comment by Van Hoang Nguyen — February 25, 2017 @ 5:44

typos: Choose $latex \epsilon

0$ which then implies is linearly independent.Comment by Van Hoang Nguyen — February 25, 2017 @ 5:52

Cool!

Comment by Ngô Quốc Anh — February 25, 2017 @ 13:11

This result can be seen as a consequence of a perturbation theorem from theory of metric regularity: roughly speaking, the metric regularity of a set-valued mapping is stable under a small linear perturbation (Since is linearly independent, the mapping . is metrically regular. So, for in X near 0, the linearly perturbed mapping is also metrically regular. The latter entails the linear independence of .

Comment by Nguyen Huy Chieu — March 6, 2017 @ 14:29