Ngô Quốc Anh

February 23, 2017

In a normed space, finite linearly independent systems are stable under small perturbations

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:21

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 (X, \|\cdot\|) is a normed space and \{x_1,...,x_n\} is a set of linearly independent elements in X. Then \{x_1,...,x_n\} is stable under a small perturbation in the sense that there exists some small number \varepsilon>0 such that for any \|y_i\| < \varepsilon with 1 \leqslant i \leqslant n, the all elements of \{x_1+y_1,...,x_n+y_n\} are also linearly independent.

We prove this result by way of contradiction. Indeed, for any \varepsilon>0, there exist n elements y_i \in X with \|y_i\| < \varepsilon such that all elements of \{x_1+y_1,...,x_n+y_n\} are linearly dependent, that is, there exist real numbers \alpha_i with 1 \leqslant i \leqslant n such that

\displaystyle \alpha_1 (x_1+y_1) + \cdots + \alpha_n (x_n+y_n) =0

with

\displaystyle |\alpha_1| + \cdots + |\alpha_n| >0.

 

Clearly,

\displaystyle \alpha_1 y_1 + \cdots + \alpha_n y_n =- (\alpha_1 x_1 + \cdots + \alpha_n x_n)

Observe that

\displaystyle \|\alpha_1 x_1 + \cdots + \alpha_n x_n \| \leqslant |\alpha_1| \|y_1\| + \cdots + |\alpha_n |\|y_n \| <\varepsilon (|\alpha_1| + \cdots + |\alpha_n| ).

Therefore, if we denote c_i = \alpha_i/(|\alpha_1| + \cdots + |\alpha_n| ), then

\displaystyle |c_1|+\cdots +|c_n| = 1, \quad \|c_1 x_1 + \cdots + c_n x_n \| < \varepsilon.

Hence, we have just shown that for any \varepsilon>0, there exist n numbers c_i with

\displaystyle |c_1|+\cdots +|c_n| = 1, \quad \|c_1 x_1 + \cdots + c_n x_n \| < \varepsilon.

Set

\displaystyle E = \{ (c_1,\dots, c_n) : |c_1|+\cdots +|c_n| = 1 \}.

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

\displaystyle f : E \to [0, +\infty), \quad f(c_1,\dots,c_n) = \|c_1 x_1 + \cdots + c_n x_n \|.

Clearly, f is continuous. We now consecutively choose \varepsilon = 1/k with k \geqslant 1. Then there exists a sequence \{c^k\}_k \subset E such that f(c^k) < 1/k. Hence

\inf_E f(c) =0.

By the Weierstrass extreme value theorem, there exists some c \in E such that f(c)=0. However, this is impossible because \{x_1,...,x_n\} is the set of linearly independent elements. The proof is complete.

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4 Comments »

  1. Define f:S^{n-1} \to [0,\infty) by f(c) = \|c_1x_1 + \cdots +c_nx_n\| with c =(c_1,\cdots,c_n). f is continous. Since x_1, \ldots, x_n is linearly independent then f(c) >0 for any c\in S^{n-1}. Hence b = \inf_{c\in S^{n-1}} f(c) > 0. By triangle inequality, we have for any c \in S^{n-1}

    \displaystyle \|\sum c_i(x_i + y_i)\| \geq \|\sum c_i x_i\| -\sum |c_i| \|y_i\| \geq b - \max_i \|y_i\|\sum |c_i| \geq b -\sqrt{n} \max_i \|y_i\|.

    Choose \epsilon < b/\sqrt{n}, then for any y_1, \ldots, y_n \in X with \|y_i\| \ne 0 which implies x_1 +y_1, \ldots, x_n +y_n is linearly independent. 🙂

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

    • typos: Choose $latex \epsilon 0$ which then implies x_1 +y_1,\ldots,x_n +y_n is linearly independent.

      Comment by Van Hoang Nguyen — February 25, 2017 @ 5:52

    • Cool!

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

  2. 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 \{x_1,...,x_n\} is linearly independent, the mapping F(x*):=(x*(x_1),...,x*(x_n)). is metrically regular. So, for y_1,...,y_n in X near 0, the linearly perturbed mapping F_y(x*):= (x*(x_1+y_1),...,x*(x_n+y_n)) is also metrically regular. The latter entails the linear independence of \{x_1+y_1,...,x_n+y_n\}).

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


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