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By Extension Theorem: you can extend list of linearly independent vectors $(v_1,v_2,\ldots,v_k)$ to a basis $(v_1,v_2, \ldots, v_k, v_{k+1}, v_n)$

How to prove: $v_i$ for $i = 1,2,...,k$ $\notin span(v_{k+1},...,v_n)$?

Thanks.

Edit: Okay, perhaps I should state the full question.

Suppose that W is a subspace of a finite-dimensional vector space V.

(a) Prove that there exists a subspace W and a function T: V → V such that T is a projection on W along W'.

I want to let W: $(v1, \ldots , vk)$ set of linearly independent vectors. W' = $span(vk+1, \ldots, vn)$

So that $V = W \oplus W'$ (direct sum)

Does this work? The definition of basis that I am using is a list of vectors $(u1,\ldots,un)$ that is linearly independent and spans V.

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3 Answers 3

This depends on what exactly you want to know, and what your definition of basis is.

Throughout my answer, I will use a well-known lemma: Let $(v_i)_{i \in I}$ be a family of vectors. Then the following are equivalent:

  1. The $v_i$ form a a linearly independent system of generators.
  2. The $v_i$ form a maximal linear independent system (i.e., adding any vector to the system makes it linearly dependent).
  3. The $v_i$ form a minimal generating system (i.e., removing any vector from it makes it cease to be a generating system). In particular, $v_i \not\in \operatorname{span}(v_1,\ldots,v_{i-1},v_{i+1},\ldots,v_n)$.

Any of these three characterizations can be used to define the basis of a vector space. Usually, one takes the first formulation to define a (Hamel) basis.

The Extension System now states that $(v_1,\ldots,v_k)$ can be extended to a basis $(v_1,\ldots,v_n)$. If you assume that the Extension Theorem holds, your question is simple to answer: since $(v_1,\ldots,v_n)$ is a basis, it is a minimal generating system, by the above characterization.

If you are instead asking about how one proves that the construction of the extension theorem guarantees that $(v_1,\ldots,v_n)$ is a basis, it is again sufficient to point to the lemma: The Extension Theorem describes the existence of a maximal linear indepdent system, and again, this system is a minimal system of generators.

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It is not clear how to interpret the question so that the answer is not obvious (since in any basis, or indeed in any independent family of vectors, none of its elements lies in the span of (any subset of) the other elements). However one could ask the following: "Given linearly independent vectors $v_1,\ldots,v_k$, if one repeatedly finds vectors $v_i$ for $i=k+1,\ldots,n$ such that each $v_i\notin\mathrm{span}(v_1,\ldots,v_{i-1})$, how can one deduce that $v_k\notin\mathrm{span}(v_{k+1},\ldots,v_n)$?".

Here's one elementary way to see this. By induction on $n$ we may assume that $v_k\notin\mathrm{span}(v_{k+1},\ldots,v_{n-1})$, and suppose for a contradiction that $v_k\in\mathrm{span}(v_{k+1},\ldots,v_n)$. Because of the inductive assumption, the expresion of $v_k$ as linear combination of $v_{k+1},\ldots,v_n$ must involve $v_n$ with nonzero coefficient. Dividing the relation by this coefficient, one can then isolate $v_n$, and obtain $v_n\in\mathrm{span}(v_i,\ldots,v_{n-1})$. But then a fortiori $v_n\in\mathrm{span}(v_1,\ldots,v_{n-1})$, contradicting the choice of $v_n$.

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Your idea works, but you do not seem to understand clearly what is given and what needs to be constructed.

You are given $W$. Let $(v_1,\dots,v_k)$ be a basis of $W$, you extend it to get a basis $(v_1,\dots,v_n)$ of $V$ and you then define $W'$ as the span of $(v_{k+1},\dots,v_n)$.

Hope this helps.

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