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.
Best Answer
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:
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.