Linearly Independent Sets and Bases

[pollytree]

Homework Statement

V is a subspace of Rⁿ and S={v₁,...,v_k} is a set of linearly independent vector in V. I have to prove that any list of linearly independent vectors can be extended to a basis for V.

Homework Equations

None that I can think of.

The Attempt at a Solution

So to be a basis, the vectors must be linearly independent (which is given) and span V. I think we also have to show that the number of elements in the basis is equal to the dimension of V. We have a hint that says we can prove the statement above by adding vectors which are not in span S one at a time until we span all of V and arguing that the result is linearly independent.

But I'm not really sure where to begin on this question. Any help would be great! Thanks :D

 

[losiu99]

Let {e₁,...,e_n} be a standard basis. Then set {e₁,...,e_n, v₁} is linearly dependent, so
\sum \alpha_i e_i+\beta v_1=0
Cetrainly \beta\neq 0, and \alpha_i\neq 0 for some i. Show that {e₁,...,e_i-1, e_i+1,..., e_n, v₁} is a base. Applying repeatedly this procedure, we may "inject" all the v's into the basis. Show that in each step we can discard one of the e's (and not the previously added v's), that's where linear independence is needed.
Good luck.

 

[JonF]

I would show that any S could be extended to be a basis for V, then show that a basis for V can be extended to be a basis for Rⁿ.

Let B be a basis for S, what would the minimal subset of B look like that spans V? Ask your self a similar question to go from V to Rⁿ

 

[HallsofIvy]

I would show that any S could be extended to be a basis for V, then show that a basis for V can be extended to be a basis for Rⁿ.

Sorry, but this makes no sense to me. The problem was to show that any such S could be extended to a basis for V and there is no mention of Rⁿ.

Let B be a basis for S, what would the minimal subset of B look like that spans V?

This also makes no sense. S is a set of vectors, not a vector space and so has no "basis". Even if S were a subspace of V, any basis for V would have to be a superset of a basis for S, not a subset.

Ask your self a similar question to go from V to Rⁿ

Unnecessary- the problem says nothing about Rⁿ.

 

[JonF]

Sorry I miss read the problem, I thought it was to prove that S could be extended to be a basis for Rⁿ. But as you stated my first step is his actual question.

Here is what I was thinking, if it's wrong please let me know!

V has a basis since it’s a subspace, and this means every vector in V can be expressed as a linear combination of the elements of that basis.

Let B be the minimal set of basis vectors of V needed to compose a linear combination of all vectors in S (we know this set exist since S is a subset of vector space V, and since V is a subspace of Rⁿ, it's basis has to be countable).

The basis of V - B union S should be a basis for V right?