• Support PF! Buy your school textbooks, materials and every day products Here!

LA - Proof Involving Linear Independence and Spanning

  • #1

Homework Statement



I was trying to prove a theorem from Axler's Linear Algebra text and my proof is different from the one in the book, and I'm wondering if someone can check whether or not my proof works, since I'm just starting to write proofs.

Theorem 2.6 (pg. 25): In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.

Homework Equations





The Attempt at a Solution



Let [itex]V[/itex] be a finite-dimensional vector space. Suppose [itex](u_1,...,u_m)[/itex], where [itex]u_1,...,u_m \in V[/itex] is a list of linearly independent vectors and [itex] (w_1,...,w_n)[/itex] where [itex]w_1,...,w_n \in V[/itex] spans [itex]V[/itex]. We need to prove that [itex]m \leq n[/itex]. Suppose [itex]m>n[/itex]. Because the list [itex] (w_1,...,w_n)[/itex] spans V, there exist scalars [itex]a_{i,j}[/itex] where [itex]i \in [1,m], j \in [1,n][/itex] such that [itex]u_i=a_{i,1}w_1+...+a_{i,n}w_n[/itex]. Thus, the vectors in [itex](u_1,...,u_m)[/itex] can be replaced by these linear combinations. However, since [itex]m>n[/itex], there exists some [itex]k[/itex] where [itex]m<k\leq n[/itex] such that [itex]u_k[/itex] can be written in terms of [itex]u_1,...,u_{k-1}[/itex], and thus [itex](u_1,...,u_m)[/itex] is a linearly dependent list, and we have reached a contradiction. Therefore, [itex]m\leq n[/itex].

If someone could please look this over that would be great. Also, I would really like any criticism on my logic or proof writing skills in general since I have learned proofs on my own and no one has really seen my proofs before.

Thanks for your help :)
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
i think that's looking good and so is the logic ;)
 
  • #3
That's good to hear I'm on the right track. Thanks for taking a look at it :)
 

Related Threads on LA - Proof Involving Linear Independence and Spanning

Replies
4
Views
2K
Replies
7
Views
603
  • Last Post
Replies
10
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
6
Views
2K
Replies
3
Views
6K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
1
Views
2K
Top