Results 1 to 4 of 4

Math Help - Simple subspace proof using induction: STUCK

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    168

    Simple subspace proof using induction: STUCK

    Hi, I'm reviewing some linear algebra this summer as a refresher. I've come across the following proof problem, and have found myself stuck. Apparently, I'm supposed to use induction to solve this, but I'm not really sure exactly WHAT to use induction on. Anyway, here's the problem:

    Prove that if W is a subspace of a vector space V and w(1), w(2),...,w(n) are in W, then a(1)w(1) + ... + a(n)w(n) is in W for any scalars a(1), ... ,a(n).

    Any help on this? :-(
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    It will not be necessary to use induction in this problem. One simply notes that a subspace is a subset of a vector space that is also a vector space on its own. One of the properties of being a vector space is that it is closed under addition and scalar multiplication. That is all that is being used here. I suppose if one really wanted to be rigorous one could potentially use induction I suppose.

    n=1
    a_1w_1 \in W by closure under scalar multiplication

    Induction hypothesis for n=k
    Assume a_1w_1+a_2w_2+...+a_kw_k \in W for any a_i \in \mathbb{F} w_i \in W.

    Show a_1w_1+a_2w_2+...+a_kw_k +a_{k+1}w_{k+1} \in W. By induction hypothesis, a_1w_1+a_2w_2+...+a_kw_k \in W so say a_1w_1+a_2w_2+...+a_kw_k =w'.
    Furthermore a_{k+1}w_{k+1} = w''\in W by closure under scalar multiplication.
    Then
    a_1w_1+a_2w_2+...+a_kw_k +a_{k+1}w_{k+1} = w' + w'' \in W by closure under addition.

    I think it a little extreme, but you asked for the induction. Hope that helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,977
    Thanks
    1121
    The reason you should use induction is that the definition of subspace (closed under addition) only refers to the fact that au+ bv is in the space for u and v in the space, a, b, scalars.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    Yeah, I guess it is just always understood that in vector spaces they are closed under linear combinations, guess I have never formally proved it because it is a pretty trivial exercise in induction. Like that's the whole point of having a basis for your vector space or free module.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stuck on proof by induction
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: October 19th 2011, 07:16 AM
  2. [SOLVED] Stuck on Subspace proof question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 8th 2010, 04:55 PM
  3. Proof by induction stuck at very end!!
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 25th 2009, 03:38 AM
  4. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 09:33 PM
  5. Induction proof...I'm stuck!
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 18th 2008, 10:24 AM

Search Tags


/mathhelpforum @mathhelpforum