Results 1 to 3 of 3

Math Help - linear algebra proof help!

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    3

    linear algebra proof help!

    Let S = {V1,ůvn} be a linearly independent set in a vector space V. Suppose w is in V, but w is not in span(S). Prove that the set T = {V1,ůVn,w} is a linearly independent set.



    since it says w is in v, does it mean that w is a subspace of v? yet w is not in span(S). I am kinda confused with what it means?

    I am thinking that suppose w is not = 0, and because w is in V.
    a1v1+a2v2+...+anvn+a(n+1)w=0 which leads to a(n+1)=-a1v1-a2v2-...anvn
    and then because S={V1,ůvn} is a linearly independent set. a1v1, a2v2... are all zeros, so {V1,ůVn, W} would be linearly indep too. then w can be write in a linear combi. thus T is a linearly indep set.

    I know it sounds bizzare, but just confused how to prove this one.



    ANy ideas? thoughts?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    hi

    hi

    It means that your vectors  [v_{j}]_{j=1}^{p} that span
    a subspace of V does not span all of V, since \vec{w} is a vector that cannot "be reached" by v1,....,vn.

    It says that \vec{w} is in V, meaning that \vec{w} is a vector in V.

    The line through the origin and w is a subspace of V, that is, all scalar multiples of \vec{w} is in the subspace spanned by \vec{w}.

    I am not sure either how to do this proof correct (I am myself currently studying linear algebra). But maybe something like this:
    (If mr_fantastic or anybody wants to correct, plz do!)

    Assume  c_{1}\vec{v_{1}} + ... + c_{n}\vec{v_{n}} = d\vec{w}

     c_{1}\vec{v_{1}} + ... c_{n}\vec{v_{n}} -d\vec{v_{w}} = \vec{0}

     c_{1}\vec{v_{1}} + ... c_{n}\vec{v_{n}} -(c_{1}\vec{v_{1}} + ... c_{n}\vec{v_{n}}= \vec{0}

     (c_{1} - c_{1})\vec{v_{1}} + ... + (c_{n} - c_{n})\vec{v_{n}}) = \vec{0}

    Giving all coefficients equaling zero.
    Last edited by Twig; April 7th 2009 at 01:59 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Say that a_1v_1+a_2v_2+...+a_nv_n + bw = 0, there are two cases. Either b=0 or b\not = 0. If b=0 then we end up with a_1v_1+...+a_nv_n = 0 and so a_1=...=a_n=0, which shows that \{v_1,...,v_n,w\} is linearly independent. Now say that b\not = 0. In this case divide by -b to get: c_1v_1+....+c_nv_n = w where c_j = -\tfrac{a_n}{b}. Thus, w\in \text{spam}(S) which is a contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear algebra proof.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 21st 2011, 05:46 AM
  2. linear algebra proof
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 30th 2010, 06:52 PM
  3. Linear Algebra Proof
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 1st 2009, 07:39 AM
  4. Linear Algebra Proof Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: June 26th 2008, 05:22 PM
  5. Linear Algebra proof
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: November 15th 2007, 05:25 AM

Search Tags


/mathhelpforum @mathhelpforum