Results 1 to 7 of 7

Math Help - linear independence: Eigenvectors/values

  1. #1
    Newbie
    Joined
    May 2012
    From
    Melbourne
    Posts
    6

    linear independence: Eigenvectors/values

    Recall that two vectors v and w are linearly independent if the only solution to the vector equation av + bw = 0
    is the solution a = b = 0.

    Let v and w be eigenvectors of a matrix M with corresponding nonzero eigenvalues λ and such that λ .
    (i) Given scalars a and b, find an expression for, M(av + bw), that does not contain M.


    (ii) Hence or otherwise, prove that the eigenvectors v and w must be linearly independent.

    Thanks
    Last edited by bluelavae; February 8th 2013 at 06:14 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,619
    Thanks
    592

    Re: linear independence: Eigenvectors/values

    Hey bluelavae.

    Hint: What is the eigenvalue equation (Remember Mx = λx)?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2012
    From
    Melbourne
    Posts
    6

    Re: linear independence: Eigenvectors/values

    Hopefully this doesn't come off really dumb:

    = λ(av + bw)
    = λ(av) + λ(bw)
    = 2λ + av + bw

    ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,382
    Thanks
    749

    Re: linear independence: Eigenvectors/values

    it's kind of on the right track, but still a miss.

    first, we use the linearity of M, to write:

    M(av + bw) = aM(v) + bM(w).

    next, we use the fact that v,w are eigenvectors with eigenvalues λ,μ to write:

    aM(v) + bM(w) = a(λv) + b(μw) <--this expression does not involve M.

    we can also write it as 6a(μv) + b(μw), given that λ = 6μ.

    now, suppose av + bw = 0. clearly, M(0) = 0, so since M(av + bw) = 6a(μv) + b(μw), we have 6a(μv) + b(μw) = 0. since μ ≠ 0, we can multiply this by 1/μ to get:

    6av + bw = 0. together with av + bw = 0, this gives us 5av = 0. since v ≠ 0 (it is an eigenvector, and eigenvectors are non-zero), we have 5a = 0, that is: a = 0. b = 0 follows quickly as a consequence, so {v,w} is a linearly independent set.

    EDIT: if we just stipulate that λ ≠ μ, we have:

    av + bw = 0
    λ(av) + μ(bw) = 0

    equivalently:

    λ(av) + λ(bw) = 0
    λ(av) + μ(bw) = 0

    thus (λ-μ)(bw) = 0, so that either λ-μ = 0, or bw = 0. since λ ≠ μ, λ-μ ≠ 0, so we must have bw = 0, so b = 0, and thus a = 0.
    Last edited by Deveno; February 8th 2013 at 07:19 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2012
    From
    Melbourne
    Posts
    6

    Re: linear independence: Eigenvectors/values

    Sorry, It was meant to come out λ ≠ , rather than λ = 6 .

    The second part looks a bit confusing.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    633
    Thanks
    255

    Re: linear independence: Eigenvectors/values

    I didn't quite follow Deveno's proof. Why is \lambda=6\mu?

    For the linear independence, suppose xv+yw=0 with x and y in the underlying field (R?), then \lambda(xv)+\mu(yv)=0

    Set c=xv and d=yw, then the equations become c+d=0 and \lambda c+\mu d=0. Multiply the first equation by \lambda and subtract from the second. We get (\mu-\lambda)d=0 and then d=0, and then also c=0. So xv=0 and yw=0. If x\ne0, we get v=0, but an eigen vector by definition is not 0. Thus x=0 and similarly y=0.

    Of course this generalizes to n eigenvectors with all different eigenvalues. If you know about the van der Monde matrix, it's easy to prove.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,382
    Thanks
    749

    Re: linear independence: Eigenvectors/values

    his post was edited after i posted my response.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Eigenvectors/values Question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 15th 2011, 01:13 PM
  2. Linear Algebra: Linear Independence question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 3rd 2011, 05:28 AM
  3. Eigenvectors and Values for T(A)=A^t
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: May 13th 2010, 03:43 PM
  4. Eigenvectors/values
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 30th 2010, 02:52 AM
  5. Linear Transformations and Linear Independence
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 6th 2008, 07:36 PM

Search Tags


/mathhelpforum @mathhelpforum