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Linear Algebra Proof- Pleaseeeee help!

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A be an n x n matrix with eigenvalue [tex]\lambda[/tex]. Prove that [tex]\lambda[/tex]^2 is an eigenvalue of A^2 and that if v is an eigenvector of A, then v is also an eigenvector for A^2.

    2. Relevant equations


    3. The attempt at a solution
    (A*A)V=([tex]\lambda[/tex] * [tex]\lambda[/tex])v
    so then v will be an eigenvector to A^2 when [tex]\lambda[/tex]^2
  2. jcsd
  3. Apr 29, 2010 #2
    Seems you have the right track, although you could be a little more precise:

    Let v be an eigenvector of the nxn matrix A.

    We have Av = [tex]\lambda[/tex]v.
    Then A2v = A(Av) = A([tex]\lambda[/tex]v) = [tex]\lambda (\lambda[/tex]v). = [tex]\lambda^2[/tex]v.

    Hence [tex]\lambda^2[/tex] is an eigenvalue of A2 and v is an eigenvector corresponding to [tex]\lambda^2[/tex].
  4. Apr 29, 2010 #3
    thank you but how come A([tex]\lambda[/tex]v) can = [tex]\lambda[/tex]([tex]\lambda[/tex]v?
  5. Apr 29, 2010 #4
    Any nonzero scalar multiple of an eigenvector is also an eigenvector, and is associated with the same eigenvalue.
  6. Apr 30, 2010 #5


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    Homework Helper

    A(λv) = (by the linearity of A!) = λA(v) = λ λv = λ^2 v.
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