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Homework Help: Eigenvectors & subspace spanning

  1. Jun 25, 2010 #1
    The question is at the end of a chapter on spanning vector spaces.

    1. The problem statement, all variables and given/known data
    Let P denote an invertible n x n matrix.
    If [tex]\lambda[/tex] is a number, show that

    [tex]E_{\lambda}(PAP^{-1}) = \left\{PX | X\;is\;in\;E_{\lambda}(A)\right\}[/tex]

    for each n x n matrix A. [Here [tex]E_{\lambda}(A)}[/tex] is the set of eigenvectors of A.]

    2. Relevant equations


    3. The attempt at a solution

    I'm having trouble understanding what the equality means, or how to read it.
    The left side looks to be the eigenvectors of a diagonal matrix, which I think are always the columns of an identity matrix.

    On the right side, I assume the matrix P would be the eigenvectors of A, so I think it's the span of the products of P multiplied by each of it's columns.

    I don't know what to make of those ideas, so I think I'm either missing something about eigenvectors or spanning (very possible), or reading the question wrong.


    Thanks
     
  2. jcsd
  3. Jun 25, 2010 #2

    vela

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    The matrix PAP-1 is diagonal if the columns of P are the eigenvectors of A, but you don't know anything about P other than it's invertible.

    Let B=PAP-1. What the question is asking you to show is that the eigenvectors of B are of the form Px and that Px is an eigenvector of B, where x is an eigenvector of A.
     
  4. Jun 26, 2010 #3
    Thanks, that helps a lot.

    Does this makes sense as an answer?


    Eigenvalues of A can be written:
    [tex](\lambda I -A)X=0[/tex]

    So the right side of the original equation is:
    [tex]P(\lambda I -A)X=0[/tex]

    Move the P inside:
    [tex](\lambda P -PA)X=0[/tex]

    Multiply by I:
    [tex](\lambda P -PA)P^{-1}PX=0[/tex]

    Move P inverse inside:
    [tex](\lambda PP^{-1} -PAP^{-1})PX=0[/tex]

    Eigenvalues of PAP-1 are of the form PX:
    [tex](\lambda I -PAP^{-1})PX=0[/tex]
     
  5. Jun 26, 2010 #4

    Office_Shredder

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    Except they're not eigenvalues, they're eigenvectors.
     
  6. Jun 26, 2010 #5
    Oh, oops. I mean to write eigenvectors.

    Just to be clear, does the method makes sense if I had written the word eigenvectors instead of eigenvalues?
     
  7. Jun 26, 2010 #6

    HallsofIvy

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    More correctly, "if X is an eigenvector of A, with eigenvalue [itex]\lambda[/itex], then
    [tex](\lambda I- A)X= 0[/tex]"

     
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