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Eigenvalues of A transpose A

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Let A be an m x n matrix with rank(A) = m < n. As far as the eigenvalues of [itex]A^{T}A[/itex] is concerned we can say that...

    2. Relevant equations

    3. The attempt at a solution

    If eigenvalues exist, then

    [itex]A^{T}A[/itex]x = λx where x ≠ 0.

    The only thing I think I can show is that 0 is an eigenvalue:

    If 0 is an eigenvalue for [itex]A^{T}A[/itex] then

    [itex]A^{T}A[/itex]x = (0)x where x ≠ 0.

    N(A) ≠ {0}, so Ax = 0 where x ≠ 0.

    Therefore [itex]A^{T}(Ax) = 0[/itex] where x ≠ 0. So λ = 0 is an eigenvalue for [itex]A^{T}A[/itex].

    Is there anything else that can be said about the eigenvalues for this matrix?
  2. jcsd
  3. Dec 7, 2011 #2

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    Did you hear about the spectral theorem for symmetric matrices?
  4. Dec 7, 2011 #3
    No, that wasn't covered in my course so I suppose that's not what the professor is looking for. Is it relatively ea
  5. Dec 7, 2011 #4


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    If A^T*A*x=lambda*x what happens if you multiply both sides on the left by x^T? No, you don't need the spectral theorem.
  6. Dec 8, 2011 #5

    I like Serena

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    Spectral theorem says that a symmetric matrix is diagonalizable.
    In particular, a real nxn symmetric matrix has n real eigenvalues.
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