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Linear Algebra

  1. Feb 22, 2006 #1
    Show that every eigenvalue of A is zero iff A is nilpotent (A^k = 0 for k>=1)
    i m having trouble with going from right to left (left to right i got)

    we know that det A = product of the eignevalues = 0
    when we solve for the eigenvalues and put hte characteristic polynomial = 0
    [tex] det (\lambda I - A) = det (-A) = 0 [/tex]
    but i have a feeling taht i am not allowed to do that last step because the characteristic polynomial need not be equal to zero.

    (Also i do not know the Cayley Hamilton theorem.)

    If A is symmetric show taht every eigenvalue of A is nonnegaitve iff A = B^2 for some symmetric matrix B

    im not even quite sure where to start with this one! Please help

    thank you for any help!
    Last edited: Feb 22, 2006
  2. jcsd
  3. Feb 22, 2006 #2


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    For the first one, what if there were an eigenvalue that were nonzero? There would be an eigenvector v for that eigenvalue. And then what would A * A * A * A * A * ... * v equal?

    For the second one, do you know that if an nxn matrix A is symmetric then A has n eigenvalues? Then simply factor A as PDP^-1, and you should be able to find B. (if you are having trouble with that then think, with A factored into PDP^-1, what is A^2?)
    Last edited: Feb 22, 2006
  4. Feb 22, 2006 #3
    then that A A ... v = 0
    and since the eigenvector is not zero (not sure) then A^ n = 0
  5. Feb 22, 2006 #4


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    No if there were an eigenvector v for a nonzero eigenvalue k then A * A * A * A * ... * v = k * k * k * k * ... * v
  6. Feb 22, 2006 #5
    but here the eignevalue is zero
    so something like taht , the right hand side is zero
    that leaves A A A A A ... v = 0

    what does this lead to , though?
    Last edited: Feb 22, 2006
  7. Feb 22, 2006 #6


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    What I was trying to show you was a proof by contradiction. What you do is, given that A^k = 0 for some k, assume that A has an eigenvalue that is NOT zero. THEN, you try to show that there is a contradiction, so you can conclude A can't have any nonzero eigenvalues.
  8. Feb 22, 2006 #7
    ok for the second one
    [tex] A = PDP^{-1} [/tex]
    [tex] A^2 = PDDP^{-1} [/tex]
    [tex] A^2 = P D^2 P^{-1} [/tex]
    not sure what happens now though
    i tried removing one of As but that just leads back to the first step
  9. Feb 22, 2006 #8


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    But what actually is D^2? Are you familiar with diagonalization? D is a diagonal matrix whose diagonal entries are the eigenvalues of A, and P is a matrix whose columns are the corresponding eigenvectors of A. You can calculate the entries of D^2 easily.

    Don't bother looking for a direct way from A^2 to the formula you are looking for, I was only saying compute A^2 as an example that would give you insight on B.
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