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Linear Algebra and Eigenvalues

  1. Nov 12, 2013 #1
    Suppose A is a diagonlizable nxn matrix where 1 and -1 are the only eigenvalues (algebraic multiplicity is not given). Compute A^2.

    The only thing I could think to do with this question is set A=PD(P^-1) (definition of a diagonalizable matrix) and then A^2=(PD(P^-1))(PD(P^-1))=P(D^2)(P^-1)

    This is how I left it on the test but I am sure this isn't right. How can you solve this without having the original matrix A or the algebraic multiplicity of the eigenvalues?
     
  2. jcsd
  3. Nov 12, 2013 #2

    Mark44

    Staff: Mentor

    Since A is diagonalizable, and its only eigenvalues are 1 and -1, then what form must D take? Even more to the point, what does D2 have to be?
     
  4. Nov 12, 2013 #3
    But if the multiplicity of either eigenvalue is more than one, D changes form. Both P and D do. That's where I got stuck.
     
  5. Nov 12, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Of course D changes form. D^2 doesn't. What is it? Once you figure out what it is, P won't matter.
     
  6. Nov 12, 2013 #5
    Ah I see it now. Thanks to all who replied.
     
  7. Nov 13, 2013 #6

    Mark44

    Staff: Mentor

    That would be all two of us: Dick and myself...
     
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