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Eigenvalue Proof

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that similar square matricies have the same eigenvalues with the same algebraic multiplicities.

    2. Relevant equations
    C^-1PC=Q



    3. The attempt at a solution
    Am I supposed to show that (P-[tex]\lambda[/tex]I)x=(C^-1PC-[tex]\lambda[/tex]I)x?
     
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  3. Jul 18, 2009 #2

    Dick

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    The eigenvalues are the roots of the characteristic polynomial. Can you show Q=C^(-1)PC and P have the same characteristic polynomial? I.e. det(Q-xI)=det(P-xI)?
     
    Last edited: Jul 18, 2009
  4. Jul 18, 2009 #3
    Well, that's equivalent to C-1PC=Q, isn't it?



    solution:
    What you want to do is to show, that they have the same char polynomial (up to scalarmultiplication). But that's easy since [tex]det(\mathbf{P} -\lambda \mathbf{I})=det(\mathbf{C^{-1}QC} -\lambda \mathbf{C^{-1}C})=det(\mathbf{C^{-1}}(\mathbf{Q} -\lambda\mathbf{I} )\mathbf{C})[/tex]
    Now use det(AB)=det(A)*det(B) and your done.
    cheers
     
  5. Jul 18, 2009 #4

    Dick

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    Hi, 1777-1855. The forum rules prohibit posting complete solutions (i.e. doing the problem for the poster). That's awfully close. Try and give a hint first and see if they can figure out the rest themselves, ok?? It's more fun that way and they learn more.
     
  6. Jul 18, 2009 #5
    So if I prove those two determinants are equal, it necessarily follows that the two matricies have the same eigenvalues and algebraic multiplicities, because the characteristic polynomials are the same?
     
  7. Jul 18, 2009 #6

    Dick

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    Yes, those two determinants are actually polynomials in x, they are the characteristic polynomials. Their roots (and the multiplicities of them) are the eigenvalues and their algebraic multiplicities.
     
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