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Prove AB and BA have the same characteristic polynomial

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that for any square matrices of the same size, A, B, that AB and BA have the same characteristic polynomial.


    2. Relevant equations



    3. The attempt at a solution

    I understand how to do this if either A or B is invertible, since they would be similar then. I saw a proof that states to take [itex]A = A + \epsilon I[/itex], so that [itex]det(A+\epsilon I) \neq 0[/itex], and then we have,

    [tex] det \left((A+\epsilon I) B - \lambda I \right) = det \left((B(A+\epsilon I) - \lambda I\right)[/tex]

    Since then we can use the similarity argument. We then take [itex]\epsilon\rightarrow 0[/itex].

    I'm wondering if someone could please provide an accessible, but sound explanation of why we are allowed to use this [itex]\epsilon \rightarrow 0 [/itex] argument.

    Thanks!
     
  2. jcsd
  3. Aug 20, 2012 #2
    Can you see that the entities on either side of the equation are multivariate polynomials with respect to epsilon and lambda?
     
  4. Aug 20, 2012 #3
    Hmm, somewhat. Could you please provide greater insight?
     
  5. Aug 20, 2012 #4
    "Somewhat" is not good. If you really don't see it, take a simple 2x2 case and work it out.

    If you do see that, then what do you know of continuity of polynomials? What is [itex]\lim_{\epsilon \rightarrow 0} P(\epsilon, \lambda)[/itex], if [itex]P(\epsilon, \lambda)[/itex] is a polynomial?
     
  6. Aug 21, 2012 #5

    Zondrina

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    Homework Helper

    Hmm are you sure that's the right question? A would need some inverse C such that :

    C-1AC = B

    Then you could start by saying :

    pB(λ) = det(B-λIn)

    Then you should use what you know about B and the identity matrix In to massage the expression into pA(λ) [ Hint : Think about inverses and how they work! ]
     
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