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What are the simplest way to show that AB and BA have the same characteristic polyno?

  1. Jun 11, 2011 #1

    td21

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  2. jcsd
  3. Jun 11, 2011 #2
    Re: what are the simplest way to show that AB and BA have the same characteristic pol

    If you're happy that Det(AB)=Det(BA), then with I the identity matrix,
    the characteristic polynomial is

    p(x) = Det(AB-xI) = Det(IAB-xI2) = Det(B-1BAB-xB-1BI)
    = Det(BABB-1-xBIB-1) = Det(BA-xI)
     
  4. Jun 11, 2011 #3
    Re: what are the simplest way to show that AB and BA have the same characteristic pol

    If A is invertible, than it is evident that AB and BA are similar matrices, therefore they have the same characteristic polynomial. Otherwise, we notice that the equation in lambda, det(A-lambda I)=0 has finitely many solutions. We can take epsilon such that, for all lambda 0<|lambda|<epsilon, A-lambda I is invertible. Therefore, (A-lambda I)B and B(A-lambda I) must have the same characteristic equation. Also, det(xI-(A-lambda I)B)=det(xI-B(A-lambda I)). For fixed x, each side of this equation is a polynomial in lambda, hence it is continuous. We can take lambda->0 and we are done.
     
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