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Property of characteristic polynomial

  1. Feb 18, 2007 #1

    Why is it that if A is m×n-matrix and B is n×m matrices such that m<n, then AB is m×m and BA is n×n matrix. Then the following is true:

    pAB(t) = t^(m-n)*pBA(t)

    where pAB(t) and pBA(t) are characteristic polynomials of AB and BA

  2. jcsd
  3. Feb 18, 2007 #2


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    What you really should be writing is:

    pBA(t) = tn-mpAB(t).

    Why is this true? Well you'll have to fill in the details, but here are some facts to consider:

    If [itex]\lambda[/itex] is a root of pAB(t), then it is an eigenvalue of AB, so there is a non-zero vector v such that ABv = [itex]\lambda[/itex]v. BA(Bv) = B(ABv) = B([itex]\lambda[/itex]v) = [itex]\lambda[/itex]Bv, so if Bv is non-zero, it is an eigenvector of BA with the same eigenvalue that AB had, [itex]\lambda[/itex]. In this case, then, [itex]\lambda[/itex] would also be a root of BA's characteristic polynomial.

    Also, since rank(A) is at most m, and likewise for B, rank(AB) and rank(BA) are at most m. But BA is nxn, so it has 0 as an eigenvalue with multiplicity at least n-m, accounting for the tn-m.

    These are vague ideas, hopefully they lead you to a proof.
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