Property of characteristic polynomial

1. Feb 18, 2007

pwhitey86

Hi,

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

thanks

2. Feb 18, 2007

AKG

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 $\lambda$ is a root of pAB(t), then it is an eigenvalue of AB, so there is a non-zero vector v such that ABv = $\lambda$v. BA(Bv) = B(ABv) = B($\lambda$v) = $\lambda$Bv, so if Bv is non-zero, it is an eigenvector of BA with the same eigenvalue that AB had, $\lambda$. In this case, then, $\lambda$ 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.