# Eigenvalue problem

1. Mar 10, 2007

### Kolahal Bhattacharya

1. The problem statement, all variables and given/known data

Two square matrices A and B of the same size do not commute.Prove that AB and BA has the same set of eigenvalues.

I did in the following way:Please check if I am correct.
Consider: det(AB-yI)*det(A) where y represents eigenvalues and
I represents unit matrix
=det[(AB-yI)A]
=det[(AB)A-(yI)A]
=det[A(BA)-A(yI)]
=det(A)*det(BA-yI)
det(A) is not equal to zero,in general.
So,if det(AB-yI)=0,det(BA-yI)=0 also.
hence, conclusion.

2. Mar 10, 2007

### Dick

That's basically it, but the argument is dubious. det(A) certainly could be zero. Try framing it this way. Let L be an eigenvalue of AB. Then ABx=Lx for some x. Act on both sides with B and conclude Bx is an eigenvector with eigenvalue L of BA. So if L is an eigenvalue of AB, it's a eigenvalue of BA.

3. Mar 10, 2007

### Kolahal Bhattacharya

Oh!It's fantastic.I salute you whole-heartedly.