Prove AB & BA Have Same Eigenvalues

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Two square matrices A and B that do not commute can be shown to have the same eigenvalues for the products AB and BA. The initial argument involves using determinants, but concerns arise regarding the potential for det(A) to be zero. A more robust approach is to consider an eigenvalue L of AB, demonstrating that acting on both sides with B leads to an eigenvector of BA with the same eigenvalue L. This establishes that if L is an eigenvalue of AB, it must also be an eigenvalue of BA. The conclusion confirms that AB and BA share the same set of eigenvalues.
Kolahal Bhattacharya
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Homework Statement




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.
 
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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.
 
Oh!It's fantastic.I salute you whole-heartedly.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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