Prove AB & BA Have Same Eigenvalues

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SUMMARY

The discussion centers on proving that two square matrices A and B, which do not commute, have the same eigenvalues for the products AB and BA. The initial approach involved determinants, specifically det(AB - yI) and det(BA - yI), to establish the relationship between the eigenvalues. However, a more robust argument was suggested, utilizing the concept of eigenvectors: if L is an eigenvalue of AB, then acting on both sides with B shows that Bx is an eigenvector of BA corresponding to the same eigenvalue L. This definitive proof confirms the eigenvalue equivalence of the two products.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly eigenvalues and eigenvectors.
  • Familiarity with matrix operations and properties, including determinants.
  • Knowledge of non-commutative matrices and their implications in linear transformations.
  • Experience with mathematical proofs and logical reasoning in the context of matrix theory.
NEXT STEPS
  • Study the properties of eigenvalues in non-commuting matrices.
  • Learn about the implications of the determinant in matrix theory, specifically in relation to eigenvalues.
  • Explore the relationship between eigenvectors and linear transformations in greater depth.
  • Investigate additional proofs related to eigenvalue equivalence for different matrix products.
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Mathematicians, students studying linear algebra, and anyone interested in advanced matrix theory and eigenvalue analysis.

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.
 

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