Composition of endomorphisms have same eigenvalues

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SUMMARY

The discussion focuses on proving that two endomorphisms, ψ and φ, on a vector space V over a field K, yield the same eigenvalues for the compositions ψφ and φψ. The key argument involves analyzing the cases where the eigenvalue λ is zero and where λ is non-zero. It is established that if λ ≠ 0, one can utilize the relationship between eigenvectors and the properties of determinants to demonstrate the equivalence of eigenvalues. For the case where λ = 0, it is concluded that if φψ is singular, then ψφ must also be singular, thus confirming the eigenvalue equivalence.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically endomorphisms.
  • Familiarity with eigenvalues and eigenvectors in vector spaces.
  • Knowledge of determinants and their properties in relation to singular matrices.
  • Basic comprehension of fields in mathematics, particularly in the context of vector spaces.
NEXT STEPS
  • Study the properties of eigenvalues and eigenvectors in linear transformations.
  • Learn about the relationship between determinants and singular matrices in linear algebra.
  • Explore the implications of similar endomorphisms and their eigenvalue characteristics.
  • Investigate the role of linear homomorphisms in vector spaces and their impact on eigenvalues.
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, eigenvalue problems, and vector space theory. This discussion is beneficial for anyone looking to deepen their understanding of endomorphisms and their properties.

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Homework Statement



For two endomorphisms ψ and φ on a vector space V over a field K, show that ψφ and φψ have the same eigenvalues. "Hint: consider the cases λ=0 and λ≠0 separately."



The Attempt at a Solution



I know that similar endomorphisms (φ and ψφ(ψ^-1)) have the same eigenvalues, so I have tried manipulating that expression with various choices for φ and ψ, but no luck. Other than that I just need a little help getting started
 
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if this is a linear? homomorphism and the values commute in a field...

[edit] working on problem, i will assume not linear...

[/2 edit] okay i see, you think will need the fact that the inverse maps have inverse eigenvalues...
 
Last edited:
hint: for λ ≠ 0, let v be an eigenvector of φψ, and consider ψφψ(v).

this argument doesn't work if the eigenvalue is 0 (why?).

all is not lost, however. note if 0 is an eigenvalue of φψ, this means φψ is singular.

all you need to do is show that ψφ is likewise singular (hint: determinants).
 

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