Proving detA = λ1...λn for Real Eigenvalues

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SUMMARY

The discussion centers on proving that the determinant of an nxn matrix A, which has real eigenvalues λ1, ..., λn, is equal to the product of its eigenvalues, expressed as detA = λ1...λn. Key approaches include utilizing the definition of eigenvalues Av = λv, understanding the relationship between similar matrices and their determinants, and applying the characteristic polynomial det(A-λI). The characteristic polynomial can be factored into (λ1-λ)(λ2-λ)...(λn-λ), allowing for evaluation at λ=0 to derive the determinant.

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Homework Statement


Let A be nxn matrix, suppose n has real eigenvalues,λ1,...,λn, repeated according to multipilicities. Prove that detA = λ1...λn.

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The Attempt at a Solution


I started by applying the definition, Av = λv, where v is an eigenvector. then I just dun know how to keep going.. is there anyone can help me out? or at least give me some hints..
thx..
 
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If matrices A and B are similar, what can you say about their determinants? Use this relationship to help you solve the problem.
 
DH offered an correct approach,but you need to know something about similarity and Jordan normal form or Schur's lemma
Here's another approach: consider the characteristic polynomial det(A-λI),by fundamental theorem of algebra, it can be factorized into (λ1-λ)(λ2-λ)...(λn-λ),then let λ=0 and see what will happen
 

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