EigenValues & EigenVectors proofs

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Homework Help Overview

The discussion revolves around eigenvalues and eigenvectors, specifically focusing on properties and proofs related to square matrices and their transposes, as well as the relationships between products of matrices and their inverses. Participants are exploring various mathematical assertions and proofs concerning these concepts.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to prove relationships between eigenvalues of matrices and their inverses, as well as the eigenvalues of transposed matrices. Questions about the validity of certain statements and the conditions under which they hold are raised. Some participants express uncertainty about specific proofs and seek clarification on the properties of eigenvalues.

Discussion Status

The discussion is ongoing, with various participants contributing different perspectives and approaches. Some have provided partial proofs or hints, while others have questioned the assumptions made in the original problems. There is a mix of attempts to clarify concepts and address misunderstandings, but no consensus has been reached on the proofs themselves.

Contextual Notes

Participants note that certain conditions, such as the non-singularity of matrices and the completeness of eigenvectors, are crucial for the validity of the discussed proofs. There is also mention of homework rules and the appropriateness of the questions being asked in the forum context.

  • #31
morphism said:
For 3, it's not too hard to prove that |det(A)| is the absolute value of the constant term of the characteristic polynomial of A. Why does this help?

Defennder said:
How do you start to prove that? Evaluating determinants by co-factor expansions for nxn matrices become very messy.
As said the first time you posted this you don't have to do any evaluation or computation. If \lambda_1, \lambda_2, ..., \lambda_n are are the eigenvalues of A, then its characteristic polynomial is (x-\lambda_1)(x- \lambda_2)\cdot\cdot\cdot (x- \lambda_n). Multiplying that out, the constant term is clearly \lambda_1\lambda_2\cdot\cdot\cdot\lambda_n|. If you have set up the characteristic equation as det(A- \lambda I)= 0, as I learned, it is obvious that taking [/itex]\lambda= 0[/itex] gives just the determinant. If, as some people learn, you set it up as det(\lambda I- A) it gives the negative of the determinant.

In either case, it is correct to say that the absolute value of the determinant is equal to the absolute value of the constant term of the characteristic equation.
 

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