EigenValues & EigenVectors proofs

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SUMMARY

This discussion focuses on proving properties related to eigenvalues and eigenvectors of matrices, specifically addressing four key questions. The first proof establishes that if λ is an eigenvalue of matrix A, then 1/λ is an eigenvalue of A^T, contingent on λ being non-zero. The second proof demonstrates that square matrices A and A^T share the same eigenvalues. The third question confirms that the absolute value of the determinant of A equals the product of its eigenvalues. Lastly, the fourth question shows that nonsingular matrices AB^-1 and B^-1A possess the same eigenvalues, relying on their characteristic polynomials.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors
  • Familiarity with matrix operations and properties
  • Knowledge of determinants and characteristic polynomials
  • Basic linear algebra concepts, including diagonalization
NEXT STEPS
  • Study the proof of the relationship between eigenvalues and the inverse of a matrix
  • Learn about the properties of the transpose of a matrix in relation to eigenvalues
  • Explore the concept of characteristic polynomials and their role in determining eigenvalues
  • Investigate the conditions under which a matrix is diagonalizable
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators seeking to deepen their understanding of eigenvalue properties and proofs.

  • #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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