Showing a 6x6 matrix has at least one positive eigenvalue

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SUMMARY

A 6x6 matrix A with a negative determinant guarantees at least one positive eigenvalue. The characteristic polynomial of A, expressed as (-λ)^6 + (tr A)(-λ)^5 + ... + det A, indicates that the polynomial's behavior at λ=0 and as λ approaches positive infinity is crucial. Specifically, the polynomial evaluates to det A at λ=0, which is negative, and approaches positive infinity as λ increases, confirming the existence of at least one positive root due to the Intermediate Value Theorem.

PREREQUISITES
  • Understanding of characteristic polynomials
  • Knowledge of eigenvalues and eigenvectors
  • Familiarity with the properties of determinants
  • Basic concepts of the Intermediate Value Theorem
NEXT STEPS
  • Study the properties of characteristic polynomials in linear algebra
  • Learn about the relationship between determinants and eigenvalues
  • Explore the Intermediate Value Theorem in mathematical analysis
  • Investigate the implications of matrix trace on eigenvalue distributions
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Students studying linear algebra, mathematicians exploring eigenvalue theory, and educators teaching matrix properties.

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Homework Statement



Show that if a 6x6 matrix A has a negative determinant, then A has at least one positive eigenvalue. Hint: Sketch the graph for the characteristic polynomial of A.

Homework Equations


Characteristic polynomial: (-\lambda)^n + (\text{tr}A)(-\lambda)^{n-1} + ... \text{det} A


The Attempt at a Solution


I'm not really sure what to do at all. I know that the characteristic polynomial for a 6x6 matrix is going to be proportional to \lambda ^6, and shifted by det(A), and that the roots of the polynomial are going to be the eigenvalues...But I don't see how this shows there will be at least one positive eigenvalue. Can anyone point me in the right direction here?
 
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What is the value of the polynomial at λ=0? For large positive λ?
 

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