Eigenvalues, eigenvectors question

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Discussion Overview

The discussion revolves around the Cayley-Hamilton theorem and its implications for a 3x3 matrix A with respect to its eigenvalues and eigenvectors. Participants explore whether the linear independence of eigenvectors is necessary for the theorem to hold and discuss the nature of the characteristic polynomial p(t) and its evaluation at the matrix A.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that if {u1, u2, u3} is a linearly independent set of eigenvectors corresponding to eigenvalues y1, y2, y3 of matrix A, then p(A) is the zero matrix, as stated by the Cayley-Hamilton theorem.
  • Another participant claims that the theorem holds true in all cases, suggesting that the linear independence of eigenvectors is not a requirement for the theorem's validity.
  • A participant expresses confusion regarding the notation p(A) and its meaning in the context of the discussion.
  • One participant argues that the special case of diagonalizable matrices implies the general case, indicating that a polynomial vanishing on a dense set of matrices also vanishes on all matrices.
  • Another participant advises against diagonalization and emphasizes that the linear independence of the eigenvectors has not been explicitly utilized in the argument, suggesting that any vector in R^3 can be expressed as a combination of the eigenvectors.

Areas of Agreement / Disagreement

Participants express differing views on whether the linear independence of eigenvectors is necessary for the Cayley-Hamilton theorem to apply. The discussion remains unresolved, with multiple competing perspectives presented.

Contextual Notes

Some participants note that the discussion hinges on the definitions and properties of diagonalizable matrices and the implications of the characteristic polynomial, but these aspects remain unresolved.

jhson114
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1). suppose that y1, y2, y3 are the eigenvalues of a 3 by 3 matrix A, and suppose that u1, u2,u3 are corresponding eigenvectors. Prove that if { u1, u2, u3 } is a linearly independent set and if p(t) is the characteristic polynomial for A, then p(A) is the zero matrix.

I thought cayley-hamilton theorem simply states that if p(t) is the characteristic polynomial for A, then p(A) is the zero matrix. do the eigenvectors have to be learly independent for this to be true? i thought it was true in all cases.
 
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It is true in all cases. The problem is a special case and is thus easier to prove. hint: the matrix is diagnalizable
 
Man... I had linear algebra last semester, but i think our notation is different, what is meant by p(A)?
 
Over the reals, the special case implies the general one, as a polynomial that vanishes on a dense set of matrices (diagonalizable ones) also vanishes on all matrices.
 
Don't diagonalize would be my advice.

I think the question is getting at this:

We know p(t)=(t-y1)(t-y2)(t-y3)

and that u1,u2, and u3 are linearly independent (so far no one has used this fact explicitly).

As they are LI in R^3 they are a basis. so we can write any v in R^3 as a combination of the u1,u2,u3, and p(A) must annihilate the vector since it annihilates each u1,u2,u3 individually.
 

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