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Eigenvalues, eigenvectors question

  1. May 26, 2005 #1
    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.
  2. jcsd
  3. May 26, 2005 #2


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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
  4. May 26, 2005 #3
    Man... I had linear algebra last semester, but i think our notation is different, what is meant by p(A)?
  5. May 26, 2005 #4


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    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.
  6. May 27, 2005 #5

    matt grime

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    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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