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Finding eigenvectors of general matrix given char. eqn.

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data This is my first post, so forgive me if anything's out of order.
    Assumean operator A satisfies the following equation:
    1+2A-A^2-2A^3=0
    Find the eigenvalues and eigenvectors for A



    2. Relevant equations



    3. The attempt at a solution
    So the eigenvalues are +1,-1, and -1/2. At least those are the values which satisfy the characteristic equation above. What I don't know is how to find the associated eigenvectors (without being given a concrete matrix). Is there a better way to solve this than making up a general 3x3 matrix and muddling through the (A-λI)v=λv equations? Thanks.
     
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  3. Jan 22, 2013 #2

    Dick

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    You can't find the eigenvectors just given the characteristic equation. They depend on the exact matrix A.
     
  4. Jan 22, 2013 #3

    Ray Vickson

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    In general, you cannot say for sure that ##p(x) = (1/2) + x - (1/2) x^2 - x^3## is the characteristic polynomial of the matrix A. You are told that p(A) = 0, so p is an annihilating polynomial of A, but one can imagine there are matrices in which the characteristic polynomial C(x) is a divisor of p(x). For example, if A is a matrix with C(x) = 1-x^2 (that is, with ##A^2 = I##, then we also have p(A) = 0, since ##p(x) = (x + 1/2)C(x)##, so
    [tex] p(A) = \left( A + \frac{1}{2}I \right) C(A) = 0,[/tex] but p is not the characteristic polynomial. In that example, A is a 2×2 matrix, and only +1 and -1 are eigenvalues!

    However, if A is 3×3 then p(x) is the characteristic polynomial, and the eigenvalues are, indeed, ±1 and -1/2.
     
  5. Jan 23, 2013 #4
    Thanks Ray, that's a good point. But the main question is how can we find general eigenvectors given nothing but this equation which the operator satisfies? This question was posed by my physics professor, so he seems convinced that it is indeed possible. Any ideas?
     
  6. Jan 23, 2013 #5

    HallsofIvy

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    Ray's point is the same as Dick's: No, you cannot determine the eigenvectors from the characteristic equation.
     
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