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How do I find the characteristic polynomial

  1. Apr 2, 2006 #1
    "Let T be a the transformation on V = C^3 given by the equation

    T(x)=-y-2z
    T(y)=3x+5y+7z
    T(z)=-2x-3y-4z

    where (x,y,z) denotes the standard basis. Find the eigenvalues of T and the corresponding eigenspaces."

    Is there a way to find the eigenvalues without solving the 3 equations? How do I find the characteristic polynomial of T without resorting to anything related to determinants?
     
  2. jcsd
  3. Apr 2, 2006 #2

    Hurkyl

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    The determinant of the matrix is one of the coefficients of the characteristic polynomial, and is also the product of the eigenvalues.

    So however you do this problem, you have to use a method that is capable of computing the determinant of the matrix.
     
    Last edited: Apr 2, 2006
  4. Apr 3, 2006 #3

    shmoe

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    You can find T^2 and T^3, then look for a linear dependance relation amongst I, T, T^2 and T^3 (this won't be difficult here). All the eigenvalues will appear as roots of this polynomial, as the minimal polynomial will divide whatever relation you find (you may in fact find the minimal polynomial this way).
     
  5. Apr 3, 2006 #4

    matt grime

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    Just row reduce the matrix to triangular form. You can find the eigenvalues in, what 3 operations.
     
  6. Apr 3, 2006 #5
    I was never actually taught HOW to compute the eigenvalue of a matrix other than solving the system of equations. How does one find the characteristic polynomial of a given matrix/transformation?
     
    Last edited: Apr 3, 2006
  7. Apr 3, 2006 #6

    HallsofIvy

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    What's wrong with that method? It certainly works here. (Yes, the characteristic polynomial is cubic but one of the roots is "trivial".)
     
  8. Apr 3, 2006 #7

    HallsofIvy

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    Matt, could you explain that more? In general, the eigenvalues of a reduce matrix are not the same as the eigenvalues of the original matrix.
     
  9. Apr 3, 2006 #8

    shmoe

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    Do you want to find the characteristic polynomial just to find the eigenvalues? This isn't necessary, each eigenvalue will be a root of the minimal polynomial, so finding this is enough to give all the eigenvalues. Even a polynomial with p(T)=0 will help, you know the minimal polynomial then divides p, so all eigenvalues will be roots of p.
     
  10. Apr 3, 2006 #9
    How did you find the characteristic polynomial so fast?
     
  11. Apr 4, 2006 #10

    matt grime

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    Sorry, you can find the determinant, is what I meant to say, so there is no reason not to resort to determinental arguments.
     
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