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Homework Help: Linear Algebra - Characteristic polynomials and similar matrices question

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    For each matrix A below, let T be the linear operator on R3 thathas matrix A relative to the basis A = {(1,0,0), (1,1,0), (1,1,1)}. Find the algebraic and geometric multiplicities of each eigenvalues, and a basis for each eigenspace.

    a) A = [tex]
    \begin{bmatrix} 8&5&-5\\5&8&-5\\15&15&-12\end{bmatrix}

    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    So I tried to find the eigenvalues normally and turns out that was pretty hard.. So I know that similar matrices have the same eigenvalues, then can I just take the eigenvalues of the matrix [tex]
    \begin{bmatrix} 1&1&1\\0&1&1\\0&0&1\end{bmatrix}

    [/tex]

    since it is similar to A? Or is it similar?
     
  2. jcsd
  3. Feb 17, 2010 #2
    The eigenvalues of similar matrices are the same but the eigenvectors may be different.
     
  4. Feb 17, 2010 #3

    vela

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    The second matrix isn't similar to A. Two matrices A and B are similar if you can write

    [tex]B=P^{-1}AP[/tex]

    for some invertible matrix P. You could use your matrix with the basis-vector columns as P.
     
  5. Feb 17, 2010 #4
    I calculated B but that doesn't seem to make finding eigenvalues any easier..?
     
  6. Feb 17, 2010 #5

    vela

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    You just need to work it out. It's only a 3x3 matrix after all.
     
  7. Feb 17, 2010 #6
    Okay nice it seems it's just because I made a mistake in calculating the inverse of P..
    turns out it's much easier to find the eigenvalues of B.

    ..or not. I don't understand why there is suddenly such a computational question when everything else is hardly as hard..
     
    Last edited: Feb 17, 2010
  8. Feb 18, 2010 #7
    Can I row-reduce a matrix before subtracting lambda and then find the determinant? Or do I have to subtract lambda first?
     
  9. Feb 18, 2010 #8

    vela

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    You have to subtract [itex]\lambda[/itex] first. Think about it. You can reduce any invertible matrix to the identity matrix. If you then subtracted [itex]\lambda[/itex], all the eigenvalues would be 1, which is obviously not the case for every invertible matrix.
     
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