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Math Help - Linear Algebra - Find a matrix P that diagonalizes A

  1. #1
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    Linear Algebra - Find a matrix P that diagonalizes A

    This question is broken up into 4 parts to help you get to the right answer. Below is where I am so far.

    Find eigenvalues of A:
    A=[2 13
    -1 -12]

    det(lI - A) = 0
    So the equation is l^2-10l - 11=0
    So lambda = -11,1

    b) Find corresponding eigenvectors
    if l=-11
    eigenvector is [-1
    1]
    if l=1
    Eigenvector is [-13
    1]

    c) Find a matrix P that diagonalizes A
    P = [-1 -13
    1 1]

    d) Verify that P^-1 A P is diagonal
    This verification does not work for me, so I assume I did something wrong above? Any idea where I went wrong?
    Last edited by mr fantastic; December 1st 2010 at 03:47 AM.
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  2. #2
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    With a 2x2, you don't have to use the det method.

    The characteristic polynomial is simply \displaystyle \lambda^2-tr(A)\lambda+det(A)\rightarrow \lambda^2+10\lambda-11=(\lambda+11)(\lambda-1)

    Your eigenvalues are correct. Looks correct to me though.
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  3. #3
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    All your work is correct. You may have made a clerical error when doing the multiplication.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by saxophonemaster View Post
    d) Verify that P^-1 A P is diagonal
    This verification does not work for me, so I assume I did something wrong above? Any idea where I went wrong?
    All is right. Verify:

    AP=\begin{pmatrix}{2}&{13}\\{-1}&{-12}\end{pmatrix}\begin{pmatrix}{-1}&{-13}\\{1}&{1}\end{pmatrix}=\begin{pmatrix}{11}&{-13}\\{-11}&{1}\end{pmatrix}

    PD=\begin{pmatrix}{-1}&{-13}\\{1}&{1}\end{pmatrix}\begin{pmatrix}{-11}&{0}\\{0}&{1}\end{pmatrix}=\begin{pmatrix}{11}&  {-13}\\{-11}&{1}\end{pmatrix}

    So, P^{-1}AP=D

    Regards.

    Fernando Revilla
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  5. #5
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    Your eigenvalues and P are correct. What did you get for P^{-1}?
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