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Diagonalization eigenvalues

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data

    I think my teacher made a mistake in his homework answer. I need to verify this for practice. The answer I got is below. The answer the teacher has is in the pdf.

    2. Relevant equations

    Please refer to attached pdf

    3. The attempt at a solution

    So there is two eigenvalues= 4 and 2
    but the eigenvalue 2 has 2 eigenvectors [-1 1 0]T and [0 0 1]T but my teacher has only one [-1 1 0]T. That's why he says A is not diagonalizable. Do you think it's correct?
     

    Attached Files:

  2. jcsd
  3. Jul 17, 2011 #2
    Re: Diagonalization

    I get the same result as your teacher for [tex] \lambda = 2 [/tex]

    [tex] A= \begin{bmatrix}
    3 & 1 & 0\\
    0 & 2 & 1\\
    1 & 1 & 3
    \end{bmatrix}
    [/tex]

    so for lamda = 2,
    [tex] (A-2I)\vec{v}=\vec{0} [/tex]
    [tex] \begin{bmatrix}
    1 & 1 & 0\\
    0 & 0 & 1\\
    1 & 1 & 1
    \end{bmatrix}
    \begin{bmatrix}
    v_1\\
    v_2\\
    v_3
    \end{bmatrix}
    = \begin{bmatrix}
    0\\
    0\\
    0\\
    \end{bmatrix}
    [/tex]
    I've personally always found it easier to not do row operations here and just jump straight in. From that you get
    [tex] v_3 = 0 [/tex] and [tex] v_1 = -v_2 [/tex]
    so therefore [tex] \vec{v} = \begin{bmatrix}
    1\\
    -1\\
    0
    \end{bmatrix}
    [/tex]

    Since it is a 3x3 matrix, it needs 3 eigenvectors to be diagonalizable.

    Hope this helps.
     
  4. Jul 18, 2011 #3
    Re: Diagonalization

    Hi, thanks for replying. Attached is how I got my vectors, do you think my steps are correct.
     

    Attached Files:

  5. Jul 18, 2011 #4

    vela

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    Staff Emeritus
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    Re: Diagonalization

    No. Why do you think (0, 0, 1) is an eigenvector? You seem to just pull that out of thin air.
     
  6. Jul 18, 2011 #5
    Re: Diagonalization

    You seem to have made a mistake in the step
    [tex] (A-2I)\vec{v}=\vec{0} [/tex]
    Why is your second row 1 0 1 instead of 0 0 1 ?
    Have you said [tex] v_2= \delta [/tex] ? I'm not sure if I have read that correctly. If it is a delta, you can't say that unless the whole row equals zero. i.e.
    [tex] \begin{bmatrix}
    0 & 0 & 0
    \end{bmatrix}
    \begin{bmatrix}
    v_1
    \end{bmatrix}
    = \begin{bmatrix}
    0
    \end{bmatrix}
    [/tex]
     
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