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Math Help - diagonalizing a matrix

  1. #1
    Member Jskid's Avatar
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    [RESOLVED]diagonalizing a matrix

    Ortogonally diagonalize the given matrix A, giving the diagonal matrix D and the diagonalizing orthogonal matrix P.



    <br /> <br />
\[A =\left[ {\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\2 & 2 & -1\\ \end{array} } \right]\]Find the eigenvalues using charactreistic polynomial det(\lambdaI_3-A)=(\lambda-3)(\lambda+3)^2For \lambda=-3

    [tex](-3I_3-A)=\left[ {\begin{array}{ccc}
    -2 & -2 & -2 \\
    -2 & -2 & -2 \\

    -2 & -2 & -2 \\
    \end{array} } \right]
    \]

    so this means the eigenvector is  ](-3I_3-A)=\left[ {\begin{array}{ccc}<br />
 1  \\<br />
 1  \\<br /> <br />
 1 \\<br />
\end{array} } \right]<br />



    for \lambda=3

    (3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc}<br />
 4 & -2 & -2  \\<br />
 -2 & 4 & -2  \\<br /> <br />
 -2 & -2 & 4 \\<br />
\end{array} } \right]

    rref(3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc}<br />
 1 & 0 & -1  \\<br />
 0& 1 & -1 \\<br /> <br />
 0 & 0 & 0 \\<br />
\end{array} } \right]<br />
so the eigvenvector for \lambda=3 is (again) ](-3I_3-A)=\left[ {\begin{array}{c}<br />
 r  \\<br />
 r  \\<br /> <br />
 r \\<br />
\end{array} } \right]<br />

    I can't diagonalize a matrix with only one eigenvector, where am I going wrong?
    Last edited by Jskid; March 30th 2011 at 03:37 PM. Reason: resolved
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Jskid View Post
    Orthogonally diagonalize the given matrix A, giving the diagonal matrix D and the diagonalizing orthogonal matrix P.



    \[A =\left[ {\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\2 & 2 & -1\\ \end{array} } \right]\]" alt="

    \[A =\left[ {\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\2 & 2 & -1\\ \end{array} } \right]\]" />Find the eigenvalues using charactreistic polynomial det(\lambdaI_3-A)=(\lambda-3)(\lambda+3)^2For \lambda=-3

    [tex](-3I_3-A)=\left[ {\begin{array}{ccc}
    -2 & -2 & -2 \\
    -2 & -2 & -2 \\

    -2 & -2 & -2 \\
    \end{array} } \right]
    \]

    so this means the eigenvector is  ](-3I_3-A)=\left[ {\begin{array}{ccc} 1 \\
    1 \\

    1 \\
    \end{array} } \right]
    " alt="
    1 \\
    1 \\

    1 \\
    \end{array} } \right]
    " />



    for \lambda=3

    (3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc} 4 & -2 & -2 \\
    -2 & 4 & -2 \\

    -2 & -2 & 4 \\
    \end{array} } \right]" alt="
    4 & -2 & -2 \\
    -2 & 4 & -2 \\

    -2 & -2 & 4 \\
    \end{array} } \right]" />

    rref(3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc} 1 & 0 & -1 \\
    0& 1 & -1 \\

    0 & 0 & 0 \\
    \end{array} } \right]
    " alt="
    1 & 0 & -1 \\
    0& 1 & -1 \\

    0 & 0 & 0 \\
    \end{array} } \right]
    " /> so the eigvenvector for \lambda=3 is (again) ](-3I_3-A)=\left[ {\begin{array}{c} r \\
    r \\

    r \\
    \end{array} } \right]
    " alt="
    r \\
    r \\

    r \\
    \end{array} } \right]
    " />

    I can't diagonalize a matrix with only one eigenvector, where am I going wrong?
    First you Latex in unreadable!

    2nd the matrix admits to eigenvalues \lambda =3 and \lambda =-3

    The repeated eigenvalue \lambda =-3 admits two linearly independent eigenvectors

    Please double check your calculation for \lambda =-3

    I get the two vectors

    \begin{bmatrix}-1 \\ 0 \\ 1 \end{bmatrix} and

    \begin{bmatrix}-1 \\ 1\\ 0 \end{bmatrix}
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  3. #3
    Member Jskid's Avatar
    Joined
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    Quote Originally Posted by TheEmptySet View Post
    First you Latex in unreadable!

    2nd the matrix admits to eigenvalues \lambda =3 and \lambda =-3

    The repeated eigenvalue \lambda =-3 admits two linearly independent eigenvectors

    Please double check your calculation for \lambda =-3

    I get the two vectors

    \begin{bmatrix}-1 \\ 0 \\ 1 \end{bmatrix} and

    \begin{bmatrix}-1 \\ 1\\ 0 \end{bmatrix}
    I have no clue how the Latex got so messed up, thanks for reading it in such a poor state.
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