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Thread: Find a basis for the eigenspace

  1. #1
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    Find a basis for the eigenspace

    B = 3 -4 4 eigenvalue = -1

    -2 1 -2

    -4 4 -5


    Find a basis for the eigenspace corresponding to the eigenvalue -1.


    So far I have:
    -1-3 -4 4
    -2 -1-1 -2
    -4 4 -1--5

    -4 -4 4 (divde by row 2)
    -2 -2 -2
    -4 4 4 (divide by row 2)

    2 2 -2
    -2 -2 -2
    2 -2 -2

    I get here and this is where I get stuck.

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  2. #2
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    Quote Originally Posted by Arita View Post
    B = 3 -4 4 eigenvalue = -1

    -2 1 -2

    -4 4 -5


    Find a basis for the eigenspace corresponding to the eigenvalue -1.


    So far I have:
    -1-3 -4 4
    -2 -1-1 -2
    -4 4 -1--5

    -4 -4 4 (divde by row 2)
    -2 -2 -2
    -4 4 4 (divide by row 2)

    2 2 -2
    -2 -2 -2
    2 -2 -2

    I get here and this is where I get stuck.

    Is B the eigenvector corresponding to the eigenvalue of -1? What was the original matrix?
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  3. #3
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    B is the matrix.

    The original matrix is:

    3 -4 4
    -2 1 -2
    -4 4 -5

    I then went to (-4, -4, 4) (-2, -2, -2)(-4, 4, 4) then reduced to (-1, -1, 1) (-2,-2,-2) (-4, 4, 4) to finally reducing to
    1 -1 1
    0 0 0
    0 0 0

    But how do I get the eigenvector
    Last edited by Arita; Jan 14th 2011 at 09:22 PM.
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  4. #4
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    Quote Originally Posted by Arita View Post
    B is the matrix.

    The original matrix is:

    3 -4 4
    -2 1 -2
    -4 4 -5
    \displaystyle\begin{vmatrix}3-\lambda&-4&4\\-2&1-\lambda&-2\\-4&4&-5-\lambda\end{vmatrix}=(3-\lambda)[-(1-\lambda)(-5-\lambda)+8]=-(\lambda+1)(\lambda^2-1)=0

    Eigenvalues

    \lambda_1=\lambda_2=-1, \ \lambda_3=1

    \displaystyle\lambda_1=\lambda_2=-1, \ \ \begin{bmatrix}4&-4&4\\-2&2&-2\\-4&1&-4\end{bmatrix}\Rightarrow\mbox{rref}=\begin{bmatri  x}1&-1&1\\0&0&0\\0&0&0\end{bmatrix}

    \displaystyle x_1=x_2-x_3\Rightarrow\begin{bmatrix}x_2-x_3\\x_2\\x_3\end{bmatrix}\Rightarrow x_2\begin{bmatrix}1\\1\\0\end{bmatrix}+x_3\begin{b  matrix}-1\\0\\1\end{bmatrix}
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  5. #5
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    Just so I understand this correctly, there are to eigenvectors for the eigenvalue of -1?
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  6. #6
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    Quote Originally Posted by Arita View Post
    Just so I understand this correctly, there are to eigenvectors for the eigenvalue of -1?
    Yes, the eigenvectors for -1 are

    \displaystyle \left\{\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{  bmatrix}-1\\0\\1\end{bmatrix}\right\}
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  7. #7
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    Thanks so much for the help!
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  8. #8
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    Quote Originally Posted by Arita View Post
    Thanks so much for the help!
    There is also an eigenvector associate with lambda equals one too. Just plug it in to

    \displaystyle<br />
\displaystyle\begin{vmatrix}3-\lambda&-4&4\\-2&1-\lambda&-2\\-4&4&-5-\lambda\end{vmatrix}

    and find the reduced row echelon form.
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