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Math Help - Eigenvalues and Eigenvectors

  1. #1
    Member mybrohshi5's Avatar
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    Eigenvalues and Eigenvectors

    Given that V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix} are Eigenvectors of the matrix

     A = \begin{bmatrix}-21&48\\-6&13\end{bmatrix}

    Find the Eigenvalues


    I know how to find eigenvalues and using those i know how to find eigenvectors, but i am not sure how to find eigenvalues if eigenvectors and a matrix are given.

    Thank you for any help
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  2. #2
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    Quote Originally Posted by mybrohshi5 View Post
    Given that V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix} are Eigenvectors of the matrix

     A = \begin{bmatrix}-21&48\\-6&13\end{bmatrix}

    Find the Eigenvalues


    I know how to find eigenvalues and using those i know how to find eigenvectors, but i am not sure how to find eigenvalues if eigenvectors and a matrix are given.

    Thank you for any help
    You need to solve

    |A - \lambda I| = 0

    \left|\left[\begin{matrix} - 21 & 48 \\ -6 & 13 \end{matrix}\right] - \left[\begin{matrix}\lambda & 0 \\ 0 & \lambda\end{matrix}\right]\right| = 0

    \left|\begin{matrix}-21 - \lambda & 48\\ -6 & 13 - \lambda\end{matrix}\right| = 0

    (-21 - \lambda)(13 - \lambda) - (48)(-6) = 0

    -273 + 21\lambda - 13\lambda + \lambda^2 + 288 = 0

    \lambda^2 + 8\lambda + 15 = 0

    (\lambda + 3)(\lambda + 5) = 0

    \lambda = -3 or \lambda = -5.
    Last edited by Prove It; April 1st 2010 at 09:37 PM.
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by mybrohshi5 View Post
    Given that V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix} are Eigenvectors of the matrix

     A = \begin{bmatrix}-21&48\\-6&13\end{bmatrix}

    Find the Eigenvalues


    I know how to find eigenvalues and using those i know how to find eigenvectors, but i am not sure how to find eigenvalues if eigenvectors and a matrix are given.

    Thank you for any help
    If I is your identity matrix, and \lambda is your eigenvalue, then you have to find the matrix A - {\lambda}I and find out \lambda by equating the determinant of the matrix |A - {\lambda}I| = 0
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  4. #4
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    Alternatively, you use the property that if

    A\mathbf{x} = \lambda \mathbf{x}

    then \mathbf{x} is the eigenvector and \lambda is the corresponding eigenvalue.

    You have been given a matrix A and two eigenvectors.


    So \left[\begin{matrix}-21 & 48 \\ -6 & 13\end{matrix}\right] \left[\begin{matrix}3 \\ 1\end{matrix}\right] = \lambda \left[\begin{matrix}3 \\ 1\end{matrix}\right]

    \left[\begin{matrix}-21 \cdot 3 + 48 \cdot 1 \\ -6 \cdot 3 + 13 \cdot 1\end{matrix}\right] = \left[\begin{matrix}3\lambda \\ \lambda\end{matrix}\right]

    \left[\begin{matrix}-63 + 48 \\ -18 + 13\end{matrix}\right] = \left[\begin{matrix}3\lambda \\ \lambda\end{matrix}\right]

    \left[\begin{matrix}-15 \\ -5 \end{matrix}\right] = \left[\begin{matrix}3\lambda \\ \lambda\end{matrix}\right]

    So that means \lambda = -5.


    Do the same for the other eigenvector.

    Note that I've edited and fixed my other post too.
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  5. #5
    Member mybrohshi5's Avatar
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    Thank you Prove It and harish21 for such fast responses i really appreciate it
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