Results 1 to 5 of 5

Thread: Eigenvalues and Eigenvectors

  1. #1
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    Eigenvalues and Eigenvectors

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

    $\displaystyle 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by mybrohshi5 View Post
    Given that $\displaystyle V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix} $ are Eigenvectors of the matrix

    $\displaystyle 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

    $\displaystyle |A - \lambda I| = 0$

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

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

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

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

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

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

    $\displaystyle \lambda = -3$ or $\displaystyle \lambda = -5$.
    Last edited by Prove It; Apr 1st 2010 at 09:37 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by mybrohshi5 View Post
    Given that $\displaystyle V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix} $ are Eigenvectors of the matrix

    $\displaystyle 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 $\displaystyle I$ is your identity matrix, and $\displaystyle \lambda$ is your eigenvalue, then you have to find the matrix $\displaystyle A - {\lambda}I$ and find out $\displaystyle \lambda$ by equating the determinant of the matrix $\displaystyle |A - {\lambda}I| = 0$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Alternatively, you use the property that if

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

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

    You have been given a matrix $\displaystyle A$ and two eigenvectors.


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

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

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

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

    So that means $\displaystyle \lambda = -5$.


    Do the same for the other eigenvector.

    Note that I've edited and fixed my other post too.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230
    Thank you Prove It and harish21 for such fast responses i really appreciate it
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Eigenvalues and Eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 9th 2011, 03:57 AM
  2. Eigenvectors/Eigenvalues help!
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Nov 16th 2008, 07:06 AM
  3. eigenvectors and eigenvalues
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Feb 6th 2008, 06:40 PM
  4. Eigenvalues and Eigenvectors, PLEASE
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Jan 23rd 2008, 01:27 AM
  5. Eigenvalues/Eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Dec 11th 2007, 06:13 PM

Search Tags


/mathhelpforum @mathhelpforum