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Thread: Eigenvector problem

  1. #1
    Senior Member
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    Eigenvector problem

    $\displaystyle \begin{bmatrix}
    -5 & 3\\
    6 & -2
    \end{bmatrix}$

    $\displaystyle \lambda = 1$ and $\displaystyle -8$

    I could find the eigenvector when $\displaystyle \lambda = 1$

    but how do i find the eigenvector when $\displaystyle \lambda = 8$?

    I came up with $\displaystyle \begin{bmatrix}
    1 & -3/13 \\
    0 & 56/39
    \end{bmatrix}$

    but the answer is $\displaystyle \frac{1}{\sqrt{2}}$ and $\displaystyle -\frac{1}{\sqrt{2}}$
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  2. #2
    MHF Contributor

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    Re: Eigenvector problem

    As you say in your first line, the eigenvalues are 1 and -8 so there is no eigenvector for $\displaystyle \lambda= 8$. Was that a typo?

    By definition, if $\displaystyle \begin{bmatrix}x \\ y\end{bmatrix}$ is an eigenvector corresponding to eigenvalue -8, then we must have
    $\displaystyle \begin{bmatrix}-5 & 3 \\ 6 & -2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-5x+ 3y \\ 6x- 2y\end{bmatrix}= \begin{bmatrix}-8x \\ -8y\end{bmatrix}$

    That is equivalent to the two equations -5x+ 3y= -8x and 6x- 2y= -8y which then give 3y= -3x and 6x= -6y both of which are equivalent to y= -x. Any vector of the form $\displaystyle \begin{bmatrix}x \\ -x \end{bmatrix}$, which, of course, includes $\displaystyle \begin{bmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{bmatrix}$, is an eigenvector corresponding to eigenvalue -8.
    Last edited by HallsofIvy; Jan 8th 2012 at 02:52 AM.
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  3. #3
    Senior Member
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    Re: Eigenvector problem

    Quote Originally Posted by HallsofIvy View Post
    As you say in your first line, the eigenvalues are 1 and -8 so there is no eigenvector for $\displaystyle \lambda= 8$. Was that a typo?

    By definition, if $\displaystyle \begin{bmatrix}x \\ y\end{bmatrix}$ is an eigenvector corresponding to eigenvalue -8, then we must have
    $\displaystyle \begin{bmatrix}-5 & 3 \\ 6 & -2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-5x+ 3y \\ 6x- 2y\end{bmatrix}= \begin{bmatrix}-8x \\ -8y\end{bmatrix}$

    That is equivalent to the two equations -5x+ 3y= -8x and 6x- 2y= -8y which then give 3y= -3x and 6x= -6y both of which are equivalent to y= -x. Any vector of the form $\displaystyle \begin{bmatrix}x \\ -x \end{bmatrix}$, which, of course, includes $\displaystyle \begin{bmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{bmatrix}$, is an eigenvector corresponding to eigenvalue -8.
    yes that was a typo.
    thanks for your answer.
    made it clear.
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