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Math Help - Eigenvector problem

  1. #1
    Senior Member
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    Eigenvector problem

    \begin{bmatrix}<br />
-5 & 3\\ <br />
6 & -2<br />
\end{bmatrix}

    \lambda = 1 and -8

    I could find the eigenvector when \lambda = 1

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

    I came up with \begin{bmatrix}<br />
1 & -3/13 \\<br />
0 & 56/39<br />
\end{bmatrix}

    but the answer is \frac{1}{\sqrt{2}} and -\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 \lambda= 8. Was that a typo?

    By definition, if \begin{bmatrix}x \\ y\end{bmatrix} is an eigenvector corresponding to eigenvalue -8, then we must have
    \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 \begin{bmatrix}x \\ -x \end{bmatrix}, which, of course, includes \begin{bmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{bmatrix}, is an eigenvector corresponding to eigenvalue -8.
    Last edited by HallsofIvy; January 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 \lambda= 8. Was that a typo?

    By definition, if \begin{bmatrix}x \\ y\end{bmatrix} is an eigenvector corresponding to eigenvalue -8, then we must have
    \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 \begin{bmatrix}x \\ -x \end{bmatrix}, which, of course, includes \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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