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Math Help - Eigenvector (easy)question

  1. #1
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    Eigenvector (easy)question

    Hi guys,

    im given this matrix here:

    |3 2|
    |3 8|

    and the eigenvalues were 2 and 9.

    I found the eigenvectors where 'lamda' = 2

    but when i use the eigenvalue for 9 my answer were (1, 1/3) but the book gave me the answer as being (1, 3).

    Could someone explain to me why they have (1,3)?

    Thanks. Your contribution is much appreciated.
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  2. #2
    Senior Member Spec's Avatar
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    Well you get 3x_1=x_2 so if you set x_1=t then x_2=3t

    So we have \left(\begin{array}{cc}x_1\\x_2\end{array}\right) =t\left(\begin{array}{cc}1\\3\end{array}\right)<br />
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  3. #3
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    thanks!

    If its not too much to ask, could you show me your working too? Still a bit confused as to why u let 3x1 = x2

    Would my original answer be correct too?
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  4. #4
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    Hi Redeemer_Pie,

    Remember that if \binom{s_1}{t_1} is an eigenvector corresponding to an eigenvalue of \lambda_1 then r\binom{s_1}{t_1} where r is some constant is also an eigenvector corresponding to an eigenvalue of \lambda_1 .

    We have M\:=\:\begin{bmatrix}3&2 \\ 3&8 \end{bmatrix}

    Note that for any eigenvalue \lambda and the corrsponding eigenvector s, (M-\lambda I)s=0 so for \lambda = 9 we have

    \:\begin{bmatrix}\text{-}6&2 \\ 3&\text{-}1 \end{bmatrix}\binom{x}{y}=0 hence

    3x-y=0 \implies y=3x . If we let x=1 then y=3 so

    s=r\binom{1}{3} where r is some constant.

    Hope this helps.
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  5. #5
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    OH!!!!!

    Got it!!! thanks Sean12345

    Forgot to mention.

    The answer i had originally, would that also be correct?
    Last edited by Redeemer_Pie; June 7th 2009 at 07:33 AM.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Any nonzero multiple of an eigenvector is correct.
    That point is it forms an eigenspace and you need to know what part of the whole space it spans.
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