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Math Help - Matrices help

  1. #1
    Senior Member chella182's Avatar
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    Matrices help

    Okay, I'll write out the question in full, but my problem is very specific to something that crops up later on in the solution I have here. Sorry if this is in the wrong subforum, but the module's called "Linear Methods", so I thought this seemed appropriate.

    Basically it's a method to compute large power of matrices, and the example gives

    \left(\begin{array}{cc}0.9 & 0.2 \\ 0.1 & 0.8\end{array}\right)^n\left(\begin{array}{c}x_1 \\ x_2\end{array}\right) or \underline{\underline{A}}^n\underline{v}, as my lecturer likes to write stuff.

    The notes then say "It would be easy to compute \underline{\underline{A}}^n\underline{v} if \underline{v} were an eigenvector i.e. \underline{\underline{A}}^n\underline{v}=\lambda^n  \underline{v}". Up to this point, I undersand completely.

    Then we start finding the eigenvectors and the calculation goes as such

    \left(\begin{array}{cc}0.9 & 0.2 \\ 0.1 & 0.8\end{array}\right)\left(\begin{array}{c}1 \\ -1\end{array}\right)=\left(\begin{array}{c}0.9\time  s1-0.2\times1 \\ 0.1\times1-0.8\times1\end{array}\right)
    =\left(\begin{array}{c}0.7 \\ -0.7\end{array}\right)=0.7\color{green}\boxed{\left  (\begin{array}{c}1 \\ -1\end{array}\right)}

    ...and he notes that the green box is always an eigenvector. So far, I still understand. It's this next bit that's confused me. He notes that 1 is always an eigenvalue, but he just seems to have plucked this vector out of the air (the one in the red box) and I really can't see where it's came from. It's probably very simple but this is my problem with the example/solution.

    \left(\begin{array}{cc}0.9 & 0.2 \\ 0.1 & 0.8\end{array}\right)\color{red}\boxed{\left(\begi  n{array}{c}2/3 \\ 1/3\end{array}\right)}\color{black}=1\left(\begin{ar  ray}{c}0.9(2/3)+0.2(1/3) \\ 0.1(2/3)+0.8(1/3)\end{array}\right)

    =1\left(\begin{array}{c}2/3 \\ 1/3\end{array}\right)

    So basically, where has the \mathbf{\left(\begin{array}{c}2/3 \\ 1/3\end{array}\right)} came from? Sorry this is a bit long winded by the way, I just thought having the whole question might help.
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  2. #2
    ynj
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    you can solve the equation AX=X\leftrightarrow (A-I)X=0, which is a linear equation groups.
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  3. #3
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    If \lambda=1 is an eigenvalue, then (A-\lambda I_2)v=0 yields our eigenvectors. Specifically, we have

    \left(\begin{array}{cc}0.9-1 & 0.2 \\ 0.1 & 0.8-1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)=\left(\begin{array}{cc}-0.1 & 0.2 \\ 0.1 & -0.2\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)=\left(\begin{array}{c}0.2y-0.1x \\ 0.1x-0.2y\end{array}\right)=\left(\begin{array}{c}0 \\ 0\end{array}\right),

    which gives us 0.1x=0.2y \Rightarrow x=2y.

    So a basis for \mathsf{E}_1 is

    \beta=\left\{\left(\begin{array}{c}2 \\ 1\end{array}\right)\right\}.

    Therefore, we have

    \frac{1}{3}\left(\begin{array}{c}2 \\ 1\end{array}\right)=\left(\begin{array}{c}2/3 \\ 1/3\end{array}\right)\in\mathsf{E}_1.
    Last edited by hatsoff; August 17th 2009 at 06:54 PM.
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  4. #4
    Senior Member chella182's Avatar
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    Okay, I'm with you right up until the \frac{1}{3} comes into it at the end... where did that come from?
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  5. #5
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    Quote Originally Posted by chella182 View Post
    Okay, I'm with you right up until the \frac{1}{3} comes into it at the end... where did that come from?
    Any linear combination of vectors in \beta is going to be in the eigenspace \mathsf{E}_1=\text{span}(\beta). So you can throw in any value you want, as long as it's in the field, a subset of which appears to be \mathbb{R}.
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  6. #6
    Senior Member chella182's Avatar
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    I still don't quite understand :\ oh well, exam in an hour and a half.
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