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Math Help - Eigenvectors

  1. #1
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    Eigenvectors

    Currently in the DE class I'm taking we're learning how to solve homogeneous linear systems. One of the steps requires you to come up with the associated eigenvector of a system. Unfortunately nowhere in our book does it discuss anything about eigenvectors and all our professor told us was that we'd find it easy if we had already taken linear algebra which I have not taken. I tried to find a good explanation of how to find eigenvectors of 2x2 and 3x3 matrices but didn't find much. For example how would I find the eigenvector of:

    \begin{vmatrix}<br />
0 & -7 & 0\\ <br />
5 & 8 & 4\\ <br />
0 & 5 & 0<br />
\end{vmatrix}<br />
\begin{vmatrix}<br />
k_{1}\\<br />
k_{2}\\<br />
k_{3} <br />
\end{vmatrix}\Rightarrow<br />
\begin{matrix}<br />
-7k_{2}=0\\<br />
5k_{1}+8k_{2}+4k_{3}=0\\<br />
5k_{2}=0<br />
\end{matrix}

    Thanks for your help.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Very briefly: an eigenvector is a vector which you multiply the matrix by, and you get a vector that's parallel to the original one.

    Hence:
    \mathbf{Ak} = \lambda\mathbf{k}

    That is:
    (\mathbf{A} - \lambda\mathbf{I}) \mathbf{k} = 0


    So what you want to do is subtract \lambda\mathbf{I} from \mathbf{A} and find \lambda by solving the determinant. It will be a cubic. Then use that to find what \mathbf{k} then has to be to make the equation zero.
    Last edited by Matt Westwood; August 14th 2009 at 12:02 PM. Reason: Hang on, that's a weird k ...
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  3. #3
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    Thanks Matt for the quick response. I've already done most of what you've told me to do... I guess I'm just a little confused how exactly to figure out what K has to be for the equation to equal zero. I've attached a PDF of what I have so far. Is there some systematic way of finding K? Thanks.
    Attached Files Attached Files
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  4. #4
    Super Member Random Variable's Avatar
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    Let  k_{3} = \alpha

    then  k_{1} = - \frac{4}{5} \alpha

    so  \begin{bmatrix}k_{1}\\k_{2}\\k_{3}\end{bmatrix} = \begin{bmatrix}-\frac{4}{5}\alpha\\0\\\alpha\end{bmatrix} = \alpha \begin{bmatrix}-\frac{4}{5}\\0\\1\end{bmatrix}

    If you let  \alpha =2, for example, you see that  \begin{bmatrix}-4\\0\\5\end{bmatrix} is an eigenvector associated with the eigenvalue  \lambda =2. There are infintely many to choose from.
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  5. #5
    Super Member Matt Westwood's Avatar
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    You have simultaneous equations in 3 variables, one of which is redundant. The others will give you one value of k in terms of the other. So make one of them an "arbitrary variable" which can be used as a scaling factor. Any vector which is a multiple of this k will be an eigenvector with the given eigenvalue.

    Hope this helps. Your working in the pdf file is exactly what I would have done. Then Random Variable finishes it off.
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  6. #6
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    Thanks Matt and Random Variable for your help... this now makes sense.
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  7. #7
    Super Member Matt Westwood's Avatar
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    Just a quick reply, seeing as you're new: feel free to press the "thanks" button against any reply that was particularly helpful.
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