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Math Help - Simple eigenvector/value problem

  1. #1
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    Question Simple eigenvector/value problem

    Attatched is the document with my question... i think the problem is that i really dont understand the steps to finding and testing eigenvalues and vectors. i use the characteristic equations but i seem to be getting alot of eigen-vectors, that, when tested are only true for the trivial case... ie. multiple of zero.

    thanks.

    **Edit**

    Just realised that if i let k be equal to -3, the eigenvector i got will be equal to the one in the solution.... does this mean that we can have k as anyy value except for zero?

    Also... if we have matrix A, eigenvector = u and eigenvalue = b, when we test to see if we have obtained the correct solution by checking if Au = bu, does it prove true if we have b = 0?
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    Last edited by Schniz2; May 23rd 2008 at 10:27 PM. Reason: New thoughts
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  2. #2
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    Quote Originally Posted by Schniz2 View Post
    Attatched is the document with my question... i think the problem is that i really dont understand the steps to finding and testing eigenvalues and vectors. i use the characteristic equations but i seem to be getting alot of eigen-vectors, that, when tested are only true for the trivial case... ie. multiple of zero.
    The eigenvector you found, \left[<br />
\begin{array}{c}<br />
-\frac34\\<br />
1\end{array}\right], is not correct. Could you show us how you arrived at that answer?

    If we call the matrix A, we have \left(\lambda I - A\right)\textbf{x} = \textbf{0}, which, for \lambda_1 = 2 gives us the system:

    \left[\begin{array}{cc|c}<br />
2 + 1 & 4 & 0\\<br />
-3 & 2 - 6 & 0\\<br />
\end{array}\right] = \left[\begin{array}{cc|c}<br />
 3 & 4 & 0\\<br />
-3 & -4 & 0\\<br />
\end{array}\right]

    which gives us

    \textbf{x} = t\left[\begin{array}{c}<br />
4\\-3<br />
\end{array}\right].

    You can see that \left[\begin{array}{c}<br />
4\\-3<br />
\end{array}\right] is an eigenvector of A because

    \left[\begin{array}{cc}<br />
-1 & -4\\<br />
3 & 6<br />
\end{array}\right]<br />
\left[\begin{array}{c}<br />
4\\-3<br />
\end{array}\right]<br />
=\left[\begin{array}{c}<br />
 8\\-6<br />
 \end{array}\right]<br />
=2\left[\begin{array}{c}<br />
  4\\-3<br />
  \end{array}\right]<br />
=\lambda_1\left[\begin{array}{c}<br />
  4\\-3<br />
  \end{array}\right]

    But be sure to see that any nonzero multiple of \left[\begin{array}{c}<br />
  4\\-3<br />
  \end{array}\right] is also an eigenvector of A. So, if you had, for example, \left[\begin{array}{c}<br />
   -\frac4{3}\\1<br />
   \end{array}\right], you would be okay.

    Quote Originally Posted by Schniz2 View Post
    Just realised that if i let k be equal to -3, the eigenvector i got will be equal to the one in the solution.... does this mean that we can have k as anyy value except for zero?
    I'm not sure what you are asking here. Please clarify.
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  3. #3
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    Quote Originally Posted by Reckoner View Post
    But be sure to see that any nonzero multiple of \left[\begin{array}{c}<br />
  4\\-3<br />
  \end{array}\right] is also an eigenvector of A. So, if you had, for example, \left[\begin{array}{c}<br />
   -\frac4{3}\\1<br />
   \end{array}\right], you would be okay.



    I'm not sure what you are asking here. Please clarify.
    sorry, the one i had was \left[\begin{array}{c}<br />
   -\frac4{3}\\1<br />
   \end{array}\right]

    At the end of an example in my notes they prove an eigenvector is correct by showing that the eigenvector multiplied by zero proves it. Is this allowed? *see zero.jpg*

    Also, the equation we were given to use was *see equation.jpg*... is this equivalent to the one you used?

    thanks.
    Attached Thumbnails Attached Thumbnails Simple eigenvector/value problem-equation.jpg   Simple eigenvector/value problem-zero.jpg  
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  4. #4
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    Quote Originally Posted by Schniz2 View Post
    sorry, the one i had was \left[\begin{array}{c}<br />
   -\frac4{3}\\1<br />
   \end{array}\right]

    At the end of an example in my notes they prove an eigenvector is correct by showing that the eigenvector multiplied by zero proves it. Is this allowed? *see zero.jpg*
    That's because 0 is an eigenvalue for that matrix. Its eigenvectors, however, cannot include the zero vector, by definition.

    Quote Originally Posted by Schniz2 View Post
    Also, the equation we were given to use was *see equation.jpg*... is this equivalent to the one you used?

    thanks.
    Multiply both sides by -1.
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  5. #5
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    ahh ok, i think i understand now....

    so the eigenvalue you get at the beginning, should be the eigenvalue you get when you check?

    thanks heaps for all your help
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