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Math Help - Finding the eigenvector

  1. #1
    Junior Member Nerdfighter's Avatar
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    Finding the eigenvector

    Given a specific Eigenvalue \lambda, how do I find an eigenvector of a Matrix A?

    Specifically, if \lambda = 4 and A = \left[\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right]

    By the way, is there a way to make my Matrix look a little neater/lined up?
    Last edited by Nerdfighter; June 10th 2009 at 02:16 PM.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    If \begin{pmatrix}x\\y\\z\end{pmatrix} is an eigenvector, then \left(\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right)\begin{pmatrix}x\\y\\z\end{  pmatrix}\ =\ \lambda\begin{pmatrix}x\\y\\z\end{pmatrix}\ =\ 4\begin{pmatrix}x\\y\\z\end{pmatrix}.

    There will be infinitely many eigenvectors corresponding the the eigenvalue 4, so just pick one of them.

    Quote Originally Posted by Nerdfighter View Post
    By the way, is there a way to make my Matrix look a little neater/lined up?
    Instead of centre-aligning them, right-align them (as I have done).
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  3. #3
    Junior Member Nerdfighter's Avatar
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    How Do I pick one? I obviously can't pick any numbers for x, y, and z.
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    I did not say pick any numbers for x,y,z. Pick x,y,z so that they satisfy the eigenvalue equation.
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  5. #5
    Junior Member Nerdfighter's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    I did not say pick any numbers for x,y,z. Pick x,y,z so that they satisfy the eigenvalue equation.
    What is a good method to find x,y,z such that x,y,z satisfies the eigenvalue equation?

    Could I simply pick the zero vector to be an eigenvector?
    Last edited by Isomorphism; June 11th 2009 at 01:49 AM.
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by Nerdfighter View Post
    Given a specific Eigenvalue \lambda, how do I find an eigenvector of a Matrix A?

    Specifically, if \lambda = 4 and A = \left[\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right]

    By the way, is there a way to make my Matrix look a little neater/lined up?
    Solve the matrix equation

    A\vec v = \lambda \vec v \iff (A-\lambda I)\vec v =0

    Since your eigen value is 4 we get

    (A-4 I)\vec v =0

    \begin{bmatrix} <br />
3-4 && 0&& -1\\<br />
2 && 3-4 && 1 \\<br />
-3 && 4 && 5-4 \\<br /> <br />
\end{bmatrix}

    \begin{bmatrix} <br />
-1 && 0&& -1\\<br />
2 && -1 && 1 \\<br />
-3 && 4 && 1 \\<br /> <br />
\end{bmatrix}

    putting this in reduced row form we get

    \begin{bmatrix} <br />
1 && 0&& 1\\<br />
0 && 1 && 1 \\<br />
0 && 0 && 0 \\<br /> <br />
\end{bmatrix}

    Solving the above homogenious equation gives

    z=t, y=-t,x=-t or

    \begin{bmatrix}<br />
-t \\ -t \\ t <br />
\end{bmatrix}=t \begin{bmatrix}<br />
-1 \\ -1 \\ 1 <br />
\end{bmatrix}

    So that is the eigen vector

    p.s if you use the \begin{bmatrix} \end{bmatrix} tags it will auto format all of your columns
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  7. #7
    Junior Member Nerdfighter's Avatar
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    Thank you, that's what I needed!
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Nerdfighter View Post
    Could I simply pick the zero vector to be an eigenvector?

    No, the zero vector is never an Eigenvector, as if it was every scalar would be an Eigenvalue. An Eigenvector for the corresponding Eigenvalue \lambda is defined to be a "non-zero vector such that Mv = v \lambda".
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  9. #9
    Junior Member Nerdfighter's Avatar
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    Quote Originally Posted by Swlabr View Post
    No, the zero vector is never an Eigenvector, as if it was every scalar would be an Eigenvalue. An Eigenvector for the corresponding Eigenvalue \lambda is defined to be a "non-zero vector such that Mv = v \lambda".
    Ooh, very good to know. Thank you.
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