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Math Help - How to find eigenvectors

  1. #1
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    How to find eigenvectors

    \begin{array}{ccc}-1&0&1\\0&-4&-3\\2&0&-2\end{array}

    I have found the eigenvalues -3 and -4 but have no idea how to find the eigenvectors. Any help? thanks.

    Ok ive found a bit more but am then stuck after

    2x1 + x3 = -3x1
    -x2 + 3x3 = -3x2
    2x1 + x3 = -3x3

    And am now stuck
    Last edited by adam_leeds; January 17th 2010 at 07:46 AM.
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Let \lambda be an eigenvalue of A. Then a vector v that satisfies Av=\lambda v \Leftrightarrow (A-\lambda I)v = 0 is an eigenvector.

    Thus if you find a vector v_1, satisfying (A+3I)v_1 = 0 and a vector v_2 with (A+4I)v_2= 0, then you're done.

    You think you can do that?
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  3. #3
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    Quote Originally Posted by Dinkydoe View Post
    Let \lambda be an eigenvalue of A. Then a vector v that satisfies Av=\lambda v \Leftrightarrow (A-\lambda I)v = 0 is an eigenvector.

    Thus if you find a vector v_1, satisfying (A+3I)v_1 = 0 and a vector v_2 with (A+4I)v_2= 0, then you're done.

    You think you can do that?
    i think thats the last bit ive done and i dont know what to do after
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  4. #4
    Senior Member Dinkydoe's Avatar
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    A+3I =

    2 0 1
    0 -1 -3
    2 0 1

    Thus we must solve the following system:

    2x+z = 0
    -y - 3z = 0

    This gives: z = -2x, -3z = y. Thus v = (1,  6 , -2)

    is an eigenvector with \lambda = -3

    Now you try the other one.
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  5. #5
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    Quote Originally Posted by Dinkydoe View Post
    A+3I =

    2 0 1
    0 -1 -3
    2 0 1

    Thus we must solve the following system:

    2x+z = 0
    -y - 3z = 0

    This gives: z = -2x, -3z = y. Thus v = (1, 6 , -2)

    is an eigenvector with \lambda = -3

    Now you try the other one.
    I understand up till you go to v from z x and y how do you do that? thanks for your help by the way its much appreciated.
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  6. #6
    Senior Member Dinkydoe's Avatar
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    So you understand how we found the relations:

    z= -2x, -3z = y.

    A vector that satisfies these relations is an eigenvector with \lambda = -3.

    Now we may choose x= c arbitrary. These relations give the values of y and z. We may choose x = c arbitrary because if v is an eigenvector, cv is an eigenvector as well.

    Let x = c then by the above relations follows, z=-2c, y = -6c. Thus v = (c,-2c,-6c) = c(1,-2,-6) is an eigenvector for any c.

    The relations only show how x,y,z relate to eachother. That's why one co-ordinate x,y or z may be freely chosen.
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  7. #7
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    Quote Originally Posted by Dinkydoe View Post
    So you understand how we found the relations:

    z= -2x, -3z = y.

    A vector that satisfies these relations is an eigenvector with \lambda = -3.

    Now we may choose x= c arbitrary. These relations give the values of y and z. We may choose x = c arbitrary because if v is an eigenvector, cv is an eigenvector as well.

    Let x = c then by the above relations follows, z=-2c, y = -6c. Thus v = (c,-2c,-6c) = c(1,-2,-6) is an eigenvector for any c.



    The relations only show how x,y,z relate to eachother. That's why one co-ordinate x,y or z may be freely chosen.
    oh i see so (2 , -4, -12) is an eigenvector for -3
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  8. #8
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    There is no eigenvector for -4

    because

    3 0 1
    0 0 -3
    2 0 2

    3x + z = 0
    -3z = 0
    2x + 2z = 0

    z = 0 and x = 0

    or is 0,1,0 an eigenvector?
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  9. #9
    Senior Member Dinkydoe's Avatar
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    There is no eigenvector for -4
    Wrong
    Indeed: x=z=0 but as we see y = c can be freely choosen

    So (0,c,0)= c(0,1,0) is an eigenvector for any c.
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  10. #10
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    Quote Originally Posted by Dinkydoe View Post
    Wrong
    Indeed: x=z=0 but as we see y = c can be freely choosen

    So (0,c,0)= c(0,1,0) is an eigenvector for any c.
    ye i understand now thank you
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