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Math Help - Eigenvalue

  1. #1
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    Eigenvalue

    I am sure there is an easier way than the one I am trying.
    Prove that the matrix has one negative and onepositive eigenvalue:

    1 a 1
    a 1 a
    1 a 1
    where a=1.0001
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  2. #2
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    Quote Originally Posted by Sabita View Post
    I am sure there is an easier way than the one I am trying.
    Prove that the matrix has one negative and onepositive eigenvalue:

    1 a 1
    a 1 a
    1 a 1
    where a=1.0001
    What is the way you tried?
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  3. #3
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    Hello,

    Quote Originally Posted by Sabita View Post
    I am sure there is an easier way than the one I am trying.
    Prove that the matrix has one negative and onepositive eigenvalue:

    1 a 1
    a 1 a
    1 a 1
    where a=1.0001
    Try to transform this matrix by combining rows with each other. I think it's called "similar matrix", am not sure...
    If you manage to find a diagonal matrix (which is the representation of the matrix in a certain basis), then the numbers in the diagonal will be the eigenvalues.

    Let e1 be the first row, e2 the second and e3 the third.

    e2 -> e2-a*e3
    Matrix is now (...)

    e3 -> e3-e1
    Matrix is now (...)

    Conclude
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  4. #4
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    I will partially guide you on your way to finding the eigenvalues of your matrix.
    Note: I know you do not need to find the actual eigenvalues here, but you should know this without a doubt.

    First give your matrix a name: A = <br />
\left( \begin{array}{ccc} 1 & a & 1 \\ a & 1 & a \\ 1 & a & 1 \end{array} \right)<br />

    Your first real step: determine the characteristic polynomial of A.

    To do this, subtract lambda from each term along the main diagonal in A.
    Like this: <br />
\left( \begin{array}{ccc} 1-\lambda & a & 1 \\ a & 1-\lambda & a \\ 1 & a & 1-\lambda \end{array} \right)<br />
.

    Compute the determinant of this modified matrix. You should get a polynomial in terms of lambda. This is the characteristic polynomial of A. Factor this and then set your factored result equal to zero and you have your characteristic equation. Solve for lambda. Have fun.

    -Andy
    Last edited by abender; May 13th 2008 at 11:58 AM.
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  5. #5
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    Thanks for the replies.
    I did try the lambda method, but I am getting an equation which I am unable to solve due to 1.0001 * 1.0001 in some terms. Hence I assumed there would be an easier way to solve it .
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  6. #6
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    Quote Originally Posted by Sabita View Post
    Thanks for the replies.
    I did try the lambda method, but I am getting an equation which I am unable to solve due to 1.0001 * 1.0001 in some terms. Hence I assumed there would be an easier way to solve it .
    The determinant is -\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1)).

    Solving 0 = -\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1)) for \lambda looks routine ........

    Clearly there's one negative eigenvalue and one positive eigenvalue. And an eigenvalue equal to zero.
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  7. #7
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    I got the equation . But how do I solve it?
    Also what is the significance of 1.0001?.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    The determinant is -\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1)).

    Solving 0 = -\lambda (\lambda^2 - 3 \lambda - 2(a^2 - 1)) for \lambda looks routine ........

    Clearly there's one negative eigenvalue and one positive eigenvalue. And an eigenvalue equal to zero.
    Quote Originally Posted by Sabita View Post
    I got the equation . But how do I solve it?
    Also what is the significance of 1.0001?.
    Either \lambda = 0 or \lambda^2 - 3 \lambda - 2(a^2 - 1) = 0.

    The latter is solved using the quadratic formula:

    \lambda = \frac{3 \pm \sqrt{8a^2 +1}}{2}.

    If a = 1.0001 then clearly \sqrt{8a^2 + 1} > 9 .......
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    If a = 1.0001 then clearly \sqrt{8a^2 + 1} > 9 .......
    Hi Sabita,

    mr fantastic means:

    If a = 1.0001 then clearly \sqrt{8a^2 + 1} > \sqrt{9} = 3 .......
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  10. #10
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    Thanks mr fantastic, Isomorphism and Andy.

    I got it!
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