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Math Help - Eigenvalues of a 3 x 3 Matrix

  1. #1
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    Eigenvalues of a 3 x 3 Matrix

    I am trying to work out some Matrix questions in advance to my course start.

    I am trying to find the eigenvalues for this 3 x 3 Matrix:

    5 3 2
    1 4 6
    9 7 3

    So far I have got to:

    L=Lambda

    5-L ... 3 ... 2
    1 ... 4-L ... 6
    9 ... 7 ... 3-L

    I am not sure where to go next - Can anybody help?

    Thanks,

    Chris
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  2. #2
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    Re: Eigenvalues of a 3 x 3 Matrix

    Well, you evaluate that determinant now! That will, of course, depend on \lambda. In fact, it will be a cubic equation.

    \left|\begin{array}{ccc}5- \lambda & 3 & 2 \\ 1 & 4- \lambda & 6 \\ 9 & 7 & 3-\lambda \end{array}\right|= 0

    Expanding on the top row, that is equal to
    (5- \lambda)\left|\begin{array}{cc} 4-\lambda & 6 \\  7 & 3- \lambda \end{array}\right| - 3 \left|\begin{array}{cc}1  & 6 \\ 9 & 3-\lambda\end{array}\right| + 2\left|\begin{array}{cc}1 & 4-\lambda \\ 9 & 7 \end{array}\right|= 0

    (5-\lambda)((4-\lambda)(3-\lambda)- 42)- 3(3-\lambda- 54)+ 2(7- 9(4- \lambda)= 0
    Multiply that out and solve for \lambda.
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    Re: Eigenvalues of a 3 x 3 Matrix

    Thanks.

    I have got:

    L + 12L + 16L - 55

    Whats the best way to solve this?

    Thanks a lot,

    Chris
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  4. #4
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    Re: Eigenvalues of a 3 x 3 Matrix

    Quote Originally Posted by RealENG View Post
    Thanks.

    I have got:

    L + 12L + 16L - 55

    Whats the best way to solve this?
    You have the correct equation -\lambda^3+12\lambda^2+16\lambda-55=0, but it does not have convenient solutions. A graph of that function shows that the roots are approximately 2.6, 1.7 and 12.9. You will have to use a numerical method if you want better approximations than that.
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  5. #5
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    Post Re: Eigenvalues of a 3 x 3 Matrix

    Thanks.

    I checked with MatLab and an online cubic calculator and got -2.568, 1.659, 12.909 as my eigenvalues, so that's all good.

    How do I go about finding the Eigenvectors for these eigenvalues?

    Any help is greatly appreciated.

    Regards,

    Chris
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    Re: Eigenvalues of a 3 x 3 Matrix

    The eigenvector corresponding to a certain eigenvalue is the vector x such that Ax=\lambda x.
    Lambda is one of the three eigenvalues.
    There are thus 3 eigenvectors in this problem.

    Using Matlab you can find the eigenvectors using: [V,L] = eig(M) where V are the vectors (1 column is 1 vector), L are the eigenvalues (lambdas) and M is your input matrix.

    The eigenvector corresponding to the eigenvalue, 12.9094 is \begin{bmatrix}0.3936\\-0.5435\\-0.7414 \end{bmatrix}.

    I hope this helps...
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  7. #7
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    Re: Eigenvalues of a 3 x 3 Matrix

    you can use this site to calculate matrix stuff:

    Calculator for Eigenvalues and Eigenvectors
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