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Math Help - Eigenvalues of a 3x3 matrix

  1. #1
    Junior Member simplysparklers's Avatar
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    Eigenvalues of a 3x3 matrix

    Hiya,
    Algebra exam coming up on Wednesday, and I'm stumped on how to get the egenvalues of a 3x3 matrix. I think I did them about 2 years ago, and they're gone completely out of my head.

    Given: A={\left(\begin{array}{ccc}2&4&0\\4&2&3\\0&3&2\end  {array}\right)}

    I know the eigenvalues comes out as \lambda={2},{-3},{7} .
    I know you subtract {\lambda} from the diagonal line at first, but then where do you go with it? I would really appreciate it if you could show step-by-step work. Thanks
    Jo
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  2. #2
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    To find the eigenvalues, find the roots of the polynomial \text{det}(A-\lambda I).
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  3. #3
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    So you are asking, basically, how to find those eigenvalues?

    The number \lambda is an eigenvalue of A if and only if Av= \lambda v, which is equivalent to Av- \lambda v= (A- \lambda I)v= 0 has "non-trivial solutions"- that is, a solution other than the obvious v= 0. If A- \lambda I is invertible, then we could multiply on both sides by its inverse to get the unique solution. But on the right we have that matrix times the 0 vector which must be 0. That is, there exist other, non-zero solutions, if and only if A- \lambda I is not invertible which means that its determinant must be 0:
    \left|\begin{array}{ccc}2-\lambda & 4 & 0 \\ 4 & 2-\lambda & 3 \\ 0 & 3 & 2-\lambda\end{array}\right|= 0
    You can expand that by the first row:
    (2-\lambda)\left|\begin{array}{cc}2-\lambda & 3 \\ 3 & 2-\lambda\end{array}\right|- 4\left|\begin{array}{cc}4 & 3 \\ 0 & 2-\lambda\end{array}\right|
    =(2-\lambda)((2-\lambda)^2- 9)- 4(4(2-\lambda))= (2-\lambda)(\lambda^2- 4\lambda- 5- 16)
    =(2- \lambda)(\lambda^2- 4\lambda-21)= 0
    =(2- \lambda)(\lambda+ 3)(\lambda- 7)= 0
    which has solutions \lambda= 2, \lambda= -3, and \lambda= 7.
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  4. #4
    Junior Member simplysparklers's Avatar
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    Oh yes! It's all coming back to me me now! Thank you so much for your explaination HallsofIvy!!
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  5. #5
    Member roshanhero's Avatar
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    How to find the eigen vector of the same question?
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  6. #6
    Super Member Gamma's Avatar
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    To find the eigenvector associated with the eigenvalue \lambda, you simply solve (A-I\lambda)\vec{v}=\vec{0}

    That is subtract the eigenvalue from the diagonal and find a basis for the null space of this transformation.
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