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Math Help - Finding Eigenvectors

  1. #1
    Junior Member
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    Unhappy Finding Eigenvectors

    Hi there,
    I have a 4 X 4 matrix,
    <br />
A=\left(\begin{array}[color=Black]{cccc}-2 & 0 & 0& 0 \\0 & 1 & 0 & 1 \\0 & 0 & 0 & 0 \\0 & 1 & 0 & 1\end{array}\right)<br />

    I've found eigenvalues of this matrix to be 2, -2, 0. I have to find the eigenvectors for each eigenvalue. I'm having great trouble in finding the eigenvectors.
    Can any one help?

    Thank you for your time.

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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    well first subtract \lambda to the diagonal and then you computed the determinant, and hence you found the eigenvalues, good, now, let's try this with the first eigenvalue so you need to solve (A-2I_4|0), do you see what you need to do? you need to solve that homogeneous system and find a basis, that's what eigenvectors are.

    do the same for each eigenvalue.
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  3. #3
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    Ah, yes, thank you very much!
    In doing that, I've calculated the eigenvectors with each eigenvalue. However, one of the eigenvectors I got for the eigenvalue 0 is different to the one in the answer. I understand the multiplicity of the eigenvalue of 0 is two. So, I think the basis for the eigenspace will contain 2 eigenvectors. I've found the eigenvectors to be \left(\begin{array}{c}0\\0\\1\\0\end{array}\right) & \left(\begin{array}{c}0\\-1\\0\\1\end{array}\right).

    The former is in the answer, the latter isn't. Instead, they have \left(\begin{array}{c}0\\-1\\1\\1\end{array}\right). I've looked over my general solution for homogeneous equation (A-0I_4|0), but I can't find where that one comes from.
    After row reducing, (A-0I_4|0) becomes \left(\begin{array}{cccc}1 & 0 & 0& 0 \\0 & 1 & 0 & 1 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{array}\right). I can't see how they got that eigenvector. Where did I go wrong?

    Thank you again, for your time.
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  4. #4
    Member alunw's Avatar
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    If there are two independent eigenvectors for a particular eigenvalue than the eigenspace has more than one dimension so there must be a variety of possible answers for the basis of eigenvectors. Your second vector is a linear combination of the two given in the official answer, so it must be correct. It's not obvious to me why their answer would be better than yours because your eigenvectors look to be orthogonal, but theirs do not.
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  5. #5
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    I understand. Thank you soo much for your time.
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