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Math Help - Max number of Linearly Independent eigenvectors

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    Max number of Linearly Independent eigenvectors

    I need help with this problem I came across. I give it in a simplified way:
    The questions asks for the maximum number of linearly independent eigenvectors for the eigenvalue 7.
    Now suppose the Matrix we are dealing with is in 3D and has eigenvectors:
    \{ e_{k_1} \otimes e_{k_2} \otimes e_{k_3} \}, the k's are natural numbers (including zero).
    Now the corresponding eigenvalues are:
    (2k_1+1)+(2k_2+1)+(2k_3+1).
    \
    \
    What I thought at first was that since the expression c_0 \mid e_0 \otimes e_0 \otimes e_0 \rangle has eigenvalue 3, adding one more term would mean adding something of the form c_1 \mid e_1 \otimes e_0 \otimes e_0 \rangle , but the added expression would have eigenvalue 8. So at most we can have 1 eigenvector.
    \
    \
    I'm not sure if just adding eigenvalues in the way I did is allowed.

    UPDATE:
    I was thinking maybe the question meant the following:
    Given the eigenvalue (2k_1+1)+(2k_2+1)+(2k_3+1), if we equate this to seven, then we obtain k_1+k_2+k_3=2, in which case there are a maximum of 6 linearly independent eigevectors:
    \mid e_1 \otimes e_1 \otimes e_0 \rangle , \mid e_0 \otimes e_1 \otimes e_1 \rangle , \mid e_1 \otimes e_0 \otimes e_1 \rangle

    \mid e_2 \otimes e_0 \otimes e_0 \rangle , \mid e_0 \otimes e_2 \otimes e_0 \rangle , \mid e_0 \otimes e_0 \otimes e_2 \rangle

    Which is Right?

    -Thanks!
    Last edited by Arturo_026; December 11th 2013 at 07:20 PM.
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