# Thread: Max number of Linearly Independent eigenvectors

1. ## 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:
$\displaystyle \{ e_{k_1} \otimes e_{k_2} \otimes e_{k_3} \}$, the k's are natural numbers (including zero).
Now the corresponding eigenvalues are:
$\displaystyle (2k_1+1)+(2k_2+1)+(2k_3+1)$.
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What I thought at first was that since the expression $\displaystyle c_0 \mid e_0 \otimes e_0 \otimes e_0 \rangle$ has eigenvalue 3, adding one more term would mean adding something of the form $\displaystyle c_1 \mid e_1 \otimes e_0 \otimes e_0 \rangle$, but the added expression would have eigenvalue 8. So at most we can have 1 eigenvector.
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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 $\displaystyle (2k_1+1)+(2k_2+1)+(2k_3+1)$, if we equate this to seven, then we obtain $\displaystyle k_1+k_2+k_3=2$, in which case there are a maximum of 6 linearly independent eigevectors:
$\displaystyle \mid e_1 \otimes e_1 \otimes e_0 \rangle$ , $\displaystyle \mid e_0 \otimes e_1 \otimes e_1 \rangle$ , $\displaystyle \mid e_1 \otimes e_0 \otimes e_1 \rangle$

$\displaystyle \mid e_2 \otimes e_0 \otimes e_0 \rangle$ , $\displaystyle \mid e_0 \otimes e_2 \otimes e_0 \rangle$ , $\displaystyle \mid e_0 \otimes e_0 \otimes e_2 \rangle$

Which is Right?

-Thanks!