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# Eigenvectors watch

1. Hello!

Just to confirm something which suddenly struck me, since in solving for , you set , thus you will never get a unique eigenvector?

In other words, you shouldn't try to solve the with a specific and expect to get unique values for x,y,z etc; for a 2 by 2 matrix, you only need either the first row or the second row, since the 2 rows just contain the same information of the sort y=f(x)?
2. Yep, if is an eigenvector of , then is an eigenvector for all k.
3. im pretty sure that constant matrices have unique eigenvalues.

why would det(A - lamdaI)=0 imply non-unique lamda?
4. (Original post by SouthernFreerider)
im pretty sure that constant matrices have unique eigenvalues.

why would det(A - lamdaI)=0 imply non-unique lamda?
No no, there are unique s, just that you can't find a unique solution since the determinant is 0, no inverse exists.

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