Is there more than one possibility for eigenvectors of a single eigenvalue

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    Eigenvalue Eigenvectors
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The discussion centers on the concept of eigenvectors corresponding to a single eigenvalue, specifically for the matrix represented by (0 -1; 1 0), which has eigenvalues i and -i. The user identifies two eigenvectors, (1, i) and (-i, 1), for the eigenvalue -i, confirming that both vectors satisfy the eigenvalue equation. This illustrates that eigenvectors are not unique and can differ by scalar multiples, a fundamental property in linear algebra. The user concludes that scalar multiples of eigenvectors are valid representations for the same eigenvalue.

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mkerikss
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I guess this is best explained with an example. The matrix (0 -1) has the eigenvalues
------------------------------------------------------------------ (1 0)
i and -i. For -i we obtain ix1-x2=0 and x1+ix2=0. I got a corresponding eigen vector (1 i), but when I controlled this result with wolfram alpha it gave me the vector (-i 1). By inserting these values in the equations both of these give 0, as they should, so my guess is they can both be used as the eigenvector for the eigenvalue -i, but I'm not sure though, as I haven't come across this kind of situation before. If somebody could help it would be most appreciated.
 
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Note that those two vectors are scalar multiples of each other, and eigenvectors are determined up to scalar multiples.
 
Ok, I noticed that now, if i multiply the first one with -i I get the second one and vice versa. Thank you :smile:
 

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