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I have a rather basic question about solving eigenvalue problems. Once you actually find all the eigenvalues for a given operator in some basis and you go about finding the respective eigenvectors through the components and run into a situation like this:
[tex]\mid \omega = 1 > [/tex] [tex]\Rightarrow \begin{bmatrix}
1 & 0 & 1\\
0 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix} \begin{bmatrix}
v_{1}\\
v_{2}\\
v_{3}
\end{bmatrix} = 0 [/tex]
and you put it into the form
[tex]v_{1} + v_{3} = 0 [/tex]
[tex]v_{2} = 0[/tex]
[tex]v_{1} + v_{3} = 0[/tex]
isn't there an ambiguity as to whether you choose to set [tex]v_{1} = -v_{3}[/tex] as opposed to [tex]v_{3} = -v_{1}[/tex] for the eigenvector(normalized or not) corresponding to the eigenvalue? Does it not make a difference regarding sign for the respective components of the eigenvector? Sorry if this is a petty question I just wasn't sure.
[tex]\mid \omega = 1 > [/tex] [tex]\Rightarrow \begin{bmatrix}
1 & 0 & 1\\
0 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix} \begin{bmatrix}
v_{1}\\
v_{2}\\
v_{3}
\end{bmatrix} = 0 [/tex]
and you put it into the form
[tex]v_{1} + v_{3} = 0 [/tex]
[tex]v_{2} = 0[/tex]
[tex]v_{1} + v_{3} = 0[/tex]
isn't there an ambiguity as to whether you choose to set [tex]v_{1} = -v_{3}[/tex] as opposed to [tex]v_{3} = -v_{1}[/tex] for the eigenvector(normalized or not) corresponding to the eigenvalue? Does it not make a difference regarding sign for the respective components of the eigenvector? Sorry if this is a petty question I just wasn't sure.