Eigenvalue problems

  • #1
WannabeNewton
Science Advisor
5,800
532
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.
 

Answers and Replies

  • #2
263
86
If there is an eigenvector then any scalar multiple of it is also an eigenvector with the same eigenvalue. In your example, one eigenvector can be obtained from another by multiplication by -1.

If you imagine that you have the matrix in its diagonal form (or "almost diagonal" form), then it becomes obvious that scalar multiplication doesn't matter.
 
  • #3
WannabeNewton
Science Advisor
5,800
532
Oh of course. Thank you very much; at least now I know why I kept getting an answer -1 times what the key had. =D
 

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