|Subject:||Re: How does Octave choose a phase when calculating eigenvectors?|
|Date:||Fri, 13 Mar 2020 19:25:18 +0200|
|User-agent:||Mozilla/5.0 (X11; Linux x86_64; rv:68.0) Gecko/20100101 Thunderbird/68.5.0|
On Fri, Mar 13, 2020 at 1:00 PM Brett Green <address@hidden> wrote:
Eigenvectors are defined up to a constant. Octave chooses the magnitude of this constant such that the eigenvectors will be normalized. However, this still leaves an arbitrary phase.
How does Octave choose this phase? It is not specified in the documentation.
- Brett Green
I'm not sure what you mean by "phase". I'm somewhat familiar with the mathematics of eigenvectors/values, but do not have any experience with the term phase, in this this context. Could you give an example of what you mean by phase? Even your claim that eigenvectors are defined "up to a constant" is not entirely accurate. Vectors (x) are either eigenvectors of a matrix (A) or not if they satisfy the equationA*x = lambda*xfor some scalar lambda, called the eigenvalue for x. Its true that scalar multiples of x are also eigenvectors of A (with the same lambda), but that has to do with the fact that linear combinations of eigenvectors with the same eigenvalue are also eigenvectors, not anything about being "defined".
James Sherman Jr.
The OP probably means polarity. I.e. if
A*x = lambda*x
is satisfied, then
A*(-x) = lambda*(-x)
is satisfied too.
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