Choosing Free Variable for Generalized Eigenvector

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Homework Help Overview

The discussion revolves around the selection of a free variable when determining generalized eigenvectors in the context of a repeated roots problem in linear algebra. The subject area includes eigenvalues, eigenspaces, and the properties of linear transformations.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • The original poster questions the implications of choosing different free variables for generalized eigenvectors, suggesting that both choices appear valid but may yield different representations. Some participants discuss the dimensionality of the eigenspace and the implications for parameter selection, while others reference the use of the standard basis.

Discussion Status

Participants are exploring the implications of their choices regarding free variables and the structure of the eigenspace. There is acknowledgment of differing perspectives on the dimensionality of the eigenspace, with some guidance offered regarding the use of the standard basis. However, no explicit consensus has been reached.

Contextual Notes

There is a mention of a potential typo affecting the eigenvectors, and a discussion about the dimensionality of the eigenspace, which is noted to be one-dimensional, impacting the choice of free variable.

rugerts
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Homework Statement
Find general solution of DE
Relevant Equations
Eigenvector and eigenvalue eqns
IMG-2049.JPG
IMG-2050.JPG
As you can see from my eigenvalues, here I've got a repeated roots problem. I'm wondering if it matters which variable I can choose to be the free variable when I'm solving for the generalized eigenvector. I think both are equally valid but they look different from one another and I'd like to know the reason behind why either choice would be fine.
Thanks for your time
 
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I think you have a typo in your last matrix at ##(1,1)## and thus wrong eigenvectors.

Anyway, we have a two dimensional eigenspace to the eigenvalue ##-1##, so the eigenvectors span the entire vector space. How you set the parameter doesn't matter, as long as you keep the linear independent.
 
If the eigenspace is the entire space, there's no reason not to use the standard basis.
 
pasmith said:
If the eigenspace is the entire space, there's no reason not to use the standard basis.
You are right and i was wrong. We have only a one dimensional eigenspace, spanned by a single vector.
The eigenspace is annihilated by ##(A+1)## whereas the other basis vector of ##\mathbb{R}^2## is only annihilated by ##(A+1)^2##.

##\operatorname{ker}(A+1)= \operatorname{span}(1,\frac{1}{2})## and ##\operatorname{ker}(A+1)^2 = \mathbb{R}^2##
 
Last edited:

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