T-Invariant Subspaces: Proving W is T-Invariant for E_{\lambda}

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Homework Help Overview

The discussion revolves around proving that a subspace W is T-invariant for the eigenspace E_{\lambda} associated with an eigenvalue \lambda of a linear operator T. Participants are exploring the properties of eigenvectors and their mappings under T.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the definition of T-invariance and how T acts on eigenvectors in the eigenspace E_{\lambda}. Questions are raised about the implications of T mapping eigenvectors to other eigenvectors and the behavior of T on scalar multiples of eigenvectors.

Discussion Status

Some participants have clarified their understanding of the eigenspace and are exploring how T operates on eigenvectors. There is an ongoing examination of specific cases, such as the null space N(T), and how it relates to T-invariance. Multiple interpretations of the problem are being discussed without a clear consensus yet.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can share or the methods they can use to prove T-invariance.

hitmeoff
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Homework Statement


Show that W is a T-invariant subspace of T for:
W = E[tex]_{\lambda}[/tex]

Homework Equations

The Attempt at a Solution



Ok, so I know that I need to show that T maps every element in E[tex]_{\lambda}[/tex] to .

E[tex]_{\lambda}[/tex] = N(T-[tex]\lambda[/tex]I)

so T must map every eigenvector related to [tex]\lambda[/tex] to another eigenvector in E[tex]_{\lambda}[/tex]

T(x) maps to zero vector, when x is an eigenvector associated with [tex]\lambda[/tex] which is in the eigenspace of [tex]\lambda[/tex], correct?
 
Last edited:
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Your post is just a huge blank space where you messed up the tex for making a lambda. Can you edit it?
 
fixed
 
Ok I get it now. [tex]E^{\lambda}[/tex] is the set of all eigenvectors with eigenvalue [tex]\lambda[/tex]. Let's say v is an eigenvector, and [tex]Tv=\lambda v[/tex]. What is [tex]T(\lambda v)[/tex] and how does this help you answer the question?
 
Office_Shredder said:
Ok I get it now. [tex]E^{\lambda}[/tex] is the set of all eigenvectors with eigenvalue [tex]\lambda[/tex]. Let's say v is an eigenvector, and [tex]Tv=\lambda v[/tex]. What is [tex]T(\lambda v)[/tex] and how does this help you answer the question?

T([tex]\lambda[/tex]v) = [tex]\lambda[/tex]2v

and this is just a multiple of T, so T(v) = [tex]\lambda[/tex]v maps to the eigenspace?
 
and another question...

Show that W is a T-invariant subspace when

W = N(T)

So N(T) : { x E W: T(x) = 0} , since W is a subspace it contains the zero vector, thus any vector in N(T) will map to zero which is in W?
 

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