Eigvector/value problem (need validation of my proof)

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Homework Help Overview

The discussion revolves around an eigenvalue and eigenvector problem involving two square matrices A and B. The original poster attempts to prove that the sum of eigenvalues from these matrices corresponds to the eigenvalue of their sum, using a specific eigenvector.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants examine the proof provided by the original poster, questioning the use of variable notation and its implications for understanding the proof's validity.

Discussion Status

There appears to be general agreement among participants regarding the correctness of the proof's concept, although some express concern over the notation used. The discussion is focused on clarifying these notational issues rather than resolving the proof itself.

Contextual Notes

Participants note the potential for typographical errors to indicate deeper misunderstandings, suggesting that the notation choice may affect the clarity of the proof.

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[SOLVED] eigvector/value problem (need validation of my proof)

Homework Statement


Show that if [itex]\lambda[/itex] is an e.value of a square matrix A with e as a corresponding e.vector, and [ites]\mu[/itex] is an e.value of the square matrix B for which e is also a corresponding e.vector,the [itex]\lambda + \mu[/itex] is an e.value of the matrix A+B with e as a corresponding e.vector


Homework Equations





The Attempt at a Solution



From the def'n of an e.vector

[tex]Ax= \lambda x[/tex]

[tex]Ax+Bx= \lambda x + \mu x[/tex]

[tex](A+B)x= (\lambda +\mu)x[/tex]

hence [itex]\lambda +\mu[/itex] is an e.value of A+B
 
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I don't see anything wrong with your proof. Although since the problem defines [tex]e[/tex] as the eigenvector, it'd probably be more prudent (albeit arbitrary) to use that instead of [tex]x[/tex].
 
foxjwill said:
I don't see anything wrong with your proof. ... it'd probably be more prudent (albeit arbitrary) to use that instead of [tex]x[/tex].
I would consider the fact he used x when he should have been using e an actual error in his proof. Yes, it might simply be a persistent typographical error (which is still an error!) -- but it might also be a symptom of a deeper misunderstanding of how to manipulate mathematical statements.

However, I do completely agreement with you that the original poster's proof idea is correct.
 
ah yeah, a typo, supposed to be e since that is the e.vector.
 
Ah, there we go. Mystery solved. :smile:
 

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