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Eigvector/value problem (need validation of my proof)

  • #1
rock.freak667
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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
 

Answers and Replies

  • #2
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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].
 
  • #3
Hurkyl
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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.
 
  • #4
rock.freak667
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ah yeah, a typo, supposed to be e since that is the e.vector.
 
  • #5
Hurkyl
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Ah, there we go. Mystery solved. :smile:
 

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