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Another Eigenvalue proof

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Let λ be an eigenvalue of A. Then λ+σ is an eigenvalue of A+σI

    2. Relevant equations

    3. The attempt at a solution

    I'm guessing I need to use the fact that λ is an e.v of A to start with. But then when I add σ to both sides somehow I feel like I'm begging the question..
  2. jcsd
  3. Apr 7, 2009 #2


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    You know that if [itex] \lambda [/itex] is an eigenvalue of A, then [itex]A\boldsymbol{v}=\lambda \boldsymbol{v}[/itex] is the appropriate eigenvalue equation with v an eigenvector. So if [itex]\lambda+\sigma [/itex] is an eigenvalue of[itex] A+\sigma I[/itex],then what would the appropriate eigenvalue equation be?
    Last edited: Apr 7, 2009
  4. Apr 7, 2009 #3
    The appropriate e.v equation would then be: Ax=(λ+σ)x..is that right?
  5. Apr 7, 2009 #4


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    No that eigenvalue does not belong to A but to [itex]A+\sigma I[/itex]. So replace A by [itex]A+\sigma I[/itex] then write out the left side of the equation and see if it checks out.
  6. Apr 7, 2009 #5
    But wouldnt that be begging the question?
  7. Apr 7, 2009 #6
    Yeah, Try adding sigma*x to both sides of equation if you're still stuck.
  8. Apr 7, 2009 #7


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    You want to solve the following equation [itex](A+\sigma I) \boldsymbol v= \eta \boldsymbol v[/itex] for [itex] \eta [/itex]. Do you understand why this is the relevant equation? Now write out the left hand side and use the information given in the exercise.
  9. Apr 7, 2009 #8
    Thanks I did that...and here's what I have

    Ax +σx=λx+σx
    Then you have (A+σI)x=(λ+σ)x
    and because x doesnt equal zero therefore (λ+σ) doesnt equal zero

    which then means that (λ,λ+σ) is an eigenpair of A+σI
  10. Apr 7, 2009 #9


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    While the first two lines you wrote are correct I am not entirely convinced. Why is [itex]A \boldsymbol{ v}+\sigma v =(A+\sigma I) \boldsymbol{v}[/itex] ?

    As for the rest. You are correct to say that x is not zero for if it was it wouldn't be an eigenvector. However 0 is a valid eigenvalue so the eigenvalue of [itex]A+\sigma I[/itex] could be zero.
    Last edited: Apr 7, 2009
  11. Apr 7, 2009 #10


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    Looks ok to me … distributive law. :smile:
  12. Apr 7, 2009 #11
    [tex] x=Ix \Rightarrow \sigma x=\sigma (Ix) = (\sigma I)x [/tex]
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