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Homework Help: Linear Proof, don't know how to start it

  1. Apr 17, 2010 #1
    Let A be nxn and let [tex]B=I-2A+A^2[/tex].
    Show that if [tex]\mathbf{x}[/tex] is an eigenvector of A belonging to an eigenvalue [tex]\lambda[/tex] of A, then [tex]\mathbf{x}[/tex] is also an eigenvector of B belonging to an eigenvalue [tex]\mu[/tex] of B. How are [tex]\lambda[/tex] and [tex]\mu[/tex] related?
     
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  3. Apr 17, 2010 #2

    nicksauce

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    Well... what happens if you act 1 - 2A + A^2 on x?
     
  4. Apr 17, 2010 #3
    I don't know what you mean by that.
     
  5. Apr 17, 2010 #4

    nicksauce

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    You want to show that x is an eigenvector B, meaning that for some constant u, Bx = ux.
    Thus you need to calculate Bx = (1 -2A + A^2)x = x - 2Ax + A^2x. Can you simplify that further now?
     
  6. Apr 17, 2010 #5
    [tex]B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mu\mathbf{x}[/tex]

    ???
     
  7. Apr 18, 2010 #6

    Dick

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    Evaluate A(lambda*x), please?
     
  8. Apr 18, 2010 #7
    [tex]B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mathbf{x}-2\lambda\mathbf{x}+(\lambda)^2\mathbf{x}=\mu[/tex]
     
    Last edited: Apr 18, 2010
  9. Apr 18, 2010 #8
    Then do the quadratic equation or factor which ever is appropriate?
     
  10. Apr 18, 2010 #9

    Dick

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    You don't have to do much of anything. Just factor x out and say how Bx is related to mu*x.
     
  11. Apr 18, 2010 #10
    Should I just solve for mu then by multiplying by 1/x?
     
  12. Apr 18, 2010 #11

    Dick

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    No, there's no such thing as 1/x if x is a vector. That's exactly the wrong answer. Just look at x-2*lambda*x+lambda^2*x=mu*x=Bx and tell me what mu is.
     
  13. Apr 18, 2010 #12
    So [tex]\mu=I-2\lambda+A\lambda[/tex].
     
  14. Apr 18, 2010 #13

    Dick

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    mu is a scalar. It's a number. It's not a matrix. It's an EIGENVALUE. Try that again. You were almost there. Now you are going backwards.
     
  15. Apr 18, 2010 #14
    [tex]\mu=1-2\lambda+A\lambda[/tex]

    But there is still A
     
  16. Apr 18, 2010 #15

    Dick

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    The A is there because once upon a time I asked you to evaluate A(lambda*x). And you did it right. And then you forgot. Why did you do that? That's why I feel things are going backwards.
     
  17. Apr 18, 2010 #16
    I did. I just decide not use it apparently.
     
  18. Apr 18, 2010 #17

    Dick

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    Ok, so 1-2*lambda+lambda^2=mu. Right? I so very hope you agree with this.
     
  19. Apr 18, 2010 #18
    I concur.
     
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