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Yet another eigenvalue proof

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Let x be a unit vector. Namely x(Transpose)*x = 1. If (A − Let x be a unit vector. If (A − λI)x = b, then λ is an eigenvalue of A − bx(transpose).

    3. The attempt at a solution

    I have no idea where to start this proof.
  2. jcsd
  3. Apr 9, 2009 #2


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    Homework Helper

    Start like this.

    (A - bx')x = (A - \lambda I + \lambda I - bx')x

    Expand the right hand side, and use facts about [tex] A, \lambda, b[/tex] and [tex] x [/tex].
    Last edited: Apr 9, 2009
  4. Apr 9, 2009 #3
    Ok...so heres what I have so far..I'm not sure if I'm on the right track..

    Since x(transpose)*x=1
    Therefore, (A-b*x(transpose))*x=λx
    So Ax - bx(transpose)x=λx
    Then Ax-λx=b
    Hence you have (A-λI)x=b

    I dont know if im begging the question by doing it this way though
  5. Apr 9, 2009 #4


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    Homework Helper

    No. Look at

    [tex] \begin{align*}
    (A - bx')x & = (A - \lambda I + \lambda I - bx') x \\
    & = (A - \lambda I) x + (\lambda I - bx')x

    What do you know about

    (A - \lambda I)x
    [/tex] ?

    use that, together with what you get when you expand

    (\lambda I - bx') x

    (don't forget that [tex] x' x = 1 [/tex])
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