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[Linear Algebra] For which a is 0 an eigenvalue?

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data

    I have to find for which "a" an eigenvalue for the following system is 0.

    The system:

    1 -1 1
    -1 2 -2
    0 a 1

    2. Relevant equations
    My characterstic equation:
    (1-λ)(2-λ)(1-λ)+2a -(1-λ) -a = 0


    3. The attempt at a solution

    I then proceed:
    (1-λ)(λ2-3λ-2+a) = 0

    but then i'm kind of clueless.. Now what?
     
  2. jcsd
  3. Aug 28, 2012 #2
    You wrote the equation for the eigenvalues of the system. Now, if you want 0 to be an eigenvalue, then it better satisfy that equation.
     
  4. Aug 28, 2012 #3
    Well I get a=2 but the answer is a=-1

    Can't seem to find my error. I've tried it a bajillion times (3 times actually) :)
     
  5. Aug 28, 2012 #4

    Hurkyl

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    Where did this come from?
     
  6. Aug 28, 2012 #5
    Can you check for the 4th time, what happens if you plug in λ=0 to (1-λ)(2-λ)(1-λ)+2a -(1-λ) -a = 0 ? :) You just calculated something wrong somewhere along the way. (I'm assuming the characteristic equation is correct)
     
  7. Aug 28, 2012 #6

    HallsofIvy

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    You really don't need to find the entire eigenvalue equation to answer this. A matrix has 0 as an eigenvalue if and only if it is NOT invertible (since there must be a non-zero v such that Av= 0) and that is true if and only if its determinant is 0. Set the determinant, which depends on a, equal to 0 and solve for a.
     
  8. Aug 28, 2012 #7
    HallsofIvy, you tha man! Solved it! :D
     
  9. Aug 28, 2012 #8

    Ray Vickson

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    The equation in (a) is not consistent with that in (b). You don't need (b); just plug λ=0 into (a).

    RGV
     
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