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Tough eigenvalue problem, trying to understand it

  1. Nov 23, 2009 #1
    Suppose P: V->V s.t. P^2 = P and V = kerP + ImP (actually not just + but a direct sum). Find all eigenvalues of P.

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    Which of the following explanations is right? (1 is an eigenvalue, but is 0 also?) Could somebody please explain?

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    First answer:

    Suppose that λ is an eigenvalue of P, with eigenvector v; then:
    Pv = λv = (P^2)v = (λ^2)v
    This gives (as v ≠ 0):
    λ(1−λ) = 0 ⇔ λ = 0 or 1
    In fact, Pv = v iff v ∈ Im(P), because:
    v ∈ Im(P) ⇔ v = Pz ⇒ Pv = (P^2)z = Pz = v
    On the other hand, if v = n + z and Pv = v, then Pv = P(n + z) = Pz ⇒ v = Pz.

    Other answer:

    P^2(v) = P(P(v)) = P(pv) = pP(v) = (p^2)v.
    But P^2(v) = P(v) = pv.
    Thus p^2 = p => p in {0, 1}.
    Where v in V is an eigenvector of P:V -> V, with associated eigenvalue p in F, the ground field (V/F).
    If P(v) = 0, then v in KerP. (Here we assume that v is the eigenvector with associated eigenvalue 0.)
    But P(v) = P^2(v) = P(P(v)) => v in ImP.
    Thus v in KerP intersection ImP.
    But V = KerP (+) ImP => KerP intersection ImP = {0}.
    Thus v = 0 => v is not a valid eigenvector (by convention).
    Thus 0 is not an eigenvalue of P.
    Thus 1 is the only eigenvalue of P.
     
  2. jcsd
  3. Nov 24, 2009 #2
    That 2nd proof certainly makes sense to me. What you have in P is a projection operator, which projects vectors of V to some subspace of V. Now, if the subspace is V itself, then you cannot find the eigenvalue zero for any vector in V. I think that's what happening here but it's certainly not obvious.
     
  4. Nov 24, 2009 #3
    Is this part true... I'm not sure it is:

    But P(v) = P^2(v) = P(P(v)) => v in ImP.
     
  5. Nov 24, 2009 #4

    Hurkyl

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    Re: Why do the propellers pull an aircraft, but push a ship?

    The subspace could be zero too -- i.e. P could be the zero operator.

    Why do you think it's true? If you can't give a precise algebraic calculation that shows it, then you're right to be skeptical about it.
     
  6. Nov 24, 2009 #5
    I was thinking that the condition

    [tex] V = \ker(p) \oplus \mathrm{im}(p) [/tex]

    might constrain the subspace to be equal to V but it seems that zero works too.
     
  7. Nov 24, 2009 #6
    I was thinking that the statement "P(v) = P^2(v) = P(P(v)) => v in ImP" is true because if P(P(v)) then it maps v so obviously v must be in the image. i thought the rest followed from it and I was thinking the same thing as clamtrox, i.e. that the condition constrained it... hm...
     
  8. Nov 24, 2009 #7

    Hurkyl

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    I can't parse this.


    Are you trying to say that you cancelled one P from each side of the equation
    Pv = PPv​
    ? If so, then that's a problem -- you can only cancel invertible matrices. P's action on all of V is not invertible.
     
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