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Generalized vectors. Eigenvalues/Eigenvectors.

  1. Nov 19, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Let [itex]A \in M_{22} (\mathbb{R})[/itex] with one single eigenvalue λ and one single eigenvector v. We denote w the generalized vector such that [itex](A - λI)w = v[/itex]. Prove that v and w are linearly independent.

    2. Relevant equations

    I know that if A has only one eigenvalue λ and one eigenvector v that the equation Av = λv is satisfied. That is, (A - λI)v = 0.

    3. The attempt at a solution

    I thought about this a bit, but I'm having trouble getting this one going. I thought about letting B = (A - λI) so that we get two equations :

    Bv = 0 and Bw = v

    Then I thought that it could be broken down into two cases, one where λ = 0 and one where λ ≠ 0, but I'm not sure this is the right path to take.

    Any pointers?
     
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  3. Nov 19, 2012 #2

    Dick

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    What's the definition of linearly independent?
     
  4. Nov 19, 2012 #3

    Zondrina

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    There are several equivalent things, but if [itex]v_1, ..., v_n[/itex] is a set of vectors, then [itex]v_1, ..., v_n[/itex] are L.I if the equation :

    [itex]c_1v_1, ..., c_nv_n = 0[/itex]

    has only a trivial solution. That is all the scalars are zero.
     
  5. Nov 19, 2012 #4

    Dick

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    That's the one. So if c1*v+c2*w=0 you want to show c1 and c2 are 0. Start by applying B to that equation. What do you conclude from that?
     
    Last edited: Nov 19, 2012
  6. Nov 19, 2012 #5

    Zondrina

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    Okay so we want to show : [itex]c_1v + c_2w = 0[/itex] has only the trivial solution.

    We know that if we multiply both sides by B = (A-λI) we get :

    [itex]B(c_1v + c_2w) = 0[/itex]
    [itex]c_1Bv + c_2Bw = 0[/itex]

    We know that Bv = 0 because v is an eigenvector for the matrix A.

    [itex]c_2Bw = 0[/itex]

    We also know that Bw = v since w is our generalized vector.

    [itex]c_2v = 0[/itex]

    Now, since eigenvectors are non-zero, it must be the case that c2 = 0.
     
  7. Nov 19, 2012 #6

    Dick

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    Sure. Now put c2=0 into your original equation. What do you conclude about c1?
     
  8. Nov 19, 2012 #7

    Zondrina

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    Now subbing c2 back into our original equation yields :

    [itex]c_1v = 0[/itex]

    Once again, since v is an eigenvector, it is non-zero. Which tells us that c1 must be zero.

    Thus, c1 = c2 = 0 which implies that v and w are linearly independent.

    EDIT : Q.E.D

    Thanks.
     
    Last edited: Nov 19, 2012
  9. Nov 20, 2012 #8

    HallsofIvy

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    That's true for any A such that [itex]\lambda[/itex] is an eigenvalue with eigenvector v. The fact that A has only that eigenvalue and one independent eigenvector, tells you that there exist a vector u such that [itex]A- \lambda I)u\ne 0[/itex] but that [itex](A- \lambda I)^2u= 0[/itex].

    (It is never true that a linear operator has "one single eigenvector" because any multiple of an eigenvector is an eigenvector.)

     
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