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Invertibility proof

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Let T be a linear operator on [tex] R^6 [/tex] whose characteristic polynomial is [tex] f(t) = (t-4)(t+1)^3 (t-2)^2[/tex]. Show that [tex] T^3 + 2T^2 -3T [/tex] is invertible.


    2. Relevant equations



    3. The attempt at a solution

    Okay, T is invertible since 0 is not an eigenvalue of T. [tex] T^3 +2T^2 -3T [/tex] is invertible if and only if zero is not an eigenvalue of T, and this means that the only vector in the null space is the zero vector. Suppose that there was a non-zero vector, [tex] x' [/tex], in the null space. Then we have [tex] T^3(x') +2T^2(x') -3T(x') = 0 [/tex] then after applying [tex] T^{-1} [/tex] to both sides, we have [tex] T^3(x') + 2T^2(x') - 3T(x') = T^2(x') + 2T(x') -3x' \Rightarrow U(x') = T^2(x') + 2T(x') = 3x' [/tex] and so we see that 3 is an eigenvalue of U.

    Now, I know from the characteristic polynomial that T's eigenvalues are [tex] \lambda_1 = 4, \lambda_2 = -1, \lambda_3 = 2 [/tex].

    For [tex] \lambda_1 = 4 [/tex] corresponding to some vector x'_1,

    [tex]U(x'_1) = T^2(x'_1) + 2T(x'_1) = T(4x'_1) + 8x'_1 = 16x'_1 + 8x'_1 = 24x'_1 [/tex].

    Through similar methods, we can show that 24, -1, and 8 are eigenvalues of U.

    Now here is where I get stuck. If this means that the ONLY eigenvalues of U is 24, -1, and 8, then since 3 isn't an eigenvalue of U, then T^3 + 2T^2 - 2T has no nonzero vector in its null-space and so it is invertible.

    But if this means that 24, -1, and 8 are might only be some of the eigenvalues of U, then I can't conclude that 3 isn't an eigenvalue of U. I was thinking if I could prove that the algebraic multiplicity's of 24, -1, and 8 add up to 6, then since this is an operator on R^6 these would be the only eigenvalues of U, concluding my proof.

    Any ideas?
     
  2. jcsd
  3. Apr 10, 2009 #2

    Dick

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    If you have (T^3+2T^2-3T)x=0, doesn't that mean (T+3)(T-1)Tx=0? Doesn't that mean T HAS to have a nontrivial eigenvector with eigenvalue either 0, 1 or -3? Aren't all of those incompatible with the characteristic polynomial? Aren't you overcomplicating this?
     
  4. Apr 10, 2009 #3
    I never even thought about factoring that. Wow
     
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