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Proof that linear operator has no square root

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose [tex]T \in L(\textbf{C}^3)[/tex] defined by [tex]T(z_{1}, z_{2}, z_{3}) = (z_{2}, z_{3}, 0)[/tex]. Prove that T has no square root. More precisely, prove that there does not exist [tex]S \in L(\textbf{C}^3)[/tex] such that [tex]S^{2} = T[/tex].

    2. Relevant equations


    3. The attempt at a solution

    I showed in a previous exercise that if T is a linear operator on a vector space V (in this case, [tex]V = \textbf{C}^3[/tex] and there exists a positive integer m and a vector v in V such that [tex]T^{m-1}v \neq 0[/tex] but [tex]T^{m}v = 0[/tex] then the vectors [tex](v, Tv, ..., T^{m-1}v)[/tex] are linearly independent.

    First notice that for there is at least one vector v such that [tex]T^{2}v \neq 0[/tex] but that [tex]T^{3}v = 0[/tex] for all v in V. Therefore i can choose a basis for V of the form [tex](v, Tv, T^{2}v)[/tex] for some v in V.

    Suppose S is a square root of T so that [tex]S^{2} = T[/tex]. then by the previous reasoning, the vector list [tex](v, S^{2}v, S^{4}v)[/tex] form a basis of V.

    However we know that [tex]S^{6} = T^{3} = 0[/tex] so we're face with two possibilities:
    either [tex]S^{5}v = 0[/tex] which would make [tex](v, Sv, S^{2}v, S^{3}v, S^{4}v)[/tex] linearly independent, but that impossible since we know that [tex](v, S^{2}v, S^{4}v)[/tex] spans V. A similar reasoning applies if we assume that [tex]S^{5}v \neq 0[/tex].

    I'm looking for someone to verify my proof because i'm not very sure about it.
     
    Last edited: Nov 15, 2008
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  3. Nov 15, 2008 #2

    morphism

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    I would say "[tex]S^5 = 0[/tex]" instead of "[tex]S^5 v = 0[/tex]" (or at least "[tex]S^5 v = 0[/tex] for all v"). And I would also expand on what would happen if [tex]S^5 \neq 0[/tex]. Other than that, this looks good.
     
  4. Nov 15, 2008 #3
    Alright, thank you :)
     
  5. Nov 15, 2008 #4

    Hurkyl

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    If you're interested in seeing other cute approaches -- what is the minimal polynomial of T look like? Using that, find a polynomial that S satisfies. What can the minimal polynomial of S be?

    Another one... can you apply anything you know about nilpotent matrices?
     
  6. Nov 16, 2008 #5
    The minimal polynomial of T is [tex]p(z) = z^{3}[/tex]. The minimal polynomial of the square root of T would have a degree greater than 3 (i guess it's degree would either be 5 or 6 but cannot be less, is there a way to tell which it would be?). But that's impossible because the degree of the minimal polynomial of an operator on a vector space V is at most dim V, and in this case dim V = 3.

    I know that nilpotent matrices have 0 as their only eigenvalue.. is this what you mean? I'm not sure how to use this to solve this problem.
     
  7. Nov 16, 2008 #6

    Hurkyl

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    Baby steps. First find any polynomial satisfied by the square root of T. Then try to find the minimal one.
     
  8. Nov 16, 2008 #7
    since [tex]T^{3} = 0[/tex], we have that [tex]S^{6} = 0[/tex]. So [tex]p(z) = z^{6}[/tex] is a polynomial that square root of T satisfies.
    Therefore we can say that the minimal polynomial of S must divide [tex]p(z) = z^{6}[/tex]. So it's either one of [tex]1, z, z^{2}, z^{3}, z^{4}, z^{5}, or z^{6}[/tex].

    Now i think it cannot be any of [tex]1, z, z^{2}, z^{3}, z^{4}[/tex] because [tex]S^{4} = T^{2} \neq 0[/tex]. so the minimal polynomial is either [tex]z^{5}[/tex] or [tex]z^{6}[/tex]. is this correct?
     
  9. Nov 16, 2008 #8

    Hurkyl

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    Exactly. And now, we can apply your argument that this is impossible, because the degree cannot exceed 3!
     
  10. Nov 16, 2008 #9
    cool, thank you :)
    could you tell me what you meant by applying something i know about nilpotent matrices? I know that 0 is the only eigenvalue of a nilpotent matrix. Is that helpful?

    I'm learning this stuff on my own and i'm interested in finding different approaches for every problem.
     
  11. Nov 16, 2008 #10

    Hurkyl

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    I hadn't thought it through -- it just seemed like a good line of attack. If that had been my first idea, I think it would probably just have led me to the minimal polynomial version.
     
  12. Nov 16, 2008 #11
    okay i see, thanks for your help anyway :)
     
  13. Jul 23, 2011 #12
    If you don't want to deal with minimal polynomials, you could also look at the problem this way: Assume there exists an S such that S^2=T. Since T^3=0,we know that S^6=0. Thus, S is nilpotent, which implies that S^(dim(C^3))=0 (this is a basic theorem in my textbook). The dimension of C^3 is 3, so S^3=0. But that means that S^4=S(S^3)=S(0)=0. Since S^4=0, we know T^2=0. But this is a contradiction because we can see that T^2(z1,z2,z3)=T(Z2,Z3,0)=(Z^3,0,0), which is not equal to 0.
     
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