(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Proof that linear operator has no square root

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