proof that linear operator has no square root

[winter85]

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

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

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, V = \textbf{C}^3 and there exists a positive integer m and a vector v in V such that T^{m-1}v \neq 0 but T^{m}v = 0 then the vectors (v, Tv, ..., T^{m-1}v) are linearly independent.

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

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

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

I'm looking for someone to verify my proof because i'm not very sure about it.

[morphism]

I would say "S^5 = 0" instead of "S^5 v = 0" (or at least "S^5 v = 0 for all v"). And I would also expand on what would happen if S^5 \neq 0. Other than that, this looks good.

[winter85]

Alright, thank you :)

[Hurkyl]

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?

[winter85]

The minimal polynomial of T is p(z) = z^{3}. 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.

[Hurkyl]

The minimal polynomial of the square root of T...
Baby steps. First find any polynomial satisfied by the square root of T. Then try to find the minimal one.

[winter85]

since T^{3} = 0, we have that S^{6} = 0. So p(z) = z^{6} is a polynomial that square root of T satisfies.
Therefore we can say that the minimal polynomial of S must divide p(z) = z^{6}. So it's either one of 1, z, z^{2}, z^{3}, z^{4}, z^{5}, or z^{6}.

Now i think it cannot be any of 1, z, z^{2}, z^{3}, z^{4} because S^{4} = T^{2} \neq 0. so the minimal polynomial is either z^{5} or z^{6}. is this correct?

[Hurkyl]

Now i think it cannot be any of 1, z, z^{2}, z^{3}, z^{4} because S^{4} = T^{2} \neq 0. so the minimal polynomial is either z^{5} or z^{6}. is this correct?
Exactly. And now, we can apply your argument that this is impossible, because the degree cannot exceed 3!

[winter85]

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.

[Hurkyl]

could you tell me what you meant by applying something i know about nilpotent matrices
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.

[winter85]

okay i see, thanks for your help anyway :)

[hello5271]

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.