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

1. Nov 15, 2008

### 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.

Last edited: Nov 15, 2008
2. Nov 15, 2008

### 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.

3. Nov 15, 2008

### winter85

Alright, thank you :)

4. Nov 15, 2008

### Hurkyl

Staff Emeritus
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?

5. Nov 16, 2008

### 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.

6. Nov 16, 2008

### Hurkyl

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

7. Nov 16, 2008

### 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?

8. Nov 16, 2008

### Hurkyl

Staff Emeritus
Exactly. And now, we can apply your argument that this is impossible, because the degree cannot exceed 3!

9. Nov 16, 2008

### 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.

10. Nov 16, 2008

### Hurkyl

Staff Emeritus
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.

11. Nov 16, 2008

### winter85

okay i see, thanks for your help anyway :)

12. Jul 23, 2011

### 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.