# Prove there is no $S$ s.t $S^2=T$

$$T \in L(\mathbb{C}^3)$$ is defined as $$T(x,y,z)=(y,z,0)$$. Prove there doesn't exist $$S$$ s.t $$S^2=T$$.

I would like to know if my proof works for this problem. Please correct me if there's anything wrong.

Proof starts.

First of all, note that $$N(T^3)= \mathbb{C}^3$$.

Now, suppose there exists $$S$$ s.t $$S^2=T$$.

Thus, $$N(S^4)=N(T^2)$$.

But, since $$N(S^6)=N(S^4)$$, $$N(T^3)=N(T^2)$$.

Thus, $$\mathbb{C}^3=N(T^2)$$.

But, $$T^2(x,y,z)=T(y,z,0)=(z,0,0)$$. This implies $$N(T^2)= \text{span} ((1,0,0),(0,1,0)) \neq \mathbb{C}^3$$ which is a contradiction.

Hence, there doesn't exist such $$S$$.

• Why is $N(S^{6})=N(S^{4})$? Aug 6, 2021 at 11:38
• That's not enough : it's not clear why if $\dim \mathbb C^3 = 3$ then the chain $N(S^k)$ must be stationary after $k=4$. It's non trivial , in my opinion. Aug 6, 2021 at 11:41
• @john It is true, but the proof of the result is not easy, I mean it's certainly not easier than the result you are proving for sure. Aug 6, 2021 at 11:45
• @john Your proof is correct, in the sense that the fact you used , and everything else you wrote is correct. The only problem is that it's a bit circular, in the sense that you've used a very strong result to prove a rather mild one (like using a hammer to swat a fly, if you like : it does the job, but simpler things also do so). For example, to prove your result, I would obviously go by contradiction, but I'd think about what $S$ must do to each of $(1,0,0),(0,1,0),(0,0,1)$, and see if I can get contradictions from there. (Side note : I'll have to go, apologies). Aug 6, 2021 at 11:49
• @john I'm really happy that there was an answer below, which used just Cayley-Hamilton. However, the point is that you have to use at least that theorem, because if you think about it, Cayley-Hamilton is the first result that allows you to relate higher powers of a matrix to lower powers, and hence understand these higher powers. So without CH, that particular step $N(S^4) = N(S^6)$ wouldn't work out. Let me see if I can provide some more intuition as well. Aug 6, 2021 at 15:08

If $$N \in L(\Bbb{C}^n)$$ satisfies $$N^k = 0$$ for any $$k \in \Bbb{N}$$, then $$N^n = 0$$.
Proof: If $$\lambda \in \Bbb{C}$$ is an eigenvalue of $$N$$ with nonzero eigenvector $$x \in \Bbb{C}^n$$, then $$0 = N^kx = \lambda^k x \implies \lambda^k =0 \implies \lambda=0.$$ Therefore, all eigenvalues of $$N$$ are equal to $$0$$ and hence the characteristic polynomial of $$N$$ is $$k_N(x) = x^n$$ so by Cayley-Hamilton we have $$0 = k_N(N) = N^n$$.
Now for your operator $$T$$ holds $$T^3 = 0$$ but $$T^2 \ne 0$$. Assume there exists $$S \in L(\Bbb{C}^3)$$ such that $$S^2 = T$$. Then $$S^6 = T^3=0$$ so our lemma above implies $$S^3 = 0$$ and therefore $$0 = S^4 = T^2$$ which is a contradiction.
Your proof is correct, but you have to justify why $$S \in L(\mathbb{C}^n) \implies N(S^n) = N(S^{n + 1}) = \dots$$. The proof is simple. If $$S$$ is invertible, then $$S^j$$ is invertible for every $$j$$, so $$N(S) = N(S^2) = \dots$$. So assume $$S$$ is not invertible. We have the nested chain of subspaces $$N(S) \subset N(S^2) \subset \dots$$. The assumption that $$S$$ is not invertible means $$\dim N(S) \geq 1$$. Note that $$N(S^j) \neq N(S^{j + 1}) \iff \dim N(S^j) < \dim N(S^{j + 1})$$. Thus we must have $$N(S^j) = N(S^{j + 1})$$ for some $$j \leq n$$. Now if $$N(S^{k - 1}) = N(S^k)$$, then $$N(S^k) = N(S^{k + 1})$$ since $$S^{k + 1}v = 0 \implies S^kSv = 0 \implies S^{k - 1}Sv = 0 \implies S^kv = 0$$. Thus by induction, $$N(S^{j}) = N(S^{j + 1}) = N(S^{j + 2}) = \dots$$. Since $$j \leq n$$, this proves the claim.