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$T \in L(\mathbb{C}^3)$ 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 $S$.

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    $\begingroup$ Why is $N(S^{6})=N(S^{4})$? $\endgroup$ Aug 6, 2021 at 11:38
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    $\begingroup$ 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. $\endgroup$ Aug 6, 2021 at 11:41
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    $\begingroup$ @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. $\endgroup$ Aug 6, 2021 at 11:45
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    $\begingroup$ @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). $\endgroup$ Aug 6, 2021 at 11:49
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    $\begingroup$ @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. $\endgroup$ Aug 6, 2021 at 15:08

2 Answers 2

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A similar idea but worded more simply is to notice this:

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.

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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.

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