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I have this problem that I am attempting, and am struggling with (b).

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Assume $\dim V = n \geq 2$ and that $T \in L(V)$ such that $\operatorname{null}T^{n-1}\neq\operatorname{null}T{^n}$

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(a) Prove that $T$ is nilpotent.

(b) Prove that $T$ does not have a square root; that is, $T\neq S^2$ for any $S \in L(V)$

(c) Prove that there exists $x \in V$ such that $(x,Tx,...,T^{n-1}x)$ is a basis for $V$.

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Here is what I have for (a), (b) and (c):

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(a) Proof:

We will first show that $\dim\operatorname{null}T^k=k$ for $k=0,1,...,n$. We see by our assumption that $\operatorname{null}T^k \neq\operatorname{null}T^{k-1}$. Thus the dimensions increase by at least $1$ per increment of $k$. But if the dimensions increase by more than $1$ at any increment, then $\dim\operatorname{null}T^n > n$ which is a contradiction. Thus, $\dim\operatorname{null}T^k = k$ and it follows that $\operatorname{null}T^n=V$. Thus, $T^n=0$, and so $T$ is nilpotent.

(b) Proof: (shaky on this, and need to verify if I am going down the correct path)

Suppose towards contradiction $\exists S \in L(V)$ such that $S^2=T$

Observe $\dim\operatorname{null}S^2 =\dim\operatorname{null}T= 1$

Then, $\dim\operatorname{null}(S)$ must be less than or equal to $\dim\operatorname{null}(S^2)$, so $\dim\operatorname{null}S$ is either $0$ or $1$. Both lead to contradiction, If it is $0$, then $\operatorname{null}S=0$, but I am unsure how to show the other case.

(c) Let $a_0, a_1, \dots, a_{n-1} \in \mathbb{F}$ such that

$$ 0 = a_0x + a_1Tx + \dots + a_{n-1}T^{n-1}x $$ for some $x \in V$

Applying $T^{n-1}$ to both sides of the equation above yields

$$ 0 = a_0T^{n-1}x, $$

which shows that $a_0 = 0$. Therefore

$ 0 = a_1Tx + \dots + a_{n-1}T^{n-1}x $

Applying $T^{n-2}$ yields

$ 0 = a_1T^{n-1}x, $

which shows that $a_1 = 0$. Continuing in this fashion, we see that $a_0 = a_1 = \dots = a_m = 0$. Thus $x, Tx, T^2x, \dots, T^{n-1}x$ is linearly independent.

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Any assistance is appreciated.

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Your proof of part a) has a major gap:

Thus the dimensions increase by at least $1$ per increment of $k$.

Unless you have proved this at some earlier point, this really requires justification.

The gap in your proof of part b) is related:

Both lead to contradiction, If it is $0$, then $null$ $(S)$ = $0$, but I am unsure how to show the other case.

Indeed if $\ker S=0$ then $\ker S^k=0$ for all $k$, but then $$\ker T^{n-1}=\ker S^{2n-2}=\ker S^{2n}=\ker T^n,\tag{1}$$ a contradiction.

If $\dim\ker S=1$ then use the fact that $\dim\ker S^2=\dim\ker T=1$, which you should have shown in part a). The proof of part a) can be imitated to prove that then $\dim\ker S^k=1$ for all $k>0$, again leading to a contradiction by $(1)$.

For part c) you have now proved that $(x,Tx,...,T^{n-1}x)$ is a basis for every $x\in V$, which is false. You write

Applying $T^{n-1}$ to both sides of the equation above yield $$ 0 = a_0T^{n-1}x, $$ which shows that $a_0 = 0$.

But this assumes that $T^{n-1}x\neq0$. So in stead of starting with an arbitrary $x\in V$, start with $x\notin\ker T^{n-1}$.

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