# Exericse about linear map $T\in L(V)$, where $\dim V=n\geq2$, with $\operatorname{null}T^{n-1}\neq\operatorname{null}T^n$

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

--

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

--

(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 $$V$$.

--

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

--

(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.

--

Any assistance is appreciated.

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 $$x\notin\ker T^{n-1}$$.