Statement: Let $ T: V \to V $ be a linear transformation over a finitely generated vector space $ V $. Suppose there exists $ k \leq n $ s.t. $\dim \ker T^{k-1} < n , \dim \ker T^k = n $ and $ \dim V=n $, show $T$ is nilpotent.
Attempt:
We know $ \forall i \in \mathbb{N}. \ker T^i \subseteq \ker T^{i+1} $.
We'll prove that $ \ker T^{k+1} \subseteq \ker T^{k} $.
Let $ v \in \ker T^{k+1} $ be arbitrary. Since $ \dim \ker T^k =n $ and $ \ker T^k \subseteq V $ then there exists a basis $ \{ v_1,…,v_n \} $ of $ \ker T^k $ and $ V $. So $ v = \alpha_1 v_1 + \cdots + \alpha_n v_n $. From the assumption that $ v \in \ker T^{k+1} $ we have that $ T^{k+1}v = TT^k v = T^k T v = T( \alpha_1 T^{k+1}v_1 + \cdots + \alpha_n T^{k+1}v_n ) = T^{k+1}( \alpha_1 Tv_1 + \cdots + \alpha_n Tv_n ) = 0 $ hence $ T^kv \in \ker T $ and $ Tv \in \ker T^k $ [ I don't know how to proceed from here, essentially I wanted to show that $ \ker T^k = \ker T^{k+1} $ and then I'd get that $ \dim \ker T^k = \dim \ker T^{k+1} $ and then I'd continue to deduce that $ T^k = 0 $ ].
Do you have please any ideas how to continue? any help would be appreciated!
Best Answer
The rank nullity theorem says that :- For any linear transformation $T:V\to W$.
($V$ is finite dimensional).
$$\text{rank}(T)+\text{null(T)}=\dim(V)$$
Here nullity just means the dimension of kernel of $T$ and the rank means the dimension of range space of $T$.
So if $\dim(\ker(T^{k}))=n$ then the rank i.e the dimension of the range space of $T$ is just the trivial subspace $\{0\}$. So $T^k(v)=0\forall\,,v\in V$.
But as $\dim(\ker(T^{i}))<n,\forall\,i<k$ , the range space has non-zero dimension and hence is not the zero transformation.
This is exactly the definition of nillpotency. That is there exists a $k\in\mathbb{N}$ such that $T^{i}$ is not the zero transformation for $i<k$ . But $T^k$ is the zero transformation.