I have been given a problem where for an algebraically closed field $\mathbb{F}$, a vector space $\dim{V}=n\in\mathbb{N}^+$ , $$ 0 <n_1<n_2<\ldots<n_{k-1},n_k=n$$ A sequence of integers, whereas $n_0\triangleq{0}$,
You'd need to prove the existence of a linear map $T:V\to V$ s.t. $n_i=\dim\ker\left(T^i\right),\ \forall{i\in\left[k\right]}$.
I have managed to find a Jordan matrix $M_n\left(\mathbb{F}\right)$, whose Jordan blocks' main diagonals consist of nothing but zeroes, for which the linear map: $$T_A:\mathbb{F^n}\to\mathbb{F^n},\ \ \vec{x}\mapsto{A\vec{x}}$$
Satisfies the requirements (including $n_i=\dim\ker\left({T_A}^i\right),\ \forall{i\in\left[k\right]}$), other than $T_A$ obviously not being $V\to{V}$.
It'd seem somewhat trivial, that you'd be able to find some nilpotent operator $T:V\to{V}$ whose Jordan normal form is $A$,
Whereas $$n_i=\dim\ker\left({T_A}^i\right),\ \forall{i\in\left[k\right]}\implies{n_i=\dim\ker\left({T}^i\right),\ \forall{i\in\left[k\right]}}$$
But I'm very much stuck and cannot figure out a way to do this?
Is it possible that I'm wrong? If not, how do I go about turning a Jordan matrix with nothing but zeroes on the diagonal into a nilpotent operator of $V$?
Thank you in advance!