Linear Algebra - Stationary Sequence

Question

Let $T\in L(\Bbb{V},\Bbb{V})$ with $\Bbb{V}$ a $\Bbb{K}$-vector space and $\dim_{\Bbb{K}}\Bbb{V}=n<\infty$. Prove that the following sequence $$\ker T\subset\ker T^2\subset \dots \subset \ker T^l \subset \dots \subset \Bbb{V}$$ of subspaces of $\Bbb{V}$ becomes stationary, i.e., $$\exists m\in \Bbb{Z}_+:\ker T^m=\ker T^{m+i}, \forall i \in \Bbb{Z}_+.$$

It seems to be a simple task, but I'm not able to see it anyway! Need some help! Grateful...

Answer 1

Note that if $A,B\subset V$ are two subspaces of $V$ such that $\dim(A)=\dim(B)$ and $A\subset B$, then $A=B$ (this can be proved by considering the fact that a maximal linearly independent subset is a basis). Therefore if $\ker(T^i)\neq \ker(T^{i+1})$, then $\dim(\ker(T^{i+1}))>\dim(\ker(T^i))$. Since $V$ is of dimension $n$, there can be at most $n$ indices $i$ such that $\dim(\ker(T^i))\neq\dim(\ker(T^{i+1}))$. Since the sequence is infinite it follows that it eventually stabilizes.

Answer 2

I have done!
Suppose that the sequence never becomes stationary, then $$\forall m\in \Bbb{Z}_+, \exists i\in \Bbb{Z}_+:\ker T^m \subsetneq\ker T^{m+i}. $$ But then is possible to construct a sequence $$\ker T \subsetneq \ker T^{m_1} \subsetneq \ker T^{m_2} \subsetneq \dots \subsetneq \ker T^{m_j}\subsetneq \dots \subsetneq \Bbb{V}, \forall j\in \Bbb{Z}_+.$$ But $\ker T^{m_{j_1}} \subsetneq \ker T^{m_{j_2}} \implies \dim _{\Bbb{K}}(\ker T^{m_{j_1}})<\dim_{\Bbb{K}}(\ker T^{m_{j_2}})$. Then we would have $$\dim _{\Bbb{K}}(\ker T)<\dim _{\Bbb{K}}(\ker T^{m_{1}})<\dots<\dim _{\Bbb{K}}(\ker T^{m_{j}})<\dots<\dim _{\Bbb{K}} \Bbb{V}, \forall j\in\Bbb{Z}_+.$$ which contradicts the assumption $\dim_{\Bbb{K}}=n<\infty$.