Prove that two kernels are equal

Question

Let $V$ be a finite dimensional linear space. How do you prove that there exists such an m that:

$$\text{ker} (T^m) = \text{ker} (T^{m+1}).$$

I have managed to prove using induction that for each $m≥1$, $\text{ker} (T^m) ⊆ ker (T^{m+1})$. So, I have a unidirectional inclusion. Instead of proving the inclusion in the other direction I think I can just prove that the dimensions of the two sets are equal. I understand that

$$\text{Dim} V = \text{Dim ker}T + \text{Dim ImT},$$

but I'm not sure how to go about it.

Answer 1

Denote $V_m=\text{ker}(T^m )$

Each time $V_m\subsetneq V_{m+1}$ then the dimension of the subspace $V_m$ increases by at least $1$.

On the other hand $|V_m|\leq |V|$ for all $m$. Therefore, $V_m$ is a strict subspace of $V_{m+1}$ finitely many times.

So, there is an $m_0$ such that for all $n\geq m_0$, we have $V_n=V_{n+1}$.

Answer 2

Let's prove by contradiction. Suppose that for all natural number $m$, we have $${ker} (T^m) \ne ker (T^{m+1})$$ which implies that $${ker} (T^m) \subsetneq ker (T^{m+1})$$ since we already have $${ker} (T^m) \subset ker (T^{m+1})$$Therefore, $$Dim \space ker(T^{m+1}) \ge Dim \space ker(T^{m})+1$$ And since $Dim \space ker(T) \ge 0$, we have $Dim \space ker(T^{m+1}) \ge m$, which contradicts the fact that $V$ is finitely dimensional.

Answer 3

Consider $\{\dim\ker(T^k):k\ge0\}$; this set of natural numbers is bounded by $\dim V$, hence there is an $m$ so that $\dim\ker(T^m)$ is maximal.

Now you know that $\ker(T^{m})\subseteq\ker(T^{m+1})$, which implies $$ \dim\ker(T^{m})\le\dim\ker(T^{m+1}) $$ By maximality of $\dim\ker(T^{m})$, you infer the two dimensions are equal.

You can also refine the result: if you take the minimum $m$ such that $\dim\ker(T^{m})$ is maximal, you have $$ \{0\}=\ker(T^0)\subsetneq \ker(T)\subsetneq\dots\subsetneq \ker(T^{m-1})\subsetneq \ker(T^{m})=\ker(T^{m+1})=\ker(T^{m+2})=\dotsb $$