Prove that two kernels are equal 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.

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.
