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.

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
$$
\{0\}=\ker(T^0)\subsetneq
\ker(T)\subsetneq\dots\subsetneq
\ker(T^{m-1})\subsetneq
\ker(T^{m})=\ker(T^{m+1})=\ker(T^{m+2})=\dotsb
$$
