Prove that the nullspaces of linear operators acting on themselves are eventually equal

Question

For a linear operator $T$ on a finite-dimensional vector space $V$ such that $dim(V)=n$, prove that $\exists k \leq n$ such that $N(T^k)=N(T^{k+1})$.

This is one of those problems where I believe it intuitively, but I am having a hard time tackling a rigorous proof. My first instinct was to start playing around with minimal polynomials and characteristic equations, but then I backed off because I was worried the problem is too general -- how do I even know that the operator has eigenvalues/eigenspaces?

Anyone have a good approach for this proof?

Answer 1

Hint: It is clear that $\ker T^k\subset\ker T^{k+1}$. So, you have an increasing sequence of subspaces of a $n$-dimensional space.

Answer 2

$N(T^{k}) \subset N(T^{k+1})$. If equality does not hold then the dimension of $N(T^{k})$ must be smaller than that of $N(T^{k+1})$. If the assertion is not true you will get a strictly decreasing sequence of positive integers all less than or equal to the dimension of the space. This is a contradiction.