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

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?

$$N(T^{k}) \subset N(T^{k+1})$$. 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.
• Doesn't this just prove that eventually $dim(N(T^k))=dim(N(T^{k+1}))$? How do I know that the contents of the nullspaces are also the same? Nov 16, 2018 at 9:47
• If $M$ and $N$ are subspaces with $M\subset N$ and if they have a the same dimension then they have to be equal. Nov 16, 2018 at 9:49
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.
• Doesn't this just prove that eventually $dim(N(T^k))=dim(N(T^{k+1}))$? How do I know that the contents of the nullspaces are also the same? Nov 16, 2018 at 9:47
• Because if you have two vector spaces $W_1$ and $W_2$ such that $W_1\subset W_2$ and $\dim W_1=\dim W_2$, then $W_1=W_2$. Nov 16, 2018 at 9:50