Q:

A linear transformation that fixes a subspace

I am struggling with the problem below: 
Let $V$ be a finite-dimensional vector space over a field $F$, and $L: V \longrightarrow V$ a linear transformation over $F$. Define a sequence of subspaces $V_0, V_1, V_2, ...$ of $V$ inductively by $V_0 = V$, and $V_{n+1} = L(V_n)$ for $n \geq 1$. Prove that there exists a positive integer $N$ such that $V_{N + 1} = V_N$. 
I observed that $V_0 = V$, $V_1 = L(V)$, $V_2 = L^2(V)$, ... That is, $V_K = L^K(V)$. Thus, to show $V_{N + 1} = V_N$, we need to show that $L^{N+1}(V) = L^N(V)$, which occurs if and only if $L^N(V) \cdot (L(V) -1) = 0$, which is true if and only if $L(V)$ is the identity operator or $L$ is nilpotent. 
Thus, I've reduced the question down to proving that $L$ is either the identity transformation or nilpotent. How can I proceed with the proof? I'm not sure where to go from here. 
Thanks! 

A:

Your last claim is false. $L^N(V) \cdot (L(V) - 1) = 0$ does not imply that $L^N = 0$ or $L = 1$. The composition of nonzero linear transformations may be zero. Note that the statement you are trying to prove is stated for an arbitrary linear transformation $L \colon V \to V$, not just nilpotent ones or the identity.
Instead, observe that
$$V \supseteq L(V) \supseteq L^2(V) \supseteq \cdots,$$
thus
$$\dim_F(V) \geq \dim_F(L(V)) \geq \dim_F(L^2(V)) \geq \cdots.$$
Can you see why there must exist $N$ such that $\dim_F(L^{N + 1}(V)) = \dim_F(L^N(V))$, which implies $L^{N + 1}(V) = L^N(V)$?

