Composition of Injective Linear Operators I'm trying to form a proof for the following:
Let $V\neq\{0\}$ be finite-dimensional and assume that $L_1,...,L_n$ are linear operators. Show that if $L_1 \circ ... \circ L_n = 0$, then at least one of the maps $L_i$ is not one-to-one.
My intuition is that the composition of one-to-one maps should also be one-to-one. I've started by considering $L_1(x_1) = 0$ with $x_1 = L_2(...(L_n)...)$. Supposing $L_1$ is one-to-one, then the only way that $L_1(x_1) = 0$ is if $L_2(...(L_n)...)$ was always equal to the same value, $y$, and hence not one-to-one. 
Continuing inductively, it can be shown that assuming the outer-most linear operator is one-to-one forces the inner composition to be not one-to-one. The approach seems awkward to me for some reason. Is it valid? Is there a more direct way to show that some $L_i$ is not one-to-one?
 A: Suppose $f$ and $g$ injective, and that $g(f(x)) = g(f(y))$ for elements $x$ and $y$ in the domain of $f$. Then $f(x) = f(y)$ because $g$ is injective. And then $x = y$ because $f$ is injective. 
Hence: any finite composition of injective maps is injective (by an induction argument that you were sneaking up on in your question itself). That's all you need to prove your theorem (since any linear map sends 0 to 0 already). If your composition sent some other item to 0, then it could not be injective. 
A: Let $L = L_1 \circ L_2 \circ \cdots \circ L_n$. If $L=0$, then $L$ cannot be injective as the zero map is not injective. If $L_1$ is not injective, we are finished. If it is, it has a left inverse $L_1^{-1}$, and $L_1^{-1} \circ L$ is our new object of study. It cannot be injective either, as applying $L_1^{-1}$ to all sides of $L = L_1 \circ L_2 \circ \cdots \circ L_n = 0$, we get $L_1^{-1} \circ 0$ which is, of course, the 0 map. 
We may repeat this process $n$ times, and if each $L_i$ is injective, we contradict the fact that the composition of injective maps is injective. So at least one of the $L_i$ was not injective. 
