Consequences of $f \neq 0, f^2 = 0$ I have been given the following problem. 
Given $f: V \to V$ linear. $f \neq 0$, $f^2 = 0$, demonstrate that $(f(u_1), f(u_2) ... f(u_m))$ is linearly independent, where $(u_1, u_2...u_m)$ is a basis of the supplementary subspace of the kernel ($U  \oplus \ker(f) = V$) 
My reasoning is the following: 
$\bar{x} \in V \implies \bar{x} = \bar{x_1} + \bar{x_2}$, $x_1 \in \ker(f), x_2 \in U$
$f(\bar{x}) = f(\bar{x_2})$ since $x_1 \in \ker(f)$
$0 = f(\bar{x_2})$ since $f(x) \in \operatorname{Im}(f) \subseteq \ker(f)$
$0 = a_1 f(u_1) + a_2 f(u_2) ... + a_m f(u_m) $
I know if $(f(u_1), f(u_2) ... f(u_m))$ was injective, then $a_1=a_2=a_m=0$ but I don't know how to demonstrate this.

Hint:
You don't really need the hypothesis $f^2=0$. Consider the restriction of $f$ to the supplementary subspace $U$. Which properties does it have?
