# 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.

• The problem statement is not mentioning injective. – Dietrich Burde Nov 7 '18 at 20:14
• Yes, It is a linear function, I have edited the post. – Zanzag Nov 7 '18 at 20:17
• Isn't injectivity needed for $f(u_1) ... f(u_m)$ being linearly independent? – Zanzag Nov 7 '18 at 20:20
• No, the definition of linear independent is needed for being linearly independent. – Dietrich Burde Nov 7 '18 at 20:23
• And how should I apply the definition here? Could you give me some clue of the path to follow? – Zanzag Nov 7 '18 at 20:27

It is true in general because if you have that $$\{u_1,\dots,u_m\}$$ is a base of the supplementary U of the kernel of $$f$$ than if you consider the linear combination

$$0=\sum_{k=1}^m a_kf(u_k)=f(\sum_{k=1}^m a_k u_k)$$

You have that

$$\sum_{k=1}^m a_k u_k\in U\cap \ker(f)=\{0\}$$

and so

$$\sum_{k=1}^m a_k u_k =0$$

then

$$a_k=0$$ for every $$k=1,\dots , n$$

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?