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.

  • $\begingroup$ The problem statement is not mentioning injective. $\endgroup$ – Dietrich Burde Nov 7 '18 at 20:14
  • $\begingroup$ Yes, It is a linear function, I have edited the post. $\endgroup$ – Zanzag Nov 7 '18 at 20:17
  • $\begingroup$ Isn't injectivity needed for $f(u_1) ... f(u_m)$ being linearly independent? $\endgroup$ – Zanzag Nov 7 '18 at 20:20
  • 1
    $\begingroup$ No, the definition of linear independent is needed for being linearly independent. $\endgroup$ – Dietrich Burde Nov 7 '18 at 20:23
  • $\begingroup$ And how should I apply the definition here? Could you give me some clue of the path to follow? $\endgroup$ – 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

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


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



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?


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