I have to say if this statement is true or false.
Let $f : V \rightarrow V$ be an endomorphism of a $\mathbb{K}$ vector space V such that $f(f(v))\neq 0,\ \forall v\neq 0$. Then $V = \ker(f) \oplus im(f)$.
I know it's related with these other posts:
Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$
When does $V = \ker(f) \oplus \operatorname{im}(f)$?
In my exercise, I haven't the hypothesis that V is finite dimensional.
I can prove that $\ker(f) \cap im(f) = \{0\}$ but I can't conclude if any vector $v$ can be written as the sum of one vector from $\ker(f)$ and another from $im(f)$.
I think, in order to conclude the proof, that $f$ must be surjective. I don't know if my proof is correct but $\ker(f) = \{0\}$.
Here what I did:
Let $v \in \ker(f)$, then $$f(v)=0 \ \Rightarrow \ f(f(v))=0 \iff v = 0.$$
So, if $V = \ker(f) \oplus im(f)$ then $v \in V$ can be written as $$ v = k + m,$$ with $k \in \ker(f)$ and $m \in im(f)$, but $k = 0$ so must be $v = m$ and therefore $v \in im(f)$ and this is that $f$ is surjective.
With this I will conclude that the statement is not true always, but if we have a surjective $f$ homomorphism, then it is true.
Thanks.
I don't understand how you can conclude $ker(f)={0}$. You can only conclude $ker(f)\cap im(f)={0}$.
But, you do need $V$ to be finite dimensional, or else the statement is false. Related to your work, any injection $f:V\rightarrow V$ that is not surjective will provide a counterexample (e.g. the right shift operator on a countable direct sum of $k$). But note counterexamples can still occur when $ker(f)$ is nonzero, for example the right shift operator follows by the projection to the complement of the second basis element.
If you have that $V$ is finite dimensional, the decomposition follows by the rank-nullity theorem.