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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:

Let $T : V → V$ be a linear transformation. Show that $T^2 = I $ if and only if $V = ker(T − I) ⊕ ker(T + I)$

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.

DrinkingDonuts
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1 Answers1

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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.

Victor Zhang
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    $(f∘f)(v)≠0$ precisely means that any nonzero image is not sent to 0 by $f$ which by definition is equivalent to $ker(f)∩im(f)=0$.

    Explicitly, a projection operator is an example of an $f$ with $(f∘f)(v)≠0, ∀v≠0$ with $ker(f)$ nonzero.

     – Victor Zhang Jul 23 '17 at 22:15
  • in the last line of your reasoning, $0=f(0)=f(f(v))$ is correct, but by definition it means $f(v)=0$ not $v=0$. – Victor Zhang Jul 23 '17 at 22:19
  • Did my response answer your question? The statement is indeed false, although the way you went about answering it isn't quite right. A proper answer consists of just the demonstration of one counterexample, for example the shift operators I mentioned. You can add that the statement is true if $f$ is surjective if you want, although there are examples where the statement is true and $f$ is not surjective. – Victor Zhang Jul 24 '17 at 03:18
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    They key point here is that if $U\cap W$ is zero, then $U+W=U⊕W$ and thus $V=U⊕W$ because the subspace $U⊕W\subseteq V$ has full dimension. – Victor Zhang Jul 25 '17 at 11:24