Property of $\ker F$ and $\operatorname{im} F$ where $F$ is an endomorphism I was working on an exercise where I have to deal with an explicit endomorphism $F: V \rightarrow V$ where $V$ is a vectorspace with dimension $n$. 
I want to show that $V$ is a direct sum of $\operatorname{im} F$ and $\ker F$. I showed already that the only vector that $\operatorname{im} F$ and $\ker F$ have in common is the zero vector.
Now I want to show that $V=\operatorname{im} F \oplus \ker F$. 
My approach: we know the following dimensional formula. $n=\dim V=\dim \operatorname{im} F+ \dim \ker F$. 
Let $v_1,...,v_s$ and $v_{s+1},...,v_{n}$ be Bases of $\operatorname{im} F$ and $\ker F$ respectively. We show that $v_1,v_2,...,v_n$ is a Basis of $V$.It is a list of $n$ vectors and since the dimension of $V$ is also $n$ it suffices to show that the vectors are linear independent. For that consider the equation  $\alpha_1 v_1 + ...+ \alpha_n v_n=0$. We rearrange this to  $ \ \ \alpha_1 v_1+...+ \alpha_s v_s=-\alpha_{s+1}v_{s+1}-...-\alpha_n v_n$.
The vector on the right hand side is in $\ker F$ because it is a lin. comb. of it's basis vectors. But it is also in $\operatorname{im} F$ for the same reason. Hence it must be the zero vector ( I already know that the intersection is zero, as stated above). So both sides of the equation above are zero and from linear independece of basisvectors we have $\alpha_1=\alpha_2=...=\alpha_n=0$. The list of vectors $v_1,...,v_n$ is linear independent. 
knowing that they form a basis , we also know that they span the space and hence $V=\operatorname{im} F\oplus \ker F$.
It would be great if you could put some light on  my proof and tell me if it is correct.
thanks in advance.
 A: The result that you want to prove is not true, so your proof is incorrect. In more details, you are right that $n = \text{dim Ker f + dim Im f}$. When you have two vector subspaces $A$ and $B$ of $V$, such that $\text{dim A + dim B = dim V}$, then you have the three equivalent properties:


*

*$A + B = V$;

*$A \cap B = \{0\}$;

*$A \oplus B = V$.
The proof of this fact is more or less contained in your question.
Your mistake is thus when you say that $Ker f \cap Im f = \{0\}$. A proof of this fact would be: let $x$ in the intersection. So we can write $x = f(y)$ and moreover $f(x) = 0$, that is $f \circ f(y) = 0$. We want to show that $f(y) = 0$, otherwise said, we have $y \in Ker (f\circ f)$ and we want $y \in Ker f$.
It is not difficult to show that you have the decomposition $V = Ker f \oplus Im f$ if and only if $Ker f = Ker (f \circ f)$.
Edit: I haven't read carefully and I have thought that $F$ was a general endomorphism. I nevertheless let my answer since I think that the general criterion is interesting.
