I read here that if $V$ is a vector space and $T : V \to V$ is a projection then $V = \text{ker}(T) \oplus \text{im}(T)$.
I came up with a counterexample of the converse:
$$S : \mathbb{K}_n[x] \to \mathbb{K}_n[x]$$
$$ p(x) \mapsto x p^{(1)}(x)$$
It's easy to see that $\text{ker}(S) = \text{span}(1)$ and $\text{im}(S) = \text{span}(x, \ldots, x^n)$, but $S \circ S \neq S$.
My question is if there is any characterization(necessary and sufficient condition which is not the definition of direct sum) of the endomorphisms $T$ such that $V = \text{ker}(T) \oplus \text{im}(T)$.
This is a wannabe proof based on the reply of Sami Ben Romdhane:
First, if $V = \text{ker}(T) \oplus \text{im}(T)$ then by definition for any $v \in V$ there exists unique $k \in \text{ker}(T) \subseteq V$ and $i \in \text{im}(T) \subseteq V$ such that $v = k + i$.
By the "unique part" the function $S : V \to V$ which send each $v$ to his $k$ is well defined. Furthermore it is a linear map since the image and the kernel of $T$ are vector spaces.
It is now trivial that $(S+T)(v) = S(v) + T(v) = k + i = v = \text{id}_V$ (so $S+T$ is invertible), and $(T \circ S)(v) = T(k) = 0$ by definition of kernel.
Now the converse.
Suppose that $S + T$ is invertible and that $T \circ S = 0$. The second condition is equivalent to say that $S(V) \subseteq \text{ker}(T)$.
$S+T$ is invertible iff it is an isomorphism, so for any $v \in V$ there exist a unique $w \in V$ such that $v = (S+T)(w) = S(w) + T(w)$, but $S(w) \in S(V) \subseteq \text{ker}(T)$ and $T(w) \in \text{im}(T)$, so for any $v$ in $V$ there exists unique $k \in \text{ker}(T)$, $i \in \text{im}(T)$ such that $v = k + i$ and therefore $V = \text{ker}(T) \otimes \text{im}(T)$.
Best Answer
Hint
Prove that :
Remark: Notice that in the case when $T$ is a projection the endomorphism $S=\operatorname{id}-T$.