Linear transformation and idempotence I was wondering... Is there any way to formally  prove that given a linear transformation $L: V \rightarrow V$ over a vector space $V$ if $\text{ker}(L^2)=\text{ker}(L)$ $\leftrightarrow$ $L^2 = L$, hence $L$ is idempotent? This reasoning can be obviously extended to the case  $\text{im}(L^2)=\text{im}(L)$. I would say that things are this way by definition... Any suggestion or comment would be greatly appreciated!
 A: No, $\ker(L^2)=\ker(L)$ just means that $\ker(L) \cap \operatorname{Im}(L)=\{0\}$. For example, if $V$ is vector space over $\mathbb{R}$, and $L:V \to V$ linear transformation such that 
$$L(x)=\lambda x,\quad \lambda \in \mathbb{R}$$
then $\ker(L)=\ker(L^2)$, but $L \neq L^2$, except when $\lambda \in \{0,1\}$.
Indeed, $L^2(x)=\lambda^2 x$. When $\lambda \neq 0$ then 
$$\ker(L)=\ker(L^2)=\{0\}, \quad \operatorname{Im}(L)=\operatorname{Im}(L^2)=V$$
and when $\lambda=0$,
$$\ker(L)=\ker(L^2)=V, \quad \operatorname{Im}(L)=\operatorname{Im}(L^2)=\{0\}$$
so in both cases $\ker(L) \cap \operatorname{Im}(L)=\{0\}$. But when $\lambda \neq 1$, transformation $L$ is not idempotent.
As it turns out, $\ker(L) \cap \operatorname{Im}(L)=\{0\}$ is necessary but not sufficient condition for idempotence of $L$. Instead, the necessary and sufficient condition is given by 
$$\ker(L) \oplus \operatorname{fix}(L)=V$$
where $\operatorname{fix}(L)=\{v \in V|L(v)=v\}$ is the space of fixed points of transformation $L$. 
Indeed, if $\ker(L) \oplus \operatorname{fix}(L)=V$, every vector $v \in V$ can be decomposed into sum $v=v_1+v_2$ of vectors $v_1 \in \ker(L)$ and $v_2 \in \operatorname{fix}(L)$, and furthermore 
$$L(v)=L(v_1)+L(v_2)=v_2,\quad L^2(v)=L(v_2)=v_2$$
which means that $L$ is idempotent. If, on the other hand, $L$ is idempotent transformation, $\operatorname{Im}(L)=\operatorname{fix}(L)$ and for any $v \in V$ 
$$L(v-L(v))=L(v)-L^2(v)=0$$
and $v=(v-L(v))+L(v)$ where $v-L(v) \in \ker(L)$ and $L(v) \in \operatorname{Im}(L)=\operatorname{fix}(L)$. Therefore 
$$\ker(L) \oplus \operatorname{fix}(L)=V$$
since $\ker(L) \cap \operatorname{fix}(L)=\{0\}$ by definition of $\operatorname{fix}(L)$.
