Let $A$ is linear transformation of finite dimensional vector space $V$. Prove equivalence
$a) \,\ker(A)\cap \operatorname{Im}(A)=\{0\}$
$b) \,\ker(A)=\ker(A^2)$
$v) \,\operatorname{Im}(A)=\operatorname{Im}(A^2) $
$a)\to b)$
First prove that $\ker(A)\subset \ker(A^2)$
Let take $x \in \ker(A)$. $A^2x=A0=0$ so for $ \forall x \in \ker(A), x \in \ker(A^2)$
then prove that $\ker(A^2)\subset \ker(A)$
If we take some $x\in \ker(A^2)$. $ A^2x=0$, then $AAx=0$, if $Ax=y$ where $y\in \operatorname{Im}(A)$ then $Ay=0$ so $y\in \ker(A)$ so then $y\in \ker(A)\cap \operatorname{Im}(A)$, since $\ker(A)\cap \operatorname{Im}(A)={0}$ then $y=0$ so that mean $Ax=0$, $x\in \ker(A)$, so we prove $\ker(A^2)\subset \ker(A)$, so $\ker(A)=\ker(A^2)$.
$b)\to v)$
$\dim\ker(A)+\dim\operatorname{Im}(A)=\dim V$
$\dim\ker(A^2)+\dim \operatorname{Im}(A^2)=\dim V$, since $\dim\ker(A)=\dim\ker(A^2)$, that means $\dim \operatorname{Im}(A)=\dim \operatorname{Im}(A^2)$ so $\operatorname{Im}(A)=\operatorname{Im}(A^2)$
$v)\to a)$
since $A=A^2$, then if we take some vector $x\in \ker(A)\cap \operatorname{Im}(A)$, so that mean $Ax=0$, and $Ay=x$, then $AAy=Ax, Ay=Ax, x=Ax,$ since $Ax=0$, that mean $x=0$ so $ \ker(A)\cap \operatorname{Im}(A)=\{0\}$.
is this ok?
$(v) \implies (a)$ uses that $A^2 = A$, but we only have $\operatorname{Im}(A) = \operatorname{Im}(A^2)$.
It might be easier to directly note that $(b) \iff (v)$ because always $\ker A \supseteq \ker A^2$ and $\operatorname{Im} A \subseteq \operatorname{Im}A^2$ and $$\dim(\ker A) = \dim(\ker A^2) \iff \dim V - \dim(\operatorname{Im} A) = \dim V - \dim(\operatorname{Im} A^2) \iff \dim(\operatorname{Im} A) = \dim(\operatorname{Im} A^2)$$
Then $(b) \implies (a)$ is as follows:
Assume that $y \in \ker A \cap \operatorname{Im} A$. Then $\exists x \in V$ such that $y = Ax$. We have $0 = Ay = A^2x$ so $x \in \ker A^2 = \ker A$. Hence $y = Ax = 0$.