Could someone tell me if my proof is right?
Problem:
Let $B^{k \times m}$ and $A^{m \times n}$ be two matrices and let $\mathrm{Ker}(B) \cap \mathrm{Ran}(A) = \{0\}$. Show that this implies that $\mathrm{Ker}(A) = \mathrm{Ker}(BA)$
My solution:
$\mathrm{Ker}(A) \subset \mathrm{Ker}(BA)$ always holds, since $$x \in \mathrm{Ker}(A) \implies Ax=0\implies BAx= B0\implies x\in \mathrm{Ker}(BA)$$ $\mathrm{Ker}(BA) \subset \mathrm{Ker}(A)$ since $$x \in \mathrm{Ker}(A) \implies BAx=0\implies Ax\in \mathrm{Ker}(A),$$ as $$\mathrm{Ker}(B) \cap \mathrm{Ran}(A) = \{0\} \implies Ax \in \mathrm{Ran}(A) \implies Ax= 0 \implies x\in \mathrm{Ker}(A)$$
2nd implication: "Ker(BA)⊂Ker(A) since x∈Ker(A) implies BA.x=0⟹A.x∈Ker(A), as Ker(B)∩Ran(A)=0⟹A.x∈Ran(A)⟹A.x=0⟹x∈Ker(A)."
Let $x\in \ker(BA)$. Then $B(Ax) = BAx=0$ and so $Ax\in\ker (B)$.
But $\ker(B)\cap range(A)=0$ and so $Ax$ cannot lie in the range of $A$. Thus $Ax=0$ and so $x\in\ker(A)$.