Ytukyg

Let $B in mathbb R^{{k}times {m}}$ and $A in mathbb R^{{m}times {n}}$. Further assume that $operatorname{ker}(B) cap operatorname{ran}(A) = {0}.$ Show that this implies $operatorname{ker}(A) = operatorname{ker}(BA).$

I have no idea how to tackle this problem.

$ker(A)subseteqker(BA)$ is obvious.

Suppose $vinker(BA)$. Then $B(Av)=0$, so $Avinker(B)capoperatorname{ran}(A)$. Hence $Av=0$, so $vinker(A)$.

It is not hard, but the structure is a bit verbose, hope that helps.
$$text{Ker}(A)={v in mathbb{R}^n ~|~Av=0}$$
$$text{Ker}(BA)={v in mathbb{R}^n ~|~BAv=0}$$
If $vintext{Ker}(A)Rightarrow Av=0 Rightarrow BAv=0Rightarrow vintext{Ker}(BA)$

or if $Av=uneq0Rightarrow (uintext{Ran}(A)$ and $B(Av)=0 iff Bu=0)$, thus:
$$text{Ker}(BA)= text{Ker}(A) cup {vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }$$
But, $text{Ker}(B)captext{Ran}(A)={0} Rightarrow (uintext{Ran}(A)/{0}Rightarrow unotintext{Ker}(B))$

If $vin mathbb{R}^n/text{Ker}(A) Rightarrow Avintext{Ran}(A)/{0}$, thus (from above) $unotin text{Ker}(B)$, so no element of $vin mathbb{R}^n/text{Ker}(A)$ satisfies $uin text{Ker}(B)$ and ${vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }={}$, finally:
$$text{Ker}(BA)= text{Ker}(A) cup {} = text{Ker}(A)$$