So for my university studies I was given this problem:
Let B ∈ R^k×m and A ∈ Rm×n . Further assume that Ker(B) ∩ Ran(A) = {o}. Show that this implies Ker(A) = Ker(BA).^
To this point I have come so far with that problem:
For showing set equality you have to show that
$Ker(A) \subset Ker(BA) \,\,\, \land \,\,\, Ker(BA) \subset Ker(A)$
I have managed to show the easier inclusion myself like this:
Show that:
$Ker(A) \subset Ker(BA)$
Let $ x \in R^n $ be an arb. vector such that $ Ax = 0$
Now look at $BAx$
$BAx = B(Ax) = B (0) = B\cdot 0 = 0$
$ \Rightarrow$ $Ker(A)$ is indeed a subset of $Ker(BA)$ Since $x$ was choosen arb.
But now im having a hard time to find a rigorous proof for the other inclusion, I started with this:
Show that:
$Ker(BA) \subset Ker(A)$
Now let $ x \in R^n$ be an arb. vector such that $BAx = 0$ Now show that also $Ax = 0$.
$BAx = 0$ $y:= Ax, \,\,\, y \in R^m$
$By = 0$
But this is it i haven't come further and I have no clue how to proceed especially how to use that one condition stated in the problem ($ Ker(B) \cap Ran(A) =$ {0}).
I would be very glad if someone could help me with this.
$\mathbf x\in\mathsf{Ker}BA$ or equivalently $BA\mathbf x=\mathbf0$ implies that $A\mathbf x\in\mathsf{Ker}B$.
Also we have $A\mathbf x\in\mathsf{Ran}A$, so $A\mathbf x\in\mathsf{Ran}A\cap\mathsf{Ker}B=\{\mathbf0\} $.
So actually we have $A\mathbf x=\mathbf0$ or equivalently $\mathbf x\in\mathsf{Ker}A$.