# Kernel Equality

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.

• Using \ker will produce $\ker$ which also has the correct spacing. It is also more appealing to the readers if you put the effort to make the entire question using $\rm\LaTeX$ and don't use "arb." as a short for "arbitrary". – Asaf Karagila Dec 2 '18 at 13:31

$$\mathbf x\in\mathsf{Ker}BA$$ 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 $$\mathbf x\in\mathsf{Ker}A$$.
• What I actually prove is that for every vector $x$ in $\mathsf{Ker}BA$ the vector $Ax$ is an element of $\mathsf{Ker}B$ and an element of $\mathsf{Ran}A$. Then we can draw the conclusion that $Ax=0$. This comes to the same as $x\in\mathsf{Ker}A$. Proved is then that every vector that is an element of $\mathsf{Ker}BA$ is also an element of $\mathsf{Ker}A$, q.e.d.. – drhab Dec 2 '18 at 13:16