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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.

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  • $\begingroup$ 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". $\endgroup$ – Asaf Karagila Dec 2 '18 at 13:31
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$\mathbf x\in\mathsf{Ker}BA$ 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 $\mathbf x\in\mathsf{Ker}A$.

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  • $\begingroup$ Could you maybe explain this a bit more in detail? $\endgroup$ – Mathmeeeeen Dec 2 '18 at 13:09
  • $\begingroup$ Do you understand and agree with the first line? $\endgroup$ – drhab Dec 2 '18 at 13:10
  • $\begingroup$ Yes I do understand it. $\endgroup$ – Mathmeeeeen Dec 2 '18 at 13:15
  • $\begingroup$ 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.. $\endgroup$ – drhab Dec 2 '18 at 13:16
  • $\begingroup$ Thank you very much that helped me a lot! $\endgroup$ – Mathmeeeeen Dec 2 '18 at 13:26

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