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Edited question.

Is it true that for $x\notin\mbox{null}(A)\cup\mbox{null}(B)$, then $ABx\neq0$? Alternately, for what x is it true that $ABx\neq0$?

Thanks!

jeremy
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2 Answers2

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If $A=B$ are nilpotent but $A^2\neq0$, then you conditions are satisfied but $\ker(A)=\ker(B)\neq\ker(AB)=\ker(A^2)$. So any $x\in\ker(A^2)\setminus\ker(A)$ will show your first guess wrong. Even for commuting $A,B$ there is way to express $\ker(AB)=\ker(BA)$ in terms of other kernels.

Marc van Leeuwen
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First of all, referring to the comments, it is not true in general that NS$(AB)=$NS($A)\cup$ NS($B$) (NS referring to null space here). A simple counter example: \begin{equation} A=\begin{bmatrix}2 &1\\2&1\end{bmatrix},\; B=\begin{bmatrix}1 &2\\3&4\end{bmatrix},\;x=\begin{bmatrix}1 \\-2\end{bmatrix}.\end{equation} You can verify that $x \in$ NS($A$), but $x \notin$ NS($AB$).

Ok - so in terms of a characterization of NS($AB$) in terms of NS($A$) and NS($B$): note that $Bx$ is in the column space (I will indicate this as CS) of $B$. So if

  • CS($B)\cap$ NS($A)=\{0\}$ we have NS($AB)=$ NS($B$).
  • CS($B)\cap$ NS($A)=$ CS($B$) we have $ABx=0$ for all $x$.

So these are the two extreme cases and there is then a range of cases in-between.

edit: ok, so I first then concluded that maybe (CS($B)\cap$ NS($A)) \cup$ NS($B)=$ NS($AB$), but as pointed out by Marc, this is also wrong since the union of subspaces is not in general a subspace (basic mistake!). My answer is based on general $A$ and $B$ and not on commuting matrices. Sorry I misunderstood the question, but hopefully there is at least some contribution in my answer.

Christiaan Hattingh
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  • Again the subspace union must certainly be replaced by a sum. But even then it fails; take $A=B$ nilpotent. – Marc van Leeuwen Sep 21 '13 at 21:43
  • hi marc. are you referring to a direct sum? Yes, it seems it is difficult to express the kernel of $AB$ directly in terms of kernel of $A$, since $A$ "acts" on the product $Bx$ and not directly on $x$. So as pointed out in your answer below any argument based on $x \in ker(A)$ is usually not valid because it might be that $Bx \notin ker(A)$ - I made that same mistake in my attempted conclusion above. So it seems CS$(B) \cap$ NS($A)$ is just an indicator of how large the dimension of $ker(AB)$ will be? do you have any comment on this Marc? – Christiaan Hattingh Sep 22 '13 at 05:44
  • Just to add that my answer does not assume $A=B$ nilpotent - I am not familiar with the notation $[A,B]=0$, so maybe I missed that...anyway I would appreciate an answer on my comment above in terms of general matrices $A$ and $B$ instead of nilpotent matrices that are equal, just for my own enrichment. Thanks in advance... – Christiaan Hattingh Sep 22 '13 at 05:50
  • I mean one should write a sum of subspaces $X+Y$ (not necessarily direct) instead of $X\cup Y$ both in the question as in your answer. The point is that $X\cup Y$ is in general not a subspace, in which case it cannot possible be a null space (kernel) of anything, nor a column space (range) for that matter. – Marc van Leeuwen Sep 22 '13 at 06:17
  • And for the dimensions, yes it does hold that $\dim\ker(AB)=\dim\ker(B)+\dim(\ker(A)\cap\operatorname{Range}(B))$, see http://math.stackexchange.com/q/202710/ . – Marc van Leeuwen Sep 22 '13 at 06:21
  • Thanks Marc, for both comments...oi, basic mistake...but ok, I'm still just a student...thanks for the input. – Christiaan Hattingh Sep 22 '13 at 09:10