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"If matrices B and AB have the same rank, prove that they must have the same null spaces."

I have absolutely NO idea how to prove this one, been stuck for hours now. Even if you don't know the answer, any help is greatly appreciated.

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    $\begingroup$ This has got to be false. Do you mean the dimensions of the null spaces are equal? Wait a minute - maybe it's true. It is easy to show the null space of $B$ is a subset of the null space of $AB$. I see someone just gave an answer, so I'll stop here. $\endgroup$ Commented Oct 10, 2012 at 21:57

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I would begin by showing that the null space of $B$ is a subspace of the null space of $AB$. Next show that having the same rank implies they have the same nullity. Finally, what can you conclude when a subspace is the same dimension as its containing vector space?

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Let $\vec x\in ker(B)$ then $B\vec x=\vec 0$, and so $AB\vec x=A(B\vec x)=A(\vec 0) = \vec 0$. Thus $\vec x \in ker(AB)$, so $ker(B)\subseteq ker(AB)$. Now, $rank(B)=rank(AB)$, so by the rank-nullity theorem, $null(B)=null(AB)$. Now $ker(B)\subseteq ker(AB)$, but $dim(ker(B))=dim(ker(AB))$, thus $ker(B)=ker(AB)$.

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Hint: $B$ and $AB$ have the same number of columns, and according to the rank–nullity theorem ...

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