Let $\operatorname{rank}(A_{3 \times 3})=\operatorname{rank}(B_{3 \times 3})=2$. I need to figure out whether $AB=0$ is possible. On the one hand, I know that $\operatorname{rank}(AB) \leq \min(\operatorname{rank}\ A, \operatorname{rank}\ B)$, so there is a possibility that $\operatorname{rank}(AB)=0$ (and thus $AB=0$). On the other hand, I have a hard time finding an appropriate example of matrices for which it holds. Maybe I'm missing something very trivial.
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1Sylvester rank inequality – sas Dec 23 '14 at 17:33
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@sas - Thanks. Didn't know about this theorem. – user202918 Dec 23 '14 at 17:37
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@sas: Why don't you put that comment as an answer? – voldemort Dec 23 '14 at 17:43
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1@voldemort: I didn't think about it as a proper answer. Well, OK, I'll add. – sas Dec 23 '14 at 17:47
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You can't find an example. Consider the endomorphisms $f$ and $g$ of $ K^3$ (K is the base field) associated with A and B respectively. AB=0 means that $f\circ g=0$, or $\mathrm{im}\: g \subset \ker f $. As $\dim\:(\mathrm{im}\: g)=2 ,$ this implies $\dim\:\ker f \geq 2$, which contradicts the rank theorem.
Bernard
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The thing is that we haven't learned about endomorphisms nor about the Sylvester rank inequality, so I thought there exists a simpler proof for why $AB \neq 0$. But thanks! – user202918 Dec 23 '14 at 17:49
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You misunderstood my answer. I meant the equivalent of the rank-nullity theorem for linear maps. The Sylvester rank inequality is of a more technical nature. – Bernard Dec 23 '14 at 17:57
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$AB$ cannot be the zero matrix. proof by contradiction. if $AB$ is the zero matrix, then every column of $B$ is in the null space of $A.$ that $rank(B) = 2$ shows that the null space of $A$ has dimension $2.$ then $rank(A) + dim(ker(A)) = 2 + 2 = 4$ but the nullity theorem implies $rank(A) + dim(ker(A)) = 3$ and that is a contradiction we sought.
abel
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