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Let $A,B,C\in M_n(\mathbb{R})$ be nonzero matrices such that $ABC=0$. How can we prove that $\operatorname{rank}(A)+\operatorname{rank}(B)+\operatorname{rank}(C)\le 2 n$ ?

I can prove this for two matrices, but in this case, i can't!

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    $\begingroup$ The non-zero hypothesis is unnecessary. If one of them is $0$, the sum of the ranks is still $\le 2n$. $\endgroup$ – user228113 Dec 30 '15 at 15:50
  • $\begingroup$ If you can prove it for two matrices, then it works for three by taking the two matrices $AB$ and $C$. $\endgroup$ – Dietrich Burde Dec 30 '15 at 16:28
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$\newcommand{\rk}{\operatorname{rk}}$ $\newcommand{\im}{\operatorname{im}}$You know that $\im(BC)\subseteq \ker A$. Hence, $n= \rk A+\dim\ker A\ge \rk A+\rk(BC)$.

Now, if you consider the restriction of multiplication by $B$ to the subspace $\im C$ and use rank-nullity theorem, you get $\rk(BC)=\rk C-\dim\ker(B|_{\im C})$. Hence \begin{align} n\ge \rk A+\rk(BC)&=\rk A+\rk C-\dim(\ker B\cap \im C)\ge \rk A+\rk C-\dim\ker B=\\&=\rk A+\rk C+\rk B-n \end{align}

Whence the thesis.

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By Silverster's rank theorem we have $$ rk(ABC)\ge rk(AB)+rk(C)-n\ge rk(A)+rk(B)+rk(C)-2n. $$ With $ABC=0$ the claim follows.

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