2

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!

SKMohammadi
  • 1,091
  • 8
  • 21
  • 1
    The non-zero hypothesis is unnecessary. If one of them is $0$, the sum of the ranks is still $\le 2n$. –  Dec 30 '15 at 15:50
  • If you can prove it for two matrices, then it works for three by taking the two matrices $AB$ and $C$. – Dietrich Burde Dec 30 '15 at 16:28

2 Answers2

4

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.

Community
  • 1
Dietrich Burde
  • 130,978
1

$\newcommand{\rk}{\operatorname{rk}}$ $\newcommand{\im}{\operatorname{im}}$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.