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Problem

Let $A, B, C \in M_n(\mathbb{R}).$ If $ABC = 0$ and rank$(B) = 1$, then either $AB = 0$ or $BC = 0$.

My idea

I only know that rank$(AB) \leq$ rank$(B)=1$ and rank$(BC) \leq$ rank$(B)=1$. However, it doesn't help. Is there any inequality between rank$(B)$, rank$(AB)$, rank$(BC)$ and rank$(ABC)?$

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    $\begingroup$ Related $\endgroup$
    – A.Γ.
    Jan 8, 2018 at 14:21

1 Answer 1

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Here's an approach I like: since $B$ has rank $1$, we can write $$ B = uv^T $$ for some non-zero vectors $u,v \in \Bbb R$. As such, we can rewrite $$ ABC = Auv^TC = (Au)(C^Tv)^T $$ It now suffices to note that for vectors $x,y \in \Bbb R^n$, we have $xy^T = 0$ if and only if $x = 0$ or $y = 0$.

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  • $\begingroup$ can't $x,y$ be orthogonal and non zero? $\endgroup$
    – clark
    Jan 8, 2018 at 14:30
  • $\begingroup$ @clark That isn't the inner product, though, that would be $x^Ty$, not $xy^T$. $\endgroup$
    – Aaron
    Jan 8, 2018 at 14:32
  • $\begingroup$ ooops sorry for the disturbance $\endgroup$
    – clark
    Jan 8, 2018 at 14:33

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