Let $A\in M_{m \times n}$ and $B\in M_{n \times k}$. Prove that

$$Rank(AB)\geq Rank(A)+Rank(B)-n.$$

I have tried to use $Im(AB) \subseteq Im(B)$ but that lead me to nowhere, how should I approach this prove?

  • 1
    $\begingroup$ I've edited your title. Please don't put all math in a title, it prevents users from right clicking and opening your question in a new page, which is a common way of browsing this site. $\endgroup$ – Jim Feb 19 '15 at 16:47
  • $\begingroup$ That's Sylvester Inequality, you can google for it first. $\endgroup$ – Vim Feb 19 '15 at 16:48
  • $\begingroup$ @Jim I will do it from now on $\endgroup$ – gbox Feb 19 '15 at 16:48
  • $\begingroup$ @gbox use null space ... $\endgroup$ – Math-fun Feb 19 '15 at 16:53
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    $\begingroup$ @Vim to be fair, you can't google for it if you don't know the name. $\endgroup$ – yurnero Feb 20 '15 at 3:29

See here for a simple proof (also my favorite one), based on the facts:
(1)Generalized elementary transformation does not change the rank of a matrix.
(2).$$r\begin{pmatrix} A & C \\ 0 & B \end{pmatrix}\ge r(A)+r(B)$$


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