4
$\begingroup$

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?

$\endgroup$
  • 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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).$$r\begin{pmatrix} A & C \\ 0 & B \end{pmatrix}\ge r(A)+r(B)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.