How to show that $Rank(AB)\geq Rank(A)+Rank(B)-n$

Question

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?

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. — Jim, Feb 19 '15 at 16:47
@Vim to be fair, you can't google for it if you don't know the name. — yurnero, Feb 20 '15 at 03:29

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

1

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

edited Apr 13 '17 at 12:21

Community

1

answered Feb 20 '15 at 03:37

Vim

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How to show that $Rank(AB)\geq Rank(A)+Rank(B)-n$

1 Answers1