Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$

I proved $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ by interpreting $AB$ as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of

loss of information.How do you manage $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$? Is there a nice interpretation like the previous one?

**Answer**

Yes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get “more stuff” in the former case, as the former is bigger than the latter.

**Attribution***Source : Link , Question Author : Community , Answer Author : Kevin H. Lin*