# How to confirm and also analyze $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$?

Allow $A$ and also $B$ be 2 matrices which can be increased. After that $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$

I confirmed $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ analyzing $AB$ as a make-up of straight maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and also making use of the kernel-image measurement formula. This additionally gives, in my point of view, a wonderful analysis : if non secure, under succeeding make-ups the bit can just grow, and also the photo can just get smaller sized, in a type of loss of details .

Just how to take care of $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$? Exists a wonderful analysis like the previous one?

0
2019-05-07 13:24:56
Source Share

Once you have actually confirmed $\operatorname{rank}(AB) \le \operatorname{rank}(A)$ , you can get the various other inequality by utilizing transposition and also the reality that it does not transform the ranking (see as an example this question).

Especially, allowing $C=A^T$ and also $D=B^T$ , we have that $\operatorname{rank}(DC) \le \operatorname{rank}(D) \implies \operatorname{rank}(C^TD^T)\le \operatorname{rank} (D^T)$ , which is $\operatorname{rank}(AB) \le \operatorname{rank}(B)$ .

0
2019-05-12 12:12:48
Source

Yes. If you consider An and also B as straight maps, after that the domain name of A is absolutely at the very least as large as the photo of B. Thus when we use A to either of these points, we need to get "extra things" in the previous instance, as the previous is larger than the last.

0
2019-05-09 08:59:46
Source

Prove first that if $f:X\to Y$ and also $g:Y\to Z$ are features in between limited collections, after that $|g(f(X))| \leq \min \{ |f(X)|, |g(Y)| \}.$

Then make use of the very same suggestion.

0
2019-05-09 08:45:30
Source