Query about another post $\operatorname{rank}(AB) \leq \min(\operatorname{rank}(A), \operatorname{rank}(B))$

Question

For, $\operatorname{rank}(AB) \leq \min(\operatorname{rank}(A), \operatorname{rank}(B))$

On this post I don't understand why it is enough to prove $\dim \operatorname{range}(AB)\leq \dim R(A)$ and $\operatorname{range}(AB)\leq \dim R(B)$. Wouldn't the "min" part play a role as well?

You don't need \operatorname{min}; you can just write \min. And in a "displayed", as opposed to "inline", context, that effects positioning of subscripts, thus: $$ \min_{x\in\mathcal X} f(x) $$ — Michael Hardy, Apr 12 '16 at 03:35

score 1 · Accepted Answer · answered Apr 12 '16 at 03:36

1

Proving both $a\le b$ and $a\le c$ is the same as proving $a\le\min\{b,c\}$.

answered Apr 12 '16 at 03:36

Michael Hardy

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Thanks. That is good to know. – MathIsHard Apr 12 '16 at 03:37

Query about another post $\operatorname{rank}(AB) \leq \min(\operatorname{rank}(A), \operatorname{rank}(B))$

1 Answers1