better lower bound on rank of product of matrices

1,158

You assume that $B$ has the size $n\times p$ and $n>p$, which implies that $\text{rank}B\le\min\{n,p\}=p<n$. Hence, it just cannot have rank $n$. The best you can get with the full column rank for $B$ is $$ \text{rank}(AB)\ge\text{rank}(A)+p-n=\text{rank}(A)-(n-p). $$

Share:
1,158

Related videos on Youtube

Annonymous
Author by

Annonymous

Updated on December 20, 2020

Comments

  • Annonymous
    Annonymous almost 3 years

    We know that if $A,B$ are $m\times n$ and $n\times p$ matrices respectively (let $n>p>m$), then $\text{rank}(AB)\geq \text{rank}(A)+\text{rank}(B)-n$ by Sylvester's inequality. Is it also true that $\text{rank}(AB)\geq \text{rank}(A)$ if B has full column rank?