Please have me with the following question:

$A$ is a $m \times n$ matrix with rank $m$, $B$ is a $n \times p$ matrix with rank $p$. Given that $p<m<n$. Is there condition of $A$,$B$ that $\operatorname{rank}(AB)=p$?

An part of answer is here Full-rank condition for product of two matrices

Many thanks!


I realize this post was made nearly 2 years ago, but I'll answer this for the sake of anyone searching the web for an answer.

Note that $\text{ker}(B) \subset \text{ker}(AB)$, so $\text{nullity}(B) \leq \text{nullity}(AB)$. Using the relationship between rank and nullity, we have that $$n - \text{rank}(B) = \text{nullity}(B) \leq \text{nullity}(AB) = p - \text{rank}(AB)$$ $$\implies \text{rank}(AB) \leq \text{rank}(B) - (n - p) < \text{rank}(B) = p$$ Hence, there is no condition which can be imposed on $A, B$ which will make $\text{rank}(AB) = p$.

  • $\begingroup$ Shouldn't rank(B) + nullity(B) = p instead of n? $\endgroup$ – is it normal Jul 4 '19 at 15:58

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