3
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $\text{rank}(AB) = p$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.