What is the rank(AB) and rank(BA)?

Question

Let $A$ and $B$ be two $n\times n$ matrices such that $\operatorname{rank}(A) =n$ and $\operatorname{rank}(B) =n-1$.

Then I know that, $\operatorname{rank}(AB) = \operatorname{rank}(BA) \leq \min\{ \operatorname{rank}(A), \operatorname{rank}(B)\} =n-1$

My question : Is it true that $\operatorname{rank}(AB) = \operatorname{rank}(BA) =n-1$?

Answer 1

If $A$ and $B$ are two matrices of the same order $n$, then $$ \operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n. $$

This implies $n-1\le \operatorname{rank} AB$, so that the answer is "yes".

Answer 2

The answer is yes. In particular, if $A$ is an $n \times n$ matrix with rank $n$ and $B$ is any $n \times n$ matrix, then $$ \operatorname{rank}(AB) = \operatorname{rank}(BA) = \operatorname{rank}(B) $$ this fails be true if $A$ has a lower rank.

Answer 3

$\DeclareMathOperator{\rk}{rank}$Consider that $A$ is invertible, so $$ \rk(AB)\le\rk B=\rk(A^{-1}AB)\le\rk(AB) $$ The information that the rank of $B$ is $n-1$ is irrelevant, so long as $\rk A=n$: $\rk(AB)=\rk(BA)=\rk B$.