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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 $\operatorname{rank}(AB) = \operatorname{rank}(BA) =n-1$?

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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".

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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.

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  • $\begingroup$ Thanks. What will be the rank($AB$), if $A$ has a lower rank? Is there any opinion? $\endgroup$ – BijanDatta Jun 3 '16 at 18:22
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    $\begingroup$ See Dietrich's answer below: $$ \operatorname{rank}(AB) \geq \operatorname{rank}(A) + \operatorname{rank}(B) - n $$ that's all we can say without knowing more. $\endgroup$ – Omnomnomnom Jun 3 '16 at 18:24
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$\DeclareMathOperator{\rk}{rank}$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 $\rk A=n$: $\rk(AB)=\rk(BA)=\rk B$.

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    $\begingroup$ I would say, the rank of $B$ is the relevant one when the rank of $A$ is $n$! $\endgroup$ – guestDiego Jun 3 '16 at 18:22
  • $\begingroup$ @guestDiego I was referring to the condition that the rank of $B$ is $n-1$, but I agree it was ambiguous. $\endgroup$ – egreg Jun 3 '16 at 19:24

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