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

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

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".

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.

• Thanks. What will be the rank($AB$), if $A$ has a lower rank? Is there any opinion? – BijanDatta Jun 3 '16 at 18:22
• 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. – Omnomnomnom Jun 3 '16 at 18:24

$\DeclareMathOperator{\rk}{rank}$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 $\rk A=n$: $\rk(AB)=\rk(BA)=\rk B$.

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