TITLE: Rank of matrices product
QUESTION [0 upvotes]: $A$ is a $2\times3$ matrix, $B$ is a $3\times2$ matrix, $\text{rank}(A)=\text{rank}(B)=2$
Does always $\text{rank}(AB)$ equal to $2$?

REPLY [0 votes]: No, a simple counter example:
$$\left(\begin{array} & 1 & 0 &0\\ 0&1&0\end{array}\right) \left(\begin{array} & 0 & 0\\ 1&0 \\ 0&1\end{array}\right) = \left(\begin{array} & 0 & 0 \\ 1&0\end{array}\right).$$

REPLY [0 votes]: One answer is no. Consider the matrices 
$$ A = \begin{pmatrix} 1 & 0 & 0\\
                       0 & 1 & 0\end{pmatrix}, \qquad B = \begin{pmatrix}1 & 0\\
                                                                         0 & 0\\
                                                                         0 & 1 
\end{pmatrix}
$$
Then $\def\r{\mathop{\rm rank}}\r A = \r B = 2$. We have 
$$ AB = \begin{pmatrix} 1 & 0\\
                        0 & 0\end{pmatrix} $$
hence $\r (AB) = 1$.
But $\r BA = 2$ (always): Note that $A$ maps $k^3$ onto $k^2$, that is the range of $BA$ equals the range of $B$, has therefore dimension $2$, hence $\r BA = 2$.
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