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$A$ 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 $2$?

3 Answers 3

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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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Rank of( AB) is less than or equal to min {rank (A),rank(B)}. So, it may not always be equal to rank of AB . Similarly for rank of BA where A and B are n×n matrix.

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    if you refer to @martini's answer, you can see that the "Similarly for rank of $BA$" is wrong. – 2017-01-11
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    Yes, it was for matrix with n×n. – 2017-01-11
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    then you should clarify that in your answer, given that the OP asks about matrices that are **not** square matrices. – 2017-01-11
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No, a simple counter example:

$$\left(\begin{array} & 1 & 0 &0\\ 0&1&0\end{array}\right) \left(\begin{array} & 0 & 0\\ 1&0 \\ 0&1\end{array}\right) = \left(\begin{array} & 0 & 0 \\ 1&0\end{array}\right).$$