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I am trying to solve the following linear algebra problem:

Let $A$ be a $3\times2$ matrix and $B$ be a $2\times3$ matrix such that:

$$ AB = \begin{bmatrix}-2 & -14 & 14 \\ 5 & 15 &-10 \\ 4 & 8 & -3\end{bmatrix}. $$

The following facts are known (and easily verified):

  1. $ \mathrm{rank}(AB) = 2 $
  2. $(AB)^2 = 5AB $

Find the rank of $BA$.

The following is the suggested answer:

$$ \mathrm{rank}(BA) \ge \mathrm{rank}(A(BA)B) = \mathrm{rank}((AB)^2) = 2 $$ $$ \text{Since } BA \text{ is } 2 \times 2, \mathrm{rank}(BA) = 2 $$

I cannot seem to understand why the following statement holds true:

$$ \mathrm{rank}(BA) \ge \mathrm{rank}(A(BA)B). $$

From what I know:

$$ \mathrm{rank}(A) \le \min(m,n) \text{ where $A$ is an } m \times n \text{ matrix } $$ $$ \mathrm{rank}(AB) \le \min(\mathrm{rank}(A),\mathrm{rank}(B)). $$

I cannot see how to relate these to the above mentioned statement. Could someone please enlighten me?

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Note that $$ \operatorname{rank}(A(BA)B) \leq \min\{\operatorname{rank}(A),\operatorname{rank}((BA)B)\} \leq \operatorname{rank}((BA)B) \leq\\ \min\{\operatorname{rank}(BA), \operatorname{rank}(B)\} \leq \operatorname{rank}(BA) $$

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  • $\begingroup$ +1! That was an extremely astute observation. Thank you! $\endgroup$
    – Donald
    Commented Apr 21, 2017 at 14:42

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