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):
- $ \mathrm{rank}(AB) = 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?
Note that $$ \operatorname{rank}(A(BA)B) \leq \min\{\operatorname{rank}(A),\operatorname{rank}((BA)B)\} \leq \operatorname{rank}((BA)B) \leq\\ \min\{\operatorname{rank}(BA), \operatorname{rank}(B)\} \leq \operatorname{rank}(BA) $$