# Finding rank of BA given rank of AB

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?

• Commented May 25, 2019 at 12:03

## 1 Answer

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

• +1! That was an extremely astute observation. Thank you! Commented Apr 21, 2017 at 14:42