# Rank of $A=BC$, when ranks of $B,C$ are given.

[NBHM-PhD Screening test-2015, Algebra(Q 1.7)]

Let $$B$$ be a $$5\times3$$ matrix and let $$C$$ be a $$3\times5$$ matrix, both with real entries. Set $$A=BC$$. Then what are the possible ranks of $$A$$ when

(1) both $$B$$ and $$C$$ have rank $$3$$

(2) both $$B$$ and $$C$$ have rank $$2$$

I know that $$\operatorname{rank}A\leq \min(\operatorname{rank}C,\operatorname{rank}B)$$. From this I can say in first case $$\operatorname{rank}A\leq 3,$$ and in second case $$\leq2$$. What more we can say about rank of $$A$$?

also is there any general method to attack such kind of problems

You can use Sylvester's rank inequality $$\operatorname{rank}(BC) \geq \operatorname{rank}(B)+ \operatorname{rank}(C)-n$$ when $$B$$ is $$m \times n$$ and $$C$$ is $$n \times k$$ to conclude that

(i) $$\operatorname{rank}(BC) \ge 3+3-3=3$$

(ii) $$\operatorname{rank}(BC) \ge 2+2-3=1$$

• One can also use that $\operatorname{rank}(BC)\le\min(\operatorname{rank}B,\operatorname{rank}C)$, and conclude that $\operatorname{rank}(BC)=3$. Dec 18, 2019 at 9:50

If both $B$ and $C$ have rank $3$, then $\operatorname{rank}A=3$. In order to see why, note that $C.\mathbb{R}^5=\mathbb{R}^3$. Therefore, since $\operatorname{rank}B=3$, $A.\mathbb{R}^5=B.(C.\mathbb{R}^5)=B.\mathbb{R}^3$, which has dimension $3$.

On the other hand, if $\operatorname{rank}B=\operatorname{rank}C=2$, $\operatorname{rank}A$ may be equal to $2$, but it may be smaller, too.

• Sir can you please explain, what is this notation is $C.\mathbb{R}^5=\mathbb{R}^3$.
– user464147
Oct 11, 2017 at 3:07
• @N.Maneesh $C.\mathbb{R}^5=\{C.v\,|\,v\in\mathbb{R}^5\}$. Oct 11, 2017 at 6:15