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[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

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2 Answers 2

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

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  • $\begingroup$ One can also use that $\operatorname{rank}(BC)\le\min(\operatorname{rank}B,\operatorname{rank}C)$, and conclude that $\operatorname{rank}(BC)=3$. $\endgroup$
    – user26857
    Dec 18, 2019 at 9:50
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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.

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

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