Trouble with proof that $C^\prime= B^\prime A^\prime$

Question

I am trying to follow a proof that if $C = AB$, then $C^\prime= B^\prime A^\prime$. I attach the image.

I do not understand the last step, since for example it was never shown that $a_{js} = b^\prime_{is}$. Is someone able to tell me what I miss?

Thank you

A is a $m \times n$ matrix, B is a $n \times r$ matrix, and $C = AB$. $A'$ is the transpose. — Cola, Dec 27 '23 at 15:47

score 1 · Accepted Answer · answered Dec 27 '23 at 15:46

Note that $$\sum_{s=1}^na_{js}b_{si}=\sum_{s=1}^nb_{si}a_{js}$$, because $a_{js}$ and $b_{si}$ are numbers (They commute). Since $b_{si}=b_{is}^\prime$ and $a_{js}=a_{sj}^\prime$, we have $$\sum_{s=1}^na_{js}b_{si}=\sum_{s=1}^nb_{is}^\prime a_{sj}^\prime.$$

score 1 · Answer 2 · answered Dec 27 '23 at 15:49

If I understand your notation correctly then I think you mean $A',B',C'$ are the transpose of $A,B,C$, respectively. Assuming that the ring is commutative, one immediately has

$\sum\limits_{s=1}^n a_{js}b_{si}=\sum\limits_{s=1}^n b_{is}'a_{sj}'$, as $a_{js}b_{si}=b_{si}a_{js}=b_{is}'a_{sj}'$.

Trouble with proof that $C^\prime= B^\prime A^\prime$

2 Answers2