rigid plover
Is this proof valid?
15.Let $M$ be a $m\times n$ matrix, and $A$ be a $n\times p$ matrix. $(MA){ij}$ is the ith element of jth column of $MA$.
$$
(MA){ij}=\sum_{k=1}^{n}M_{ik}A_{kj}=\sum_{k=1}^{n}M_{ik}\left( \sum_{l=1}^{p}c_{l}A_{kl} \right)=\sum_{l=1}^{p}c_{l}\left( \sum_{k=1}^{n}M_{ik}A_{kl} \right)=\sum_{l=1}^{p}c_{l}(MA)_{il}
$$
fervent owlBOT
fervent owlBOT
noble forgeBOT
anonymous190
primal pier
Is this proof valid?
15.Let $M$ be a $m\times n$ matrix, and $A$ be a $n\times p...
I believe one can write this in a better way using the fact that $(MA){*, j}=MA{*, j}$
noble forgeBOT
Lana
rigid plover
I believe one can write this in a better way using the fact that $(MA)_{*, j}=MA...
Is it M times Aij?
primal pier
Is it M times Aij?
if you mean A_{*, j}, then it means the j-th column viewed as a (number of rows)x1 matrix vector
rigid plover
$$
(MA){*j}=MA{*j}=M\sum_{l=1}^{p}c_{l}A_{*l}=\sum_{l=1}^{p}c_{l}MA_{*l}=\sum_{l=1}^{p}c_{l}(MA)_{*l}
$$
noble forgeBOT
anonymous190
rigid plover
@primal pier
primal pier
<@1427076721508876390>
It's correct 👍
fervent owlBOT
Thank you for your feedback! Lana has been awarded 1 :helper_points:. They now have 7 :helper_points:. They have 3 :helper_points: daily left for today.