i know it hasn't been demonstrated by Sal, but any matrix product can be represented as a sum of outer products (column-row products). all vectors by default are column vectors regardless of how I label them:
(AB)^T = [Acol1*(Brow1)^T + Acol2*(Brow2)^T +...+ Acoln*(Brown)^T]^T =
[Acol1*(Brow1)^T]^T + [Acol2*(Brow2)^T]^T +...+ [Acoln*(Brown)^T]^T =
Brow1*(Acol1)^T + Brow2*(Acol2)^T +...+ Brown*(Acoln)^T =
(since now all the rows of B are represented as columns we might as well say...)
(B^T)*(A^T)
alkalait 2 years ago
I cant find the proof of (AB)^T=(B^T)*(A^T) in previous videos.
prolarka 2 years ago
i know it hasn't been demonstrated by Sal, but any matrix product can be represented as a sum of outer products (column-row products). all vectors by default are column vectors regardless of how I label them:
(AB)^T = [Acol1*(Brow1)^T + Acol2*(Brow2)^T +...+ Acoln*(Brown)^T]^T =
[Acol1*(Brow1)^T]^T + [Acol2*(Brow2)^T]^T +...+ [Acoln*(Brown)^T]^T =
Brow1*(Acol1)^T + Brow2*(Acol2)^T +...+ Brown*(Acoln)^T =
(since now all the rows of B are represented as columns we might as well say...)
(B^T)*(A^T)
alkalait 2 years ago
I cant find the proof of (AB)^T=(B^T)*(A^T) in previous videos.
prolarka 2 years ago