Transpose of a vector | Matrix transformations | Linear Algebra | Khan Academy

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Transpose of a column vector. Matrix-matrix products using vectors
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[@vigneshkumarreddyvennapusa4867]

congratulations khan sir on getting 8M subscribers

[@alkalait]

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)

[@kiob]

Have to ask, how is [v1w1+v2w2+....vnwn] is a 1x1 matrix? Isn't it a 1xn matrix? shouldn't a 1x1 matrix be just one #? I'm quite confused.

[@nacholopez7348]

Didn't watch the video and didn't get too deep into the topic but I guess it's because you can ad those up to one number/one value in the matrix.

[@prolarka]

I cant find the proof of (AB)^T=(B^T)*(A^T) in previous videos.

[@dawnfantasy]

It is in 'Transpose of a matrix product'.

[@rajkhushkumar7560]

Nicely Explained

[@debendragurung3033]

11: 04 bookmark for later project.

[@mirafzalmirjaloliy2rug44]

anyone coming from econometrics?