Talk:Calculus/Derivatives of multivariate functions

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On the matrix of the first section, The matrix of a linear transformation, shouldn't the matrix go all the way to $$a_{m,n}$$ instead of $$a_{n,n}$$? As we're talking about a transformation $$L:\R^n\to\R^m$$?

I am really not very confident at Math, so I got confused. If I multiply an $$N \times N$$ matrix by an $$N$$-dimensional vector, the result will be still an $$N$$-dimensional vector right? Is this my mistake or a mistake on the page?

Raideluz (discuss • contribs) 13:30, 18 January 2021 (UTC)


 * I believe you are correct @Raideluz - the theorem is for R^n --> R^m but the proof mostly seems to apply to R^n --> R^n. This needs correcting, I'll look at trying to correct it but like you I'm not very confident with Maths! Also there is a superscript 't' in the proof which I don't understand. Matthew.howey (discuss • contribs) 12:05, 15 May 2021 (UTC)