Definition talk:Inner Product

Maybe there is need for a proof that $=$? --Espen180 22:25, 22 October 2009 (UTC)

I would say so, definitely. Feel free to write such a page. --Matt Westwood 05:28, 23 October 2009 (UTC)

It's done. I didn't know what to call it though. Espen180 05:51, 23 October 2009 (UTC)

Complex and reals?
How important is it to specify "or $\R$" in that first sentence? A subfield of $\R$ is by definition also a subfield of $\C$ is it not? And if you have the language of abstract algebra under your belt enough to understand what a subfield is, you'll know that. Furthermore, the concept of an "inner product" is fairly well advanced down the route of abstract algebra (coming as it does after vector spaces) that such an interpolation would seem clumsy. --prime mover 04:43, 29 April 2012 (EDT)
 * Well, in my class we're doing inner product spaces, and we haven't been introduced to fields yet. But maybe my class's curriculum is pathological. Fraleigh has inner product spaces as chapter $3.5$ and fields as $9.2$. We're introduced to it as a specific type of a regular vector space, and answering questions like "is $\left \langle {\cdot, \cdot} \right \rangle : \mathbf M_2 \left({\R}\right) \times \mathbf M_2 \left({\R}\right) \to \mathbb \R, \left \langle {\begin{bmatrix} a_1 & a_2 \\ a_3 & a_4 \end{bmatrix}, \begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \end{bmatrix}} \right \rangle = a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4$ an inner product?" --GFauxPas 07:46, 29 April 2012 (EDT)


 * IMO Fraleigh *is* pathological. OK so what's anyone else think? --prime mover 09:51, 29 April 2012 (EDT)