Talk:Geometric algebra: Difference between revisions

:::::::::::::::The proof that the geometric product defines a scalar product is not too hard. First you define the inner product of two vectors <math>\mathbf{a}\cdot\mathbf{b} = (\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{a})/2</math>. It's straightforward to see that it is a symmetric and bilinear. What's not so obvious is that it is a form, that is that it returns a scalar. To see it you just notice the equality : <math>\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{a} = \mathbf{a}^2 - \mathbf{a}^2 + \mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{a} + \mathbf{b}^2 - \mathbf{b}^2 = (\mathbf{a} + \mathbf{b})^2 -\mathbf{a}^2 - \mathbf{b}^2</math>, and this is a real number because it's a linear combination of real numbers.

:::::::::::::::It's not very complicated a proof, but it's quite irrelevant to the task and putting it in the description would spam it imho. I won't add it unless other people complain.--[[User:Grondilu|Grondilu]] ([[User talk:Grondilu|talk]]) 11:21, 20 October 2015 (UTC)

 

== "Orthonormal basis" ==