Talk:Geometric algebra: Difference between revisions
→A possible source of confusion
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:::::::::::::::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)
:::::::::::::::: You might want to take a look at the wikipedia writeup on [[wp:Vector_space|vector spaces]]. At the very least you will have to agree that some people find that concept to be reasonable. More than that, though, the axioms themselves do not specify what kind of product they use - which means that the reader should be able to determine that. Of course, other statements will constrain this, but baring considerable familiarity with the topic, we are left with the sort of trial and error (or hypothesis and test) that leads to talk pages as large as this one. --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 11:
== "Orthonormal basis" ==
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