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June 24

Clifford algebra vs. geometric algebra

Judging from the articles, Clifford algebras and geometric algebras are defined almost exactly the same way, but then the articles don't seem very specific about the relationship between the two. Is one a variation/generalization of the other? --RDBury (talk) 03:42, 24 June 2018 (UTC)[reply]

My understanding is a geometric algebra is a real Clifford algebra with a non-degenerate quadratic form. So Clifford algebras are a generalisation of geometric algebras. The name "geometric algebra" emphasises the geometric nature of these algebras, as most interesting applications of Clifford algebras arise from such geometric algebras.

I know though that this is not universal – my favourite reference on geometric algebra, Lounesto’s "Clifford Algebras and Spinors" call such algebras Clifford algebras. Other sources call them geometric algebras – Hestenes e.g..--JohnBlackburne^words_deeds 04:21, 24 June 2018 (UTC)[reply]

I gather it depends on context as well; the videos I'm watching on geometric algebra talk about blades and oriented areas but I don't think you'd see that much in a discussion of Clifford algebras even if the underlying structure is the same. Your answer clears things up a lot, but I was already a bit familiar with Clifford algebras in the context of representation theory, so perhaps 'interesting' is in the eye of the beholder. --RDBury (talk) 10:34, 24 June 2018 (UTC)[reply]

Trouble understanding article on Multilateration

Question on Multilateration

I have asked this question before, but I got a totally confusing answer - it implied that the underlying maths of the provided solution were incorrect. I don't believe this likely. It is almost certainly my misunderstanding, so I'll ask again.

I have been having great difficultly implementing equation 7 in the article.

I have 3 receivers. I have modified equation 7 by simply removing the z terms (and Cm), as I can only do a 2D fix with 3 receivers. So I have receivers P0, P1 and P2, i.e. 0 <= m <= 2 as per Fig 2.

When you substitute the values into equation 7, I only have a single set of data m=2 to substitute. This gives me 1 equation with 2 unknowns (x,y). There will be an infinite number of solutions.

What am I missing? — Preceding unsigned comment added by Mhillman (talk • contribs) 14:13, 24 June 2018 (UTC)[reply]

Courtesy link: Wikipedia:Reference_desk/Archives/Mathematics/2017 November 8. Since I was the one who wrote the "confusing answer" before I'll refrain from further comment. --RDBury (talk) 17:14, 24 June 2018 (UTC)[reply]

Transforming N nonlinear equations into N linear equations is generally not possible. But you may be able to transform them into N equations of which one is nonlinear and the others are linear. If there are more equations than unknowns, then you may discard the nonlinear one and solve the linear ones. I think that is what happens here. Bo Jacoby (talk) 14:37, 25 June 2018 (UTC).[reply]