@nairalos ...and so on, so that the nth edge lies in the x1...xn hyperspace
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Replying to @St_Rev
@nairalos now convince yourself that if the (n-1)-volume of the paralelliped generated by the first n-1 edges is V, and the nth edge...
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Replying to @St_Rev
@nairalos ...has coords xn = (xn1,xn2,...,xnn,0,...,0), that the n-volume of the paralleliped generated by the first n edges is Vxnn
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Replying to @St_Rev
@nairalos basically the geometry of rotating the paralleliped so the edges lie in the x, xy, xyz etc. spaces corresponds to putting the...
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Replying to @St_Rev
@nairalos ...corresponding matrix in lower-triangular form. and it makes it easy to see that the sub-volumes correspond to the product...
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Replying to @St_Rev
@nairalos of the diagonal entries. but that IS the determinant when the matrix is triangular.
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Replying to @fire__exit
@nairalos Yeah you will need to walk it through to understand it but I think this is the most compact 'intuitive' way to see it.
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Replying to @St_Rev
@nairalos I had someone ask me the exact same question a few months ago which is why I had the explanation ready to bust out :)
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@nairalos That too. It would just have taken a little while longer to think it through.
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. Banned in Sweden. SubGenius, Zhuangist, white-hat troll. Defrocked mathematician. Brain problems.