tgt
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Why aren't there any legit operation for division of two vectors (any kind of vectors)?
Division of Vectors: Legit Operations Explained
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maze
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In a geometric algebra it makes sense to divide vectors. If a and b are vectors, then the result of a/b would be interpreted as the geometric object such that (a/b)*b = a, where * is the geometric product.
tgt
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what is a geometric product? It is a rotation, reflection or translation?
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He's not talking about "geometry", he's talking about geometric algebra- a lot more complicated.
http://en.wikipedia.org/wiki/Geometric_algebra
http://en.wikipedia.org/wiki/Geometric_algebra
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Ask about multiplication first!tgt said:
I'm skeptical about maze's comment, because geometric algebra has too many zero-divisors; there are generally lots of solutions to equations like bx = a. Furthermore, it's noncommutative, so a solution to bx=a might not be a solution to xb=a. I expect it to be hard to define any sort of coherent division operation.
maze
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Hurkyl said:
Can you please elaborate? The solution to b*x = a is x=b-1*a where b-1=b/||b||. The solution to x*b = a is x = a*b-1 = 1/||b||(a.b+a^b) = 1/||b||(b.a-b^a) so it is almost the same as the solution to b*x = a, except the 2-form portion has the opposite sign.
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I was thinking of the algebra as a whole, rather than just the vectors.maze said:
Specific example: if v is a unit vector, then (1+v)(1-v) = 0, so neither 1+v nor 1-v are invertible in the geometric algebra.
For the opening poster: if you're just working in the vector space, expressions like 1+v are nonsense. They only have meaning in a structure that supports such an operation, like a geometric algebra.
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adriank
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Well, this thread got a bit sidetracked. :)
Before it makes sense to talk about division of vectors, it better make sense to talk about multiplication of vectors. (The geometric algebra that was being discussed talks about one such way.) The natural way to define division from multiplication is this: if c = ab (where a, b, and c are just some abstract things) and you can find an inverse of, say, b (call it b-1), then you should have a = cb-1. (You might write a = c/b.)
You're aware, I hope, of two ways to combine two vectors (in [itex]\mathbb{R}^3[/itex]): the dot product and the cross product. The dot product takes two vectors and gives you a scalar, so that won't help you in trying to divide two vectors. The other one, the cross product, does give you a vector from two vectors, but the main issue is that if you are given vectors c and b and are told that c = a × b, then there isn't a unique solution for a. (For example, given 0 = a × b, you could take a = kb, where k is any scalar.)
Pretty much what this amounts to is this: in order for it to make sense to divide vectors, you probably need to find another way to multiply them first.
Before it makes sense to talk about division of vectors, it better make sense to talk about multiplication of vectors. (The geometric algebra that was being discussed talks about one such way.) The natural way to define division from multiplication is this: if c = ab (where a, b, and c are just some abstract things) and you can find an inverse of, say, b (call it b-1), then you should have a = cb-1. (You might write a = c/b.)
You're aware, I hope, of two ways to combine two vectors (in [itex]\mathbb{R}^3[/itex]): the dot product and the cross product. The dot product takes two vectors and gives you a scalar, so that won't help you in trying to divide two vectors. The other one, the cross product, does give you a vector from two vectors, but the main issue is that if you are given vectors c and b and are told that c = a × b, then there isn't a unique solution for a. (For example, given 0 = a × b, you could take a = kb, where k is any scalar.)
Pretty much what this amounts to is this: in order for it to make sense to divide vectors, you probably need to find another way to multiply them first.
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I should point out that a useful product doesn't have to be of two vectors. e.g. he should know of the product
{scalar} * {vector} = {vector}
For some pairs of vectors, one can solve for a scalar x that satisfies xv=w (and the solution is unique). Of course, most pairs of vectors do not admit a quotient in this manner.
{scalar} * {vector} = {vector}
For some pairs of vectors, one can solve for a scalar x that satisfies xv=w (and the solution is unique). Of course, most pairs of vectors do not admit a quotient in this manner.
maze
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Hurkyl said:I was thinking of the algebra as a whole, rather than just the vectors.
Specific example: if v is a unit vector, then (1+v)(1-v) = 0, so neither 1+v nor 1-v are invertible in the geometric algebra.
For the opening poster: if you're just working in the vector space, expressions like 1+v are nonsense. They only have meaning in a structure that supports such an operation, like a geometric algebra.
ahh thanks for the explanation. For vectors though there are no such issues.