Why can't we divide two vectors?
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in general a vector space supports only addition and scalar multiplication so the answer would be no.That being said their other algebraic structures in which division makes sense. To divide you first need to multiply so your vector space also have to be an algebra.
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That's an interesting question. What does "divide" mean? We say that x=a÷b if b•x=a, where "•" is some sort of multiplication. If lots of different "x's" satisfy b•x=a then we can't get a unique result for a÷b. For ordinary numbers, either real or complex, we do have a unique solution x for b•x=a, where "•" is ordinary multiplication, so long as b isn't zero. So we can divide numbers by numbers, as long as we don't divide by zero.
So what sort of "•" could be involved with vectors? The most obvious "•" is the vector dot product, which gives an ordinary number for the dot product of two vectors of the same dimension. The problem is this: if the dimension is two or bigger, you can always find various x's with b•x=0, vectors at right angles to b. You can add those x's to any solution to b•x=a and get other solutions. So there's no unique answer for a÷b where a is a number and b is a vector.
You might wonder about the vector cross product for 3D vectors, instead of the dot product. We run into a similar problem: any x parallel to b gives (b cross x) = 0, etc.
Now you might get fancier and think about how the product of a matrix and a vector is another vector. You end up with the same problem. If an answer exists, it's not unique.
So what sort of "•" could be involved with vectors? The most obvious "•" is the vector dot product, which gives an ordinary number for the dot product of two vectors of the same dimension. The problem is this: if the dimension is two or bigger, you can always find various x's with b•x=0, vectors at right angles to b. You can add those x's to any solution to b•x=a and get other solutions. So there's no unique answer for a÷b where a is a number and b is a vector.
You might wonder about the vector cross product for 3D vectors, instead of the dot product. We run into a similar problem: any x parallel to b gives (b cross x) = 0, etc.
Now you might get fancier and think about how the product of a matrix and a vector is another vector. You end up with the same problem. If an answer exists, it's not unique.
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