[ravi-delia]

So you can of course multiply some vectors- the real numbers are a vectorspace, and they have a perfectly nice product. Even R^2 can have a nice multiplication- (a,b)(c,d) = (ac - bd, bc + ad), the product of the complex numbers. However, it's much trickier to define a multiplication on an arbitrary vectorspace V such that

1) For any vectors u,v in V, the product u*v is in V. (this rules out the dot product as a general product)

2) For u,v in V, u*v = v*u

3) For u,v,w in V, v*(u+w) = (v*u) + (v*w)

4) For u,v in V and s in F (the field V is a vectorspace over), s(u*v) = (su)*v = s*(uv)

Under these restrictions you can still cook up products, but fewer and sadder ones. In general you will not have a multiplicative inverse, for instance.

[ngcc_hk]

Any proof you cannot have inverse.

If so, may I wonder if your drop rule 2 and insist on having for most vector u there exist a v so that u*v = 1. As an icing let us say having a 0 vector. That could be something.

[gpm]

This is a theorem that under approximately those rules* in R^n you cannot have inverses except in the cases of the reals, complexes, and quaternions: https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_divi...

*It doesn't rely on rule 2, it does however rely on (u*v)*w = u*(v*w). Adding rule 2 excludes the quaternions.

[gpm]

By the way, the definition of a vector space means we have a 0 vector, and the distribute property means that multiplication has to respect it.

v * w = v * (w + 0) # Vector space property that w + 0 = w

= v * w + v * 0 # Distributive property

Another vector space property guarantees that there exists -(v * w) such that -(v * w) + (v * w) = 0. Add it to both sides

-(v * w) + v * w = -(v * w) + v * w + v * 0

0 = 0 + v * 0 # And then reduce based on the definition of -(v * w)

0 = v * 0 # Vector space property that 0 + x = x

So you have your icing :)