# Why is the × operator not defined for vector × vector?

For a vector space, the + operator maps two vectors to another vector while the × operator maps a scalar and a vector to another vector.

To me, it seems strange that scalars are seen as separate to vectors when defining a vector space, with the x operator being specially created to map a scalar and vector to vectors.

Why can't instead the × operator map two vectors to another vector while remaining consistent with the scalar × vector operation?

For example, $(a, b, c) × (d, e, f) = (ad, ae, af) + (bd, be, bf) + (cd, ce, cf)$ where $ad$ scales $d$ by a factor of $a$; $bd$ rotates $d$ by an amount $b$ about some axis, $cd$ rotates $d$ about another axis by $c$.

This way there would be no need to bring in an additional set of scalars, providing it was consistent.

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There are many different possible vector multiplications. For vectors in $\mathbb{R}^3$ there is the cross product. For all $\mathbb{R}^n$ there is a componentwise product: $(a,b,c)(x,y,z)=(ax,by,cz)$. The problem is these products don't obey all of the usual algebraic laws you are used to. For example, the cross product is not commutative and not associative. The componentwise product allows for zero divisors. In general a vector space equipped with a multiplication is called an algebra. –  Bill Cook Oct 29 '11 at 22:06
Don't use $\times$ to represent the scalar multiplication, since it is usually reserved for the cross product –  Arturo Magidin Oct 29 '11 at 22:06
It seems that the way you defined $\times$, we have $(a,b,c) \times (d,e,f) \stackrel{(def)}{=} (a+b+c) (d, e, f)$ which is again a scalar multiplication. Are you sure this is what you want? –  Srivatsan Oct 29 '11 at 22:07
Something might work out along your lines for $\mathbb{R}^3$, but there seems to be nothing natural for arbitrary vector spaces. –  André Nicolas Oct 29 '11 at 22:09
Is there an easy way to see that $R^n$ can't be equipped with a ring structure, for $n>2$? –  user7530 Oct 29 '11 at 22:10