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.

  • $\begingroup$ 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. $\endgroup$
    – Bill Cook
    Oct 29, 2011 at 22:06
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    $\begingroup$ Don't use $\times$ to represent the scalar multiplication, since it is usually reserved for the cross product $\endgroup$ Oct 29, 2011 at 22:06
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    $\begingroup$ 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? $\endgroup$
    – Srivatsan
    Oct 29, 2011 at 22:07
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    $\begingroup$ Something might work out along your lines for $\mathbb{R}^3$, but there seems to be nothing natural for arbitrary vector spaces. $\endgroup$ Oct 29, 2011 at 22:09
  • $\begingroup$ Is there an easy way to see that $R^n$ can't be equipped with a ring structure, for $n>2$? $\endgroup$
    – user7530
    Oct 29, 2011 at 22:10

1 Answer 1


The concept of a vector space isn't so much defined as recognized. All over Mathematics there are sets of things that have a natural addition, and a natural multiplication by real or complex numbers, but no natural way to multiply two things in the set to get another thing in the set. Once you've seen enough of these things, you accept that that's what Mathematics is giving you, and you make your definitions accordingly.

The definitions commonly in use lead to a huge number of useful concepts and results. Linear combination, span, linear dependence/independence, basis, dimension for starters - does your proposed multiplication lead to any of those concepts, or to any other concepts even half as useful?

  • $\begingroup$ Yes, and I wish I could delete this question because it's rather stupid upon reflection. $\endgroup$
    – user10389
    Oct 30, 2011 at 0:21
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    $\begingroup$ Don't be so hard on yourself. It's quite natural to ask for a (useful) way to multiply two vectors to get a vector, and many good things have come from that question (e.g., quaternions). $\endgroup$ Oct 30, 2011 at 2:35

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