If I have a real vector space, when are left and right scalar multiplication identical?

I'm coming from the angle that the quaternions form a 2 dimensional vector space over $\mathbb{C}$ and yet left and right scalar multiplication are not the same in this space.

So I guess my question is, if I refer to a real vector space, do I have to state whether I'm talking about left scalar multiplication or right scalar multiplication?

Thanks for any replies.


The phenomenon you're observing arises from the facts that the quaternions are noncommutative and that the left (right) scalar multiplication on the quaternions as a 2-dim'l vector space over $\mathbb{C}$ is "the same as" multiplication in the quaternions with the left (right) factor restricted to $\mathbb{C}$, which do not commute with the quaternions.

I'd have to double-check, but if I recall correctly the reals do commute with the quaternions, so this would go away if you took the quaternions as a 4-dim'l vector space over $\mathbb{R}$.

In general, yes, you should specify whether scalar multiplication is on the left or on the right (or if both are legal and distinct from each other) whenever it matters. Normally, as far as examples I've seen, we take vector spaces to have scalar multiplication on just one side for convenience, but bimodules are a thing, so I don't see why that concept couldn't be applied to vector spaces as well. Nothing's stopping you from being as wacky as defining a vector space with its left scalars coming from one field and its right scalars coming from a smaller field, for instance, though I'm not sure why you would do that.

  • $\begingroup$ Thanks for the reply! Ok so the left multiplication isn't even necessarily related to the right multiplication, it's just that in this case it's because the scalar multiplication is just a subset of the ring multiplication? What if in a general vector space I have an expression like $(a_1+b_1\mathbf{v_1})(a_2+b_2\mathbf{v_2})$ and I expand it to get $(a_1a_2+a_1b_2\mathbf{v_2}+b_1\mathbf{v_1}a_2+b_1\mathbf{v_1}b_2\mathbf{v_2}$. Don't the third and fourth terms require definition of right scalar multiplication? $\endgroup$ – James Machin Jan 7 '14 at 18:57
  • $\begingroup$ @JamesMachin: I don't quite know what you mean by $\textbf{v}_1$, $\textbf{v}_2$, but you cannot "multiply" two vectors in a general vector space $\endgroup$ – zcn Jan 7 '14 at 19:03
  • $\begingroup$ In a general vector space, that expression is ill-defined, requiring the multiplication of two vectors and the addition of a scalar and a vector. $\endgroup$ – Nick Jan 7 '14 at 19:04
  • $\begingroup$ Oh no sorry, I mean an algebra. I'm such an idiot. $\endgroup$ – James Machin Jan 7 '14 at 19:05
  • $\begingroup$ @Nick But I still have the same question, just with vector space replaced with algebra! My apologies! $\endgroup$ – James Machin Jan 7 '14 at 19:14

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