# Real Vector Space Scalar multiplication

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 $\mathbb{R}$.
• 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? – James Machin Jan 7 '14 at 18:57
• @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 – zcn Jan 7 '14 at 19:03