Right and left multiplication by scalar on a vector space This is a really basic question but I always wanted to know the answer to it. If we have a vector space $V$ over a field $K$, then scalar multiplication is usually defined by taking $\alpha v$, $\alpha \in K$ and $v \in V$ and the scalar always come on the left side. I know that in this case one can freely change the order of multiplication $\alpha v = v\alpha$. However, what axiom of vector space justifies this exchange? Since scalar multiplication is always introduced by placing the scalar on the left, how can we even start to consider a scalar on the right? Shouldn't we have some property about, say, $1v = v1 = v$? I have never found a reference in which some property is explicitly stated.
 A: Well, in some sense the order doesn’t matter, because these are the objects of different nature! When we are working with operation $*:A\times A\to B$, the order is important because both $\forall a_{1},a_{2} \in A: a_{1}*a_{2}\in B \text{ and } a_{2}*a_{1}\in B$ by definition.
In your case if $V$ is a vector space over $\mathbb{K}
$ field, then multiplication by number is defined like
\begin{gather}
\mathbb{K}\times V \to V\\
(\alpha
,v)\mapsto \alpha v.\end{gather}
The $v\alpha$ notation is in some sense meaningless, because $\mathbb{K}$ and $V$ are principally different sets. You can always define operation
\begin{gather}
\mathbb{K}\times V \to V\\
(\alpha
,v)\mapsto v\alpha.\end{gather}
But it is some different operation. If it does have the same properties as $\alpha v$, then there is an isomorphism of structures, which allows you to rearrange vectors and scalars.
A: $v\alpha = \alpha v$ is a definition. In "$\alpha v$", we view $v$ as an element of $V$, but in $v\alpha$, we view $v$ as a linear map from $K$ to $V$ defined by $v\alpha := \alpha v$, i.e. an element of $L(K, V)$. This identification is an isomorphism between $V$ and $L(K, V)$. The inverse of this isomorphism is the map taking $\omega \in L(K, V)$ to $\omega 1$.
A: Interesting question. I suspect that if we define a left and a right scalar multiplications, with the same properties, should be possible to prove that they give the same result, i.e. $\alpha\mathbf{v}=\mathbf{v}\alpha$ for each scalar $\alpha$ and for each vector $\mathbf{v}$. For example, it is easy to prove for positive integer $\alpha$, say $\alpha=2$:
$$
2\mathbf{v} = (1+1)\mathbf{v} = 1\mathbf{v}+1\mathbf{v}=\mathbf{v}+\mathbf{v}=\mathbf{v}1+\mathbf{v}1=\mathbf{v}(1+1)=\mathbf{v}2
$$
But I don't have a general proof.
