How can a vector space be closed under scalar multiplication?

Question

A set is closed under an operation if performing the operation on members of the set always produces a member of the set. In fact, any algebraic structure must be closed under the operations that define it. So if a given vector space $V_1$ is closed under a binary operation $T$, it must be $T: V_1 \times V1 \to V_1$.

It is said that vector spaces are closed under the operations of scalar multiplication (i.e., multiplication with an element of a field $F$).

A scalar multiplications on $V_1$ would be a map $T_1: V_1 \times F \to V_1$. This is not how we stated closed binary operations should work like, since it’s not $T: V_1 \times V_1 \to V_1$.

There are two ways to solve this:

1. You could consider scalar multiplication to be a unary operation taking its input from the vector space (intuitively, we “absorb the scalar inside the function”). In this case, any vector space would have only one binary operation (the addition), and a whole set of unary operations: the scalar multiplications. For each given scalar a of the field, there would be a unary function $F: V_1 \to V_1$ such that $F(v) = a \times v$. It works, but the cost is that you cannot say anymore that a general vector space has only 2 operations, and you cannot say anymore that scalar multiplication is a binary map.

2. You could still consider the scalar multiplication as a binary map, but in order to satisfy the closed-operation requirement stated at the beginning, you must re-define the structure as the union of the set of vectors and of the set of scalars. Is this what we do? If yes, consider a set of vectors $V_1$ defined on a field $F$: when we say “vector space” are we referring to the set $V_1$ or the union set ($V_1$ and $F$)? And with which title can we refer to the other set?

So which one is it? Is scalar multiplication a unary-map, or is the actual algebraic structure (closed under scalar multiplication) the set of vectors + field elements? And most importantly: do you have sources?

Answer 1

If you insist that binary operations should have both arguments in the same set, then scalar multiplication is not a binary operation. Just give it another name if you want. It's often referred to as an external operation. It truly doesn't matter, there is actually no rule that all algebraic operations should be internal to a single set (a counter-example being... vector spaces).

Answer 2

If you really want to shoehorn in a semantically consistent interpretation, then you can say that scalar multiplication is a unary operation on the vector space, parameterised by a value taken from the underlying field. In other words, the operation is given by the family of functions $f_\alpha : V \rightarrow V$ with the parameter $\alpha \in \mathbb{F}$.

Under this interpretation, given any $\alpha$, the vector space $V$ is closed under $f_\alpha$.

As it so happens, you can directly link the family of unary operators $\{f_\alpha\}$ to the binary operation of scalar multiplication as it is typically defined. In fact, the map $\alpha \rightarrow f_\alpha$ is both bijective and linear in all the ways we'd want it to be.