# Does the multiplicative identity have to be 1?

I am just starting out with vector spaces and I am having a hard time understanding them. One of the requirements states that $1\mathbf{v}=\mathbf{v}$ where $1$ is the multiplicative identity.

Does 1 have to be the identity? Or is that whatever is the multiplicative identity is labelled 1?

I ran across a question where $a(x,y)$ was defined as $(a x/3, a y/3)$. It was not specified but I guess the question also implied that scalar a belongs to real numbers. Why has 1 got to be the identity here? Why can't I define 3 to be the identity here?

Edit: I realise that I didn't describe the question in enough detail. $\mathbf{v}$ here is the set of all ordered pairs $(x,y)$ where $x, y \in\mathbb{R}$. Addition is defined as $(x_1,y_1) + (x_2,y_2) = (2x_1-3x_2,y_1-y_2)$. Now addition obviously violates many axioms but I am interested in scalar multiplication, which is defined as given above. The question is a simple example question asking us to list down all axioms violated.

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Run through the vector space axioms again. Try to find one which is violated in this case. – Qiaochu Yuan Feb 15 '11 at 15:01
When one writes down the axioms of a vector space formally, the convention is to denote the multiplicative identity (which lies in the "field" of scalars) by 1. Without more information, I'm not sure what to make of the second question. Maybe you could explain in more detail what that question asked? – Dan Ramras Feb 15 '11 at 15:01
Probably Qiaochu is right, and your second question is about an example of a non-vector-space. – Dan Ramras Feb 15 '11 at 15:02
I am sorry, I should have defined the question in more detail. – user7118 Feb 15 '11 at 15:08

Let's ignore the axiom that requires $1\mathbf{v}=\mathbf{v}$ to hold for a moment.

Let's call an element $\alpha$ of the field with the property that $\alpha\mathbf{v}=\mathbf{v}$ for every $\mathbf{v}$ an "identity". First, let's note that we really want to have at most one of them, not multiple ones. Otherwise, we could just define scalar multiplcation by "do nothing", and we don't really have a vector space structure (the scalars are just dummys, not doing anything). And also, because otherwise we're going to lose the notion of linear independence (which is very important later on): if $\alpha$ and $\beta$ both "work" as the identity and $\alpha\neq\beta$, then for every vector you are going to have $$(\alpha-\beta)\mathbf{v} = \alpha\mathbf{v}-\beta\mathbf{v}=\mathbf{v}-\mathbf{v}=\mathbf{0},$$ but with $\alpha\neq\beta$. This will mean that no collection of vectors is linearly independent. So we don't want more than one identity.

One axiom is the associativity of scalar multiplication: $$(\alpha\beta)\mathbf{v} = \alpha(\beta\mathbf{v}).$$ Why can't you define the "identity" to be $3$? Because then you would have $3\mathbf{v}=\mathbf{v}$, hence $$9\mathbf{v} = 3(3\mathbf{v}) = 3\mathbf{v}=\mathbf{v},$$ so that now $9$ is also an identity, even though $3\neq 9$, and we already said that is not a good idea.

So, whatever the identity is, you want it to equal its square, for precisely this reason. Otherwise, you're going to have lots of identities, and hence lots of troubles.

But in a field, there are only two solutions to $x^2=x$: namely, $x=0$ and $x=1$. So the "identity" has to be either $0$ or $1$. But the fact that $0\mathbf{v}=\mathbf{0}$ holds follows without the identity axiom ("$1\mathbf{v}=\mathbf{v}$"), so unless every vector is $\mathbf{0}$, you cannot have $0$ be the identity, because then we would have $$\mathbf{v} = 0\mathbf{v}=\mathbf{0}.$$

So, really, the only reasonable choice for identity is $1$.

So, then, why do we make it an axiom? Because otherwise, we can define scalar multiplication by "any scalar times any vector is $\mathbf{0}$". This definition satisfies all the axioms of a vector space except the existence of the identity, but it's a "silly" vector space (and one where the notion of linear independence is also ultimately completely screwed up). So we want to exclude it. And the way to exclude it is precisely by requiring that there be an element of the field that does not map everything to $\mathbf{0}$ through scalar multiplication, so it makes sense to kill both birds with one stone and ask for $1$ to not send everyone to $\mathbf{0}$ by multiplying each vector to itself.

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Thanks, that made sense! – user7118 Feb 15 '11 at 16:35