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I want to determine if the following set of vectors is a vector space. I know the answer is it's not a vector space because it violates the property $(c + d)\vec{u} = c\vec{u} + d\vec{v}$ for scalars c,d and vectors u and v.

The set is all vectors in $\mathbb{R}^2$ with the usual addition but scalar multiplication defined as: $$c\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}cx \\ y\end{bmatrix}$$

However there is a theorem that says if W is a nonempty subset of a vector space V, then W is a suspace of V if and only if we have closure in scalar multiplication (cu in W) and addition (u + v in W).

If W is a subspace of V then by definition it is a vector space. So my set which I know is not a vector space (which is nonempty subset of $\mathbb{R}^2$ for example $\begin{bmatrix}2\\2\end{bmatrix}$) must violate either closure in addition or closure in scalar multiplication.

My question is does my set violate closure in addition or closure in scalar multiplication (or both).

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    $\begingroup$ What vector space is your $W$ a subset of? $\endgroup$
    – Xander Henderson
    Dec 21, 2023 at 17:29
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    $\begingroup$ I don't understand the property you cite. The left hand only refers to the vector $\vec u$, but the right hand refers to two vectors, $\vec u, \vec v$. $\endgroup$
    – lulu
    Dec 21, 2023 at 17:59
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    $\begingroup$ @lulu I think that aweller3 has figured that out, but is trying to apply a particular theorem, and can't figure out where it goes wrong. My first comment points in the direction of figuring that out. Though it would be nice if aweller3 would, like, respond to the comments. :/ $\endgroup$
    – Xander Henderson
    Dec 21, 2023 at 18:03
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    $\begingroup$ @XanderHenderson given that it's literally (equal to) the whole set, and that there is only one sense in which objects are subsets of one another (being subobjects in SET), this is most likely a confusion arising from not observing the distinction between a (fully specified) structure and its carrier/underlying set $\endgroup$
    – ac15
    Dec 22, 2023 at 21:01
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    $\begingroup$ @XanderHenderson: I think ac15's point was not that it is confusing to them, but rather that it could be confusing to someone who is not experienced with linear algebra. For what it's worth, I agree with ac15: I don't think it really makes sense to speak of whether something is a subset in the "required sense" – if $A$ and $B$ are sets, then either $A\subseteq B$ holds, or it doesn't. What you really mean is that $W$ is not closed under addition and multiplication, which is a separate issue to whether $W\subseteq V$. $\endgroup$
    – Joe
    Dec 25, 2023 at 19:16

2 Answers 2

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The theorem you have in mind says that if $W$ is a subset of $V$ and the restrictions of all the pertinent operations $f: V^n \rightarrow V$ to $f|_W:W^n \rightarrow V$ are such that the direct image $f|_W[W^n]$ ends up being a subset of $W$, so that one may see the operations as being in fact $f|_W: W^n \rightarrow W$, then $W$ is a substructure of $V$

The case for vector spaces is a particular instance of this by considering $+: V^2 \rightarrow V$ and, for each fixed $\alpha$ in your field $k$, "scalar multiplication by $\alpha$" as a function $\alpha \cdot (-):V \rightarrow V$

Problem is, in the example you want to use the theorem, the 'new' scalar multiplication is not the restriction of the original scalar multiplication, so one cannot really apply the theorem to it

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It's helpful to ask what exactly a vector space is in the first place. We often come across phrases like "Consider the vector space $\mathbb{R}^2$", but what exactly does that mean?

Strictly speaking $\mathbb{R}^2$ is just a set that we might write as follows: $\mathbb{R}^2 = \{(x_1, x_2) \mid x_1, x_2 \in \mathbb{R}\}$. So it is really just a bunch of points like $(1, \sqrt 2)$ and $(-1/2, 13)$.

A vector space, on the other hand, is kind of algebraic structure: a set together with some operations that are required to satisfy certain rules. So until one has specified not just the set, but the additional operations, and checked that they satisfy the rules, one hasn't specified an algebraic structure.

In the case of a vector space, there are two operations: addition of vectors (elements of the set) and some notion of multiplication of vectors by scalars, along with some rules they must satisfy. So when we say "Consider the vector space $\mathbb{R}^2$", that's really shorthand for the set $\mathbb{R}^2$ together with the standard rules for addition and multiplication.

The rule for multiplication, for example, is $c \cdot (x_1, x_2) := (cx_1, cx_2)$ where the terms $cx_1$ and $cx_2$ are understood with respect to the usual multiplication of one real number by another. But this is so standard that we just don't bother saying it explicitly when we refer to $\mathbb{R}^2$ as a vector space.

Now you are perfectly free to consider other ways in which scalars and $\mathbb{R}^2$ could interact. For example, you could define a notion of multiplication of a vector by a scalar by $c \cdot (x_1, x_2) := (cx_1, x_2)$, and investigate its properties.

But as you have noticed, that operation does not satisfy the rules required for scalar multiplication in a vector space. So $\mathbb{R}^2$, equipped with the usual notion of addition of vectors and this alternative notion of scalar multiplication, is not a vector space. Thus no theorem that begins with "Suppose that $V$ is a vector space" is going to apply to it.

In particular, even though $\mathbb{R}^2$ is closed under the usual notion of addition and your alternative notion of scalar multiplication, and is (rather trivially) a subobject of itself, it does not follow that it is a vector space.

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