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).
Best Answer
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