Proving V is a Vector Space: Simplifying the Process with Axiom Lemma

[Bachelier]

Though this may be related to lin. alg. but it deals with Analysis.

There are 8 axioms for Vector Spaces. To prove a space $V$ is a VS, one must check all 8 axioms (i.e. closure under addition, scalar multi. etc...)

My question is this, it seems cumbersome to have to do this every time. Would it be better to use the lemma that states: "A non∅ subset $W$ or a VS $V$ is a subspace $iff$

$\alpha v + \beta w \in W, \forall \ v, w \in W, \ \alpha , \beta \in \mathbb{F} \ \ (1)$​

where $\mathbb{F}$ is the field of scalars.

First, would it be correct to use this "Lemma"?

And second, what should the encompassing VS be: because the lemma states $W$ is a subspace of a VS $V$, but if I want to prove a set

$G \subsetneq \mathbb{R \times R}$ is a VS, I guess I should aim to show it is subspace of the Vector Space $\mathbb{R \times R}$

 

[WannabeNewton]

If you know that a given set $V$ is already a vector space, and if $W\subseteq V$, then all you need to check is that $W$ is closed under scalar multiplication and vector addition, which is what you wrote above. As for subspaces of $\mathbb{R}^{2}$, if this is over the field $\mathbb{R}$ and under the usual scalar multiplication as well as vector addition then one can easily classify all non-trivial subspaces of $\mathbb{R}^{2}$ as lines through the origin. All you need to do is check that $G$ is a line through the origin or is a trivial subspace of $\mathbb{R}^{2}$.

 

[Tac-Tics]

For sure, it is tedious having to go through all 8 axioms. In any class beyond an intro to linear algebra class, you could simply say "it's easy to see that W is a subspace".

You can also phrase things in a slightly clever way (again, if your audience has the background):

* W is the image of a linear map f : X -> V (for some vectorspace X). It's easy to prove the image of a linear map is a subspace of the codomain.

* W is the kernel of a linear map f : V -> X (for some vectorspace X). Again, kernels are always subspaces.

* If you have a basis B = b1, b2, ..., bn for V, you can always take some subset B' of those b_i's and take W to be the span of B'. Spans are basically a way to satisfy all the closure requirements.

 

[Bachelier]

Cool. Thank you.