I'm studying for the Math GRE subject test and I'm currently going over Linear Algebra. I'm carefully re-reading my course book "Linear Algebra Done Right" By Sheldon Axler (Third Edition). I have a few questions in regards to the fundamental concepts:
Going over the definition of a Vector Space, it doesn't seem obvious to me why we need the scalars, adjoined to the vector set, to be members of a Field. If we loosen this definition of a Vector Space a little bit I think we can get algebraic structures that are analogous to that of a Vector Space and thus meaningful. Hence my first question: Is the Real Number Line a Vector Space?.
If so, then this would imply that numbers themselves can be interpreted as vectors. Then, in trying to consider "small" Vector Spaces (other than the trivial case of the singleton zero vector set $\{0\}$) I wonder, in this sense, my second question: can there be Vector Spaces contained inside the Real Number Line?
It is here where loosening the definition of a vector space can be of merit. If we allow our scalars to be Integers and our vector set to be all the multiples (positive and negative) of a Natural Number then that vector set adjoined with the operation of vector addition and scalar multiplication would manifest all the characteristics that define a Vector Space. Try using the set that contains all the multiples of three – $\{x | x = 3\alpha \text{ where } \alpha \in \mathbb{Z}\}$.
Thus why would such a structure not be a vector space. Is it missing something? Or why do we demand that we use scalars from a field? I'm interested in knowing what you think.
Best Answer
This is a pretty broad question, so I'm not optimistic it'll stay open for long. That said, it's a great observation!
First, I'll answer something you implicitly mentioned before your first question:
What if we don't require scalars to form a field?
If we instead just ask them to form a commutative ring, (i.e. we can do everything you're used to in a field except for division) then the structure we get isn't a vector space any more, but it's a slightly different structure called a module.
There's a lot of interesting things to be said about modules, but the main point is that without division, scaling down isn't generally possible.
This has some surprising differences from a vector space. For instance, in a vector space if you have a spanning set which is not independent, you can throw away useless vectors and make a basis. This isn't always possible in a module, and the notion of 'dimension' is less clearly defined for a module.
For instance, consider the module $\mathbb{Z}$ with the base ring $\mathbb{Z}$. The set $\{2,3\}$ is linearly dependent (in the way that you're used to) and spans $\mathbb{Z}$ since they are coprime, but neither $\{2\}$ nor $\{3\}$ span $\mathbb{Z}$.
In particular, when dealing with vector spaces you'll likely use the fact that if a finite set is linearly dependent, then you can write one of the vectors in terms of the others. How? Well, suppose $$\sum_{i=1}^n \lambda_i v_i = \lambda_1 v_1 + \dotsb +\lambda_n v_n = 0$$ for $\lambda_i \in \mathbb{F}$ and $v_i \in V$. Since the vectors are linearly dependent, one of the $\lambda_j \neq 0$. Then we can just rearrange to get
$$\lambda_j v_j = -\sum_{i=1\\i\neq j}^n \lambda_i v_i \implies v_j = -\sum_{i=1\\i\neq j}^n \frac{\lambda_i}{\lambda_j} v_i$$ by dividing through by $\lambda_j$. But in a module, we can't divide by scalars. So, such an expression isn't always possible, creating things that can be surprising when you're used to vector spaces.
Is the real number line a vector space?
Yes! In a few ways, actually. First, it's a vector space over itself: take the base field to be the reals, and you get the reals as a one dimensional subspace.
Another way you can make the reals a vector space is to define the base field as the rationals. Then, the real numbers form a vector space over the rationals, and it's not difficult to see it's infinite dimensional. What is difficult however, is to write down a basis, which requires the Axiom of Choice.
Can there be Vector Spaces contained inside the Real Number Line?
Yes! Going in this direction pushes you into Galois theory. For example, taking $\mathbb{Q}$ as the base field again, the vector space with basis $1,\sqrt{2}$ is a ($2$ dimensional) vector space purely contained in the reals (we call it $\mathbb{Q}(\sqrt{2})$. It has elements of the form $a + b \sqrt{2}$, where $a,b$ are rational numbers. Alternatively, you can put $\sqrt[3]{2}$ in your vector space instead, and give the (3 dimensional) vector space with basis $1, \sqrt[3]{2}, \sqrt[3]{2}^2$, also contained in the reals.
You can get even bigger ones by considering the algebraic numbers, which has countably infinite dimension over $\mathbb{Q}$.
You can make the base field larger too, for instance you could use a base field $\mathbb{Q}(\sqrt{2})$, and consider the numbers of the form $a + b \sqrt{3}$, where $a,b \in \mathbb{Q}(\sqrt{2})$, which has dimension 2 over $\mathbb{Q}(\sqrt{2})$ (and dimension 4 over $\mathbb{Q}$).
Finally, as mentioned earlier you can take $\mathbb{R}$ over $\mathbb{Q}$, which has uncountable dimension.