The element of a field scalar or vector?

Question

A vector space is comprised of a set, a field and two binary operations. The element of the field is called scalar. But a field over itself is also a vector space, therefore the element of the field is also a vector? The element of a field can be both a scalar and a vector? Plus, in a vector space the scalars from the field are often numbers, can there be other kinds of scalars for this vector space? Such as the element of the field is 2*1 matrix?

Answer 1

A vector space is comprised of a set, a field and two binary operations. The element of the field is called scalar. But a field over itself is also a vector space, therefore the element of the field is also a vector? The element of a field can be both a scalar and a vector?

It is possible for the field and the vector space to coincide. It's also possible to consider a field as a vector space over a subfield (for example, $\Bbb{C}$ over $\Bbb{R}$, or $\Bbb{R}$ over $\Bbb{Q}$), in which case every scalar will be a vector, but not vice-versa.

It's also possible for the vector space to be a proper subset of the field (i.e. every vector will be a scalar). The simplest way is to pick a point (typically $0$) from a field, and define the trivial vector space on this set. Slightly less trivially, one can define a vector space on the positive real numbers $(0, \infty)$, defining vector "addition" by $u \oplus v = uv$ (i.e. regular real multiplication), and defining scalar multiplication by $a \cdot v = v^a$, i.e. using exponentiation. Here, the vector space $(0, \infty)$ is actually a subset of the field $\Bbb{R}$.

But any overlap (or not) between the sets is not usually particularly important. Indeed, the exact makeup of elements of the vector space and the field tend not to be of great interest to mathematicians. What matters is structure. Two spaces may have complete different elements in their underlying set, but they may still be "the same" in all the relevant ways.

You might notice, for example, that the space of real ordered pairs $$\Bbb{R} \times \Bbb{R} = \{(a, b) : a, b \in \Bbb{R}\},$$ equipped with the usual is awfully similar to, say, the space of $2 \times 1$ matrices $$M_{2 \times 1}(\Bbb{R}) = \left\{\pmatrix{a \\ b} : a, b \in \Bbb{R}\right\},$$ even though they have not even one element in common! The idea that formalises this is the notion of an isomorphism, which you may or may not have encountered before. But, essentially, the two spaces are basically the same: anything important that you can do in one, you can do in the other as well. In this way, a vector space is almost independent of what makes up its set of vectors. So, if some of these vectors happen to fall into the field of scalars, it's not a big deal.

Plus, in a vector space the scalars from the field are often numbers, can there be other kinds of scalars for this vector space?

Absolutely! The idea of a number is pretty nebulous from the start, but no, you don't need your field to fit into $\Bbb{R}$, $\Bbb{C}$, or $\Bbb{Q}$. There are plenty of other fields one can choose from.

There are, for example, finite fields, the simplest ones of which take the form $\Bbb{Z} / p\Bbb{Z}$ where $p$ is a prime, i.e. the classes used in arithmetic modulo $p$ (sometimes denoted $\Bbb{Z}_p$). These fields have exactly $p$ elements in them.

The vector space $\Bbb{Z} / 2\Bbb{Z}$ in particular, and vector spaces made from it, are particularly useful in the field of computer science. In graph theory, there's some study of a vector space called the cycle space of a graph. Also, here's an old answer of mine that outlines a sort of fun use of vector spaces over $\Bbb{Z}/ 2\Bbb{Z}$.

On top of this, without going into detail, many fields are defined by a sort of modulo arithmetic on polynomials, whose coefficients lie in the field. The complex numbers are frequently defined this way. This means that the individual elements of the field are sets of infinitely many polynomials. This sounds daunting, but it's a completely standard construction in abstract algebra. On a similar tack, you can also define fields of formal rational functions (i.e. quotients of polynomials).

Such as the element of the field is 2*1 matrix?

Ah, that one is easy! Consider the set $M_{2 \times 1}$ as above. Define addition on $M_{2 \times 1}(\Bbb{R})$ as it normally is: $$\pmatrix{a \\ b} + \pmatrix{c \\ d} = \pmatrix{a + c \\ b + d},$$ but define multiplication like so: $$\pmatrix{a \\ b} \times \pmatrix{c \\ d} = \pmatrix{ac - bd \\ ad + bc}.$$ Why is this a field? Well, I've basically just rewritten the complex numbers. Rather than writing $a + ib$, I'm writing $2 \times 1$ matrices $\pmatrix{a \\ b}$ instead. The same calculations that show $\Bbb{C}$ is a field work to show that $M_{2 \times 1}(\Bbb{R})$ is a field under these operations. And yes, you can build vector spaces with this field as well, in much the same way that you would with complex numbers.

It's hardly a creative solution. Once again, this new field has the same structure as $\Bbb{C}$. They are isomorphic as fields. The fact that they have no individual elements in common is of no particular importance.

Answer 2

Your question is pretty open ended, and these questions abound. This is all part of doing mathematics.

Let me just give one example. The vector space $\Bbb R^{\Bbb R}$ of functions $f:\Bbb R\to \Bbb R$, over $\Bbb R$. This interesting for several reasons. For one thing it's infinite dimensional, as is often the case with function spaces.

For another, it's vectors are functions. These types of vector spaces are studied in functional analysis.

There are all kinds of possibilities. I've heard good things about Halmos' book, Finite Dimensional Vector Spaces.

As you noted any field is a vector space over itself. So the scalars can come from any field, and there's lots of different fields to choose from.

Fields are a tricky subject which I encourage you to explore. One example, Galois theory, exploits a connection between certain fields and groups (another interesting subject).

Finite fields are vector spaces over their prime subfield, and for that reason have order $p^n$.

A vector space is in some sense just an abelian group with some extra structure.

Finally, an interesting and important generalization of vector spaces (over fields) is modules over rings.