# Are co-ordinate vectors for abstract vector spaces elements of $\mathbb{R}^n$??

Recently I've been looking into the idea that tuples/column vectors/can be referred to as elements of $$\mathbb{R}^n$$ to represent vectors in the following questions, as part of the Euclidean Space:

What is the difference between a column vector and tuple?

Row and column vector representation for Euclidian vector spaces

Is a 'column vector' actually a vector or a matrix?

Is a 'column vector' actually a vector or a matrix?

How about the situation where we get an abstract vector space like the set of polynomials with real coefficients, we often will express this in a basis such as $$e_1=x^2$$, $$e_2=x$$, $$e_3=1$$. We will use a change of basis matrix, to determine the co-ordinate vector like $$x^2+2x+1$$ as $$(1,2,1)$$, this is a tuple of real numbers, can we consider this as an element of $$\mathbb{R}^n$$ if we consider the equivalent column vector as part of $$\mathbb{R}^n$$? Even though we are modelling an abstract vector space with a co-ordinate vector and not the Euclidean space, we receive something which should be an element of that vector space under a specific basis.

Is it that we use the elements of Euclidean vector space to help us to work with polynomials as there is an equivalence between them?

• Let $V$ isa vector space over $F$ of dimension $n$ and let $\mathcal{B}=\{e_i:1\le i\le n\}$ be a basis. Then $\forall x\in V$ , $x=\sum_{i}^{n}c_ie_i$ . $(c_1, c_2, \ldots, c_n)$ is called coordinate of $x$ relative to the basis $\mathcal{B}$. It's clear that $(c_1, c_2, \ldots, c_n) \in F^n$ Aug 6, 2022 at 13:33
• If your tuple adds like a vector and has the scalar multiplication properties of a vector, then it is a vector. Formally there is a little more to meet the definitions of a vector space, but this is the abbreviated definition. What this tuple represents is irrelevant. If the algebra is the same, what we can prove for vectors applies to all. Regarding one of the linked questions, a matrix is a vector (we can add matrices of the same shape and scalar multiply). Aug 6, 2022 at 17:21
• Suppose that $V$ is a vector space over $\mathbb R$ and $V$ has dimension $n$. Let $\beta$ be an ordered basis for $V$. If $v \in V$, then the coordinate vector of $v$ with respect to $\beta$, denoted $[v]_\beta$, is an element of $\mathbb R^n$. Aug 6, 2022 at 18:38

Whoa whoa, there are actually quite a number of things which are going on with your question. I'll address your questions/statements individually.

How about the situation where we get an abstract vector space like the set of polynomials with real coefficients, we often will express this in a basis such as $$e_1=x^2$$, $$e_2=x$$, $$e_3=1$$.

The vector space of all real polynomials with real coefficients does not have $$\{e_1,e_2,e_3\}$$ as a basis. This is an infinite-dimensional vector space. However, the set $$\{e_1,e_2,e_3\}$$ is certainly a linearly independent set in this vector space. Now, I don't think this is your main question or even your main confusion but I just wanted to point this out.

We will use a change of basis matrix, to determine the co-ordinate vector like $$x^2+2x+1$$ as $$(1,2,1)$$, this is a tuple of real numbers, can we consider this as an element of $$R^n$$ if we consider the equivalent column vector as part of $$R^n$$?

No, you may not. I see what you're saying. The idea here is that if you have the polynomial $$p(x) = x^2+2x+1$$, then this polynomial is determined entirely by its coefficients and so, I could just specify this polynomial by talking about the triple $$(1,2,1)$$. Equivalently, any $$n$$-tuple in $$\mathbb{R}^n$$ could be identified, in precisely the same way, as being the coefficients of some polynomial.

If we let $$\mathcal{P}_n$$ of all real polynomials of degree $$n$$ or less, then the entire discussion above is encoded by the following linear map: $$T: \mathcal{P}_n \to \mathbb{R}^{n+1}$$ $$a_0 +a_1 x+\ldots+a_n x^n \mapsto (a_0,a_1,\ldots,a_n)$$ You can show that this function is a linear isomorphism of both vector spaces and this is what people usually are referring to when they say that a given polynomial can be "identified" as some $$n$$-tuple.

The problem with being too taken in by this philosophy is that you start to lose sight of the main point of it. The main point of it is for us to place linear isomorphisms at the center of linear algebra so that we can understand how one vector space behaves in terms of a possibly more familiar vector space.

On the other hand, the space $$\mathcal{P}_n$$ is a space of functions. It's a function space. You can differentiate functions, integrate functions and do all sorts of things with functions that you can't quite do in the same way with $$n$$-tuples of real numbers.

So, yes, in the sense of linear algebra, both $$\mathcal{P}_n$$ and $$\mathbb{R}^{n+1}$$ are isomorphic to each other and could be seen as being "equivalent". That is, as vector spaces, they really are not so different from each other. But as sets with possibly additional structure, they are not the same and shouldn't be treated as being the same.

Is it that we use the elements of Euclidean vector space to help us to work with polynomials as there is an equivalence between them?

Yes, I think that's a good way of thinking about it. If $$V$$ is a finite-dimensional vector space over a field $$k$$, then given any (ordered) basis $$\{v_1,\dots,v_n\}$$ of $$V$$, there is an isomorphism of vector spaces $$V\to k^n$$ that takes the vector $$\sum_{i=1}^n a_i v_i\in V$$, where the $$a_i$$ are scalars in $$k$$, to the vector $$(a_1,\dots,a_n)\in k^n$$. In a sense, if we understand how to work with vector spaces like $$k^n$$, then we understand how to work with any finite-dimensional $$k$$-vector space $$V$$: just pick a basis of $$V$$ and take coordinates with respect to that basis. This translates a linear algebra problem in $$V$$ to an equivalent linear algebra problem in $$k^n$$. Your question is specifically about the case where $$k=\mathbb{R}$$, but it works for any other scalar field as well. (Of course, as this isomorphism is merely an isomorphism of vector spaces, it will not necessarily preserve other structure associated with these vector spaces, as Mordeus Morgenstern's more in-depth answer highlights.)

In your example the field is $$F = \textrm{span}(1, x, x^2)$$. So we construct the field by adjoining an algebraic element to $$R$$. What this means is that for any element of $$R(c)$$ where $$c$$ is the root of $$x^2 + x + 1$$, that $$R(c) \cong R[x] / \langle x^2 + x + 1 \rangle$$.

We also see that $$R(c) = span(1, x, x^2) = \{ a_0 + a_1 x + a_2 x^2 : a_i \in R \}$$.

So there are actually 2 ways to represent an element of this extended field. One is the actual element $$a \in R(c)$$, another is the tuple of coefficients $$(a_0, a_1, a_2)$$ so you can build the polynomial that $$a$$ is a root of.

• What? The span of $1,x,x^2$ is a subspace of the space of polynomials. It's not even a field, so it's certainly not "the field". Aug 6, 2022 at 15:04
• @DavidC.Ullrich It is a vector space though, isn't it? Aug 14, 2022 at 11:31