Can all vectors be described with a tuple?

Question

Vector Space Definition

As I understand, an vector space is defined by:

1. A field $\mathbb{F}$ (e.g. $\mathbb{R}$, $\mathbb{C}$) which is a set whose elements $a \in \mathbb{f}$ are numbers that follow standard algebra rules (see this). We call the elements of this set scalars.
2. A set $V$, whose elements $\vec{v} \in V$ we call vectors.
3. A scalar multiplication operation $a \cdot \vec{u} = \vec{v}$ such that $a \in \mathbb{f}$ and $\vec{v}, \vec{u} \in V$.
4. A vector addition operation $\vec{u} + \vec{v} = \vec{w}$ such that $\vec{u}, \vec{v}, \vec{w} \in V$.
5. The 8 axioms are satisfied.

As such, our definition of a vector is simply element in $V$ (notice, at this point, there is no notion of a vector in the conventional understanding as a points in space, or an arrow with magnitude and direction, or a list of numbers, etc ...). As you'll see below, the number $5$, the list $(1, 2, 3)$, the matrix $\begin{bmatrix} 1&2\\3&4 \end{bmatrix}$, and the polynomial $2x^3 - x^2 + 2$ can all be vectors.

Some Examples

If we define our field as $\mathbb{R}$ and we use standard element-wise definitions for vector addition and scalar multiplication, the following sets form vector spaces:

• Real numbers ($\mathbb{R}$), complex numbers ($\mathbb{C}$
• The set of all n-tuples with real numbered elements ($\mathbb{R^n}$), the set of all n-tuples with complex numbered elements $\mathbb{C^n}$
• The set of all polynomials with degree at most $n$ ($\mathbb{P}_n$)
• The set of all $m \times n$ matrixes with real numbered elements ($\mathbb{M}^{\mathbb{R}}_{m,n}$)

Matrices as Tuples

To prove $\mathbb{M}^{\mathbb{R}}_{m,n}$ forms a vector space (assuming our field is real numbers, and the standard element-wise definitions for addition and scalar multiplication), we could one-by-one prove the 8 axioms.

An alternative line of thinking however would be to "co-erce" a matrix to an mn-tuple (say, by "stacking" the columns of the matrix ontop of each other, starting from the left and moving to the right). In this way, we can see that $\mathbb{M}^{\mathbb{R}}_{m,n}$ is secretly "the same" as $\mathbb{R}^mn$, and we already know that $\mathbb{R}^mn$ is a vector space, so $\mathbb{M}^{\mathbb{R}}_{m,n}$ must also be a vector space.

Now, our definition for vector is any element $\vec{v} \in V$, so in this case we would consider the matrix $\begin{bmatrix} 1&2\\3&4 \end{bmatrix}$ to be a vector under $\mathbb{M}^{\mathbb{R}}_{m,n}$. We also just saw that we can express such a matrix as a tuple.

Similarly, the vector space $\mathbb{R}$ can be thought of as $\mathbb{R}^1$ (i.e. 1-tuples). So here's my question:

Question

1. Can all vectors (defined by an arbitrary vector space) be "convert" or "thought of" as an n-tuple?

A follow up question:

2. What is an example of a vector space that is "weird" or "unintuitive" vector space (i.e. if the answer to the first question is no, what would such a vector space look like)?
3. If the answer to question 1 is yes, do the elements of the tuples need to be a field? Do they even need to be numbers?

EDIT: I'm primarily curious for cases with finite-dimension vector spaces (but also, to a lesser degree, I'd be curious understand these questions from the perspective of infinite-dimension vector spaces)

Answer 1

It's easy to prove that every finite-dimensional vector space is isomorphic to $\mathbb{R}^n.$

Proof. Let $\dim V = n$ with basis $\{ v_1, \ldots, v_n \}$ so that each $v$ is expressible as $a_1v_1 + \ldots + a_nv_n.$ Define the function $f: V \to \mathbb{R}^n$ by $f(v) = (a_1, \ldots, a_n)$ where each $a_i \in \mathbb{R}.$ We'll go through the usual steps.

$f$ is well-defined: if $v = v'$, then $f(v) = (a_1, \ldots, a_n)$ and $f(v') = (a_1', \ldots, a_n').$ But we must have $a_i = a_i'$ for each $i.$ Thus $f$ is well-defined.

$f$ is one-to-one: if $f(v) = f(v')$ then each $a_i = a_i'.$ Then $v = v'.$

$f$ is onto: Given $(a_1, \ldots, a_n) \in \mathbb{R}^n,$ we can construct $v \in V$ so that $f(v) = (a_1, \ldots, a_n)$ in the following manner: since we have a basis for $V$ we can write any $v$ as above, and such a $v$ gets mapped to its "coordinates."

$f$ is linear: Suppose $f(v) = (a_1, \ldots, a_n)$ and $f(w) = (b_1, \ldots, b_n).$ For scalars $\lambda, mu$ we have

\begin{align*} f(\lambda v + \mu w) &= (\lambda a_1 + \mu b_1, \ldots, \lambda a_n + \mu b_n) \\ &= (\lambda a_1, \ldots, \lambda a_n) + (\mu b_1, \ldots, \mu b_n) \\ &= \lambda (a_1, \ldots, a_n) + \mu (b_1, \ldots, b_n) \\ &= \lambda f(v) + \mu f(w) \end{align*}

Thus $f$ is an isomorphism. $\square$ (You can replace $\mathbb{R}^n$ with $\mathbb{C}^n$ or, as Axler does, with $\mathbb{F}^n.$

You can't do this for infinite dimensional vector spaces, but you can attempt to mimic the argument using the space $\mathbb{F}^\infty$ (again, as Axler writes), the space of infinite sequences in the field $\mathbb{F}$.

Answer 2

If you're willing to extend the definition of a polynomial, all vector spaces can be seen as weird polynomials.

As you said $P_n$, the set of polynomials of degree at most $n$, is a vector space. But, you can view these as tuples of the coefficients. So, $3x - \pi x^4$ can be seen as $(0, 3, 0, 0, -\pi)$.

Now, given any set, $X$, we can imagine the "polynomials" with exponents from that set and coefficients from your field. For example, if $X = \mathbb{R}$, we get "polynomials" like $3x^e - \frac{1}{2}x^6 + x^{-\pi}$. If your set is finite, then this is the same as $P_n$, for some $n$. If the set is not finite, then the idea of a tuple might not make as much sense.

More exotically, if your set $X$ is the set of all subsets of natural numbers, you'd get things like $x^{\{1, 2\}} - 9x^{\{2, 4, 6, \ldots\}}$.