Do we really need to specify a basis to describe a tuple?

Question

Say there is a two dimensional vector space $\mathbf{V}$, when I represent a vector $x\in\mathbf{V}$ using a tuple $(1,2)$, then clearly it's not clear about which vector I really meant because it's actually

$$x=1\cdot u+2\cdot v,$$

which uses an ordered basis, $\{u,v\}$ which I haven't say what they are, of $\mathbf{V}$ in this describing. Now considering $\mathbf{R}^3$ as example, when I say

$$(1,2,3)\in\mathbf{R}^3,$$

which basis did I just implicitly use? It seems like I don't have to use the idea of basis to make it clear about which one of $\mathbf{R}^3$ I'm describing. Even you say I'm using the standard ordered basis $\{e_1,e_2,e_3\}$, then it is just because the tuple under this basis is also $(1,2,3)$! Which part did I miss, it seems like something is wrong in my reasoning?

Another example is that when I want to describe a vector in a two dimensional plane, I "have to" decompose it into that "x-component, y-component" to describe it. Once it's decomposed, I have to specify the basis I'm using, and yes implicitly I should mean the standard ordered basis. But any element of $\mathbf{R}^3$ is of its essence already be decomposed into three part, so no need the idea of basis when I just want to say "that $(1,2,3)$"?

Answer 1

More or less by definition, tuples are elements of $\mathbb{R}^n$. What's being glossed over is the following.

Selecting an ordered basis $\mathcal{B} = (u,v)$ for a two-dimensional vector space $V$ amounts to the same thing as choosing a linear isomorphism $T : \mathbb{R}^2 \to V$, defined by

$$ T(a,b) = a \cdot u + b \cdot v $$

When you say "the element of $V$ represented by $(1,2)$", what you really mean is "the vector $T(1,2)$".

Another notation people sometimes write is, for $x \in V$, to use the notation $[x]_\mathcal{B}$ to mean the value $T^{-1}(x)$, which is a tuple. This notation is meant to be read as "the coordinates of $x$ relative to $\mathcal{B}$". So the statement you make in the OP would be written

$$ [x]_\mathcal{B} = (1,2) $$

There's a similar notation for linear transformations; given a choice of bases for the input and output spaces $[A]_{\mathcal{B}'}^{\mathcal{B}}$ means the matrix whose entries are the coordinates of $A$ relative to the two choices of bases.

So, for example, using matrix arithmetic to compute linear transformations boils down to the identity

$$ [A]^\mathcal{B}_{\mathcal{B}'} \cdot [x]_\mathcal{B} = [A(x)]_{\mathcal{B}'}$$

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As an aside, when interpreting vectors of $\mathbb{R}^n$ as matrices, there are reasons why it's most natural to consider them as $n \times 1$ matrices (i.e. "column vectors") rather than as $1 \times n$ matrices (i.e. "row vectors").

Answer 2

To answer your question, you don't need to specify a basis to define a tuple, since a tuple is just a finite sequence that needn't have anything to do with vectors or with linear algebra for that matter. You do need to specify a basis in order for your tuple to refer to a specific vector. However, for $\Bbb{R}^n$ and $\Bbb{C}^n$ and $\Bbb{Q}^n$ the standard orthonormal basis is understood. For most other vector spaces (and even for those if you aren't using inner products in the problem) it usually won't matter which basis you choose, so people often don't bother to specify, but it's important that you are using the same basis when you refer to (2,2,5) in the next sentence as you are when you referred to (1,2,3). In other words, the exact basis doesn't generally matter as long as you hold it fixed. (The exceptions are because we already have a specific bilinear form (an inner product) we like using on those vector spaces, so that makes different bases inequivalent. For other vector spaces, typically if one is going to use a bilinear form, one can specify it while setting up the notation.

If they really wanted students to learn this stuff, they'd include linear algebra over the quaternions, just so you'd have to keep track of these things more carefully. But I digress.

Answer 3

No, you do not need a basis to specify a tuple - but your ability to specify tuples means there is an obvious choice of basis to pick.

By definition, $\mathbb R^n$ is the set of $n$-tuples of real numbers. So, it is absolutely true that $(1,2,3)$ refers to a particular element of $\mathbb R^3$ - and it makes perfect sense to apply the vector space operations to that element, to say, for instance $2\cdot (1,2,3)=(2,4,6)$ just because that's how we defined things. There is no reference to bases here because this treats $\mathbb R^n$ as a set-theoretic object, on which we have defined additional vector space structure.

It is worth noting that, if you have a vector space $V$ and an ordered basis $\{b_1,\ldots,b_n\}$, then there is an isomorphism $\mathbb R^n\rightarrow V$ taking $(\alpha_1,\ldots,\alpha_n)$ to $\alpha_1b_1+\ldots+\alpha_n b_n$, which means, basically, that tuples can represent the position of a vector with regards to a basis - but that this is not, intrinsically, what tuples are.

It's also worth noting that, even though there always exists such an isomorphism from $\mathbb R^n$ to a real vector space $V$ of dimension $n$ (since there always exists a basis), it is not unique - and it is often counterproductive to fix a basis, meaning that you'd have to leave tuples out of reasoning on $V$.

Answer 4

You are implicitly talking about coordinate of a vector with respect to a given basis.

Standard basis $E$ is orthonormal so finding projections are easy.

Coordinates are projections so standard basis are used.

For example $(3,5)$ means $$3(1,0)+5(0,1)$$ and $(1,3,5)$ means $$ 1(1,0,0)+3(0,1,0)+5(0,0,1)$$

Now if I choose a different basis , say $$ B=\{ (1,2)\text { , } (2,1)\}$$ then

$(3,5)_B $ means $3(1,2)+5(2,1)=(13,11)_E$

As you see coordinates are not easily seen in the new basis.

You will enjoy learning about Change of Basis matrix in linear algebra.