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They say (here, for instance) that you can represent a vector, $\vec v$ as coordinate vector, $[v]_B$, in base, $B$,

$$\vec v = v_1 \vec b_1 + v_2 \vec b_2 + \cdots = \begin{bmatrix}\vec b_1 & \vec b_2 & \cdots \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \end{bmatrix} = B\, [v]_B.$$

That is, $B^{-1}$ can serve as a coordinate map to translate a vector into coordinates:

$$[v]_B = B^{-1}\vec v.$$

Everything is fine: we had a vector and got its coordinates in basis B. There is only one thing that I do not understand: what is B? Is it a matrix or operator?

If abstract operator B is not a matrix and v is not a tuple then how do we get column of numbers, $[v]_B$, multiplying them? I know how to get a column of numbers as result of matrix multiplication only when multiplying a matrix of numbers with tuple of numbers. However, if B and $\vec v$ are matrices right away, then, we already have the coordinates of $\vec v$ and the question is why to muliply it with $B^{-1}$ ever then? To get just another coordinates of $\vec v$?

Because the way the topic is always exemplified, I suppose that $B$ and $\vec v$ are provided as matrices in some another basis. But what is that basis? Why not to use $[v]_{ANOTHERBASIS}$ instead of deceptive $\vec v$? Can this help me to answer the difference between components and coordinates?

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You should forget about expressions treating a list of abstract vectors, like $[\vec b_1~~\vec b_2~~\ldots~~]$ as a matrix. Almost everything one does with matrices assumes they have entries in a ring, not in a vector space. While one can (ab)use matrix product notation as a shorthand for (systems of) sums-of-products of things that are not scalars (such as a single linear combination in your example, which is a sum of vector$\times$scalar products), the "gain" of doing so is negligeable with respect to the confusion it causes. Notably, there is hardly any sense one can make in the example of $B^{-1}$: a list of vectors is just not something that allows a reasonable interpretation of "inverse".

What you can do, given an ordered basis$~B$, is associate to it a linear map (in fact isomorphism) $\phi_B:\Bbb R^n\to V$ that associates to a list of scalars the corresponding linear combination of vectors of$~B$. This map has an inverse map $\psi_B:V\to\Bbb R^n$ that associates to a vecor its coordinates with respect to the basis$~B$.

So to resume: you are entirely justified in you incomprehension of the symbol $B^{-1}$. What is meant is the inverse isomorphism $\psi_B$ of the "linear combinations of$~B$" isomorphism $\phi_B$.

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Not sure I understood the question. Please ask for clarification if I can elaborate on something for you.

It is sometimes convenient to deal with other more abstract vector spaces (not just $\mathbb{R}^n$) like the space of all square-integrable functions $$ L^2[a,b] = \left\{f:[a,b] \to \mathbb{R} \left| \int_a^b f(x)^2 dx < \infty \right. \right\}, $$ and talk about bases and coordinates there with respect to these bases. You use the basis that is convenient for you (e.g. in such spaces, Fourier series can be viewed as coordinates for such a basis).

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  • $\begingroup$ "It is sometimes convenient..." No! Linear algebra is about abstract vector spaces. That a lot of attention is given to spaces $\Bbb R^n$ is only because thanks to the choice of bases, one can find isomorphisms of any finite dimensional vector space with some space$~\Bbb R^n$. $\endgroup$ – Marc van Leeuwen Dec 12 '13 at 14:49
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$B$ is a linear transformation from a general vector space $V$ into $\mathbb{R}^n$ or something similar like $\mathbb{C}^n$. It can itself be represented as a matrix if $V$ happens to also be $\mathbb{R}^m$, but that need not be the case.

Example: $V=P_2(t)$, polynomials in degree at most 2, in the variable $t$. We take a basis $B=\{1,t,t^2\}$ for $V$. Now, $B$ maps the polynomial $a+bt+ct^2$ to the vector $(a,b,c)$ in $\mathbb{R}^3$. $B(2+3t)=(2,3,0),\ B(t-t^2)=(0,1,-1)$. In this case $V$ doesn't have explicit coordinates, so $B$ isn't a matrix.

You may point out that $V$ has implicit coordinates, i.e. we can call the constant term the "first" coordinate, and so on. That's true, and that observation is exactly what $B$ is implementing -- it is making implicit coordinates (as described in $B$) explicit.

There's another natural choice for $B$, namely $B'=\{t^2,t,1\}$, which is $B$ in a different order. This leads to different outcomes in $\mathbb{R}^3$; $B'(2+3t)=(0,3,2),\ B'(t-t^2)=(-1,1,0)$. Which one is right? Whichever you want, that's why we specify a basis $B$ so that we can choose how to impose coordinates from $V$.

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There seems to be some notational confusion; in your question, $B$ is being used in two different ways.

A basis is a collection of vectors: ${\cal{B}}=\{ b_1 , \ldots , b_n \}$.

Given a basis, you can form a matrix whose columns are these basis vectors: $B = [ b_1 \ b_2 \ \cdots \ b_n]$.

If $v \in \mathbf{R}^n$ is a vector written in standard coordinates, and if $[v]_{\cal{B}}$ denotes $v$ written in ${\cal{B}}-$coordinates, then we have the relation $v = B [v]_{\cal{B}}$, or equivalently $B^{-1}v=[v]_{\cal{B}}$.

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  • $\begingroup$ Do you mean that $\vec v$ simply stands for $[v]_I$ (here I use $I$ = identity matrix to signify the standard basis). $\endgroup$ – Val Dec 12 '13 at 15:15

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