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Some mathematitians told me that vector components and coordinates are different things. They say that vector $F^n$ always has N components but coordinates depend on chosen basis and, therefore, it is meaningless to talk about coordinates when basis is unspecified. But don't components depend on basis? Why can you talk about components regardless of the basis? Do they just demonstrate their domination by being too picky and making the difference between the things, which are the same, and, thus, causing confusion? Later they mock my definition of vector space, where I say that it is a collection of vectors. They say that it is wrong since vector space an Abelian group whose elements can be scaled. But, I see no difference between these definitions.

edit If this answer and Wikipedia are correct then $\psi_i = \langle \psi|i\rangle$ in

$$\vec \psi = \begin{bmatrix}|1\rangle |2\rangle \cdots |n\rangle\end{bmatrix} \begin{bmatrix} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_n \end{bmatrix}$$

must be coordinates in the basis $\begin{bmatrix}|1\rangle |2\rangle \cdots |n\rangle\end{bmatrix}$, which I believe is different from the standard basis. Why do quantum mechanics say that $\psi_i$ are components then rather than coordinates? Is it incorrect?

Miami Operators and Matrices also say

Now you see where the defining equation for operator components comes in. Eq. (7.7) is $$\sum_k u_k \vec e_k = \sum_i v_i \sum_k f_{ki} \vec e_k$$

Actually, I was motivated to ask this question when tried to represent these equations, which involve integration, in vector form and do not understand why everybody calls the coefficients $u_k$ in $\sum u_k \vec e_k$ components rather than coordinates. It seems to me that matematicians use these terms interchangably and troll the noobs with immaterial difference.

PS Dec 2013 I see here

enter image description here

that components, $a_i \psi_i$ are contrasted with coefficients $a_i$. Are components just another names for coordinates or we have a 3rd type of object?

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"Later they mock my definition..." that's just mean. Especially if they give you an effectively useless (for your situation) definition. Don't take that too much to heart. But mathematically they have a point. –  Willie Wong Jun 26 '13 at 7:37
    
To be honest, this question seems very confusing to me. Positions have coordinates, but I've never heard of other, more general vectors having coordinates. –  Muphrid Jun 26 '13 at 18:17
    
en.wikipedia.org/wiki/Coordinate_vector –  Val Jun 26 '13 at 19:06
    
Do they just demonstrate their domination by being too picky and making the difference between the things, which are the same, and, thus, causing confusion?... The same mathematical question can be asked without such futile expressions of hostility. –  Did Jul 2 '13 at 5:41
    
Re: Dec 4 edit, this looks like physics. Physicists and mathematicians use different language for the same things all the time, so it helps to be comfortable with one system before making sense of the other. That said, it does look like your author is using "component vectors" in contrast with "coefficients" (scalars) and is conflating/confusing "component vectors" with "coordinate vectors". This is an extremely common abuse of terminology and you're best off to roll with it. –  Eric Stucky Jan 8 at 23:17
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3 Answers

This is a tremendously common confusion to have, and in my experience, people are notoriously bad at explaining this concept. I'm sorry that you had to deal with people who were abrasive in addition to poor expositors.

In an arbitrary vector space, you cannot talk about components. They actually don't exist. Now, you can impose them on a finite-dimensional space by providing a bijective linear transformation from the arbitrary vector space to $F^n$, but then they're just that: an imposition, because any other bijective linear transformation will choose different would-be "components".

Components exist in $F^n$ because of the actual nature of the objects involved. So you don't need a basis, you can just look at an arbitrary object $(a,b,\dots,n)$, and find any of its components, because they're built into the object. This can be confusing because we also write coordinate vectors in this way, and when the basis is the standard basis, there is no difference between the components and the coordinates. However, in any other basis, there will be a difference.

(Edit: Val made an important point in the comments. I should have been more careful when I said there was "no difference". The fact is that coordinates and components are never conceptually the same, but I meant to say that in the standard basis case they will be numerically equal.)

Lacking a basis at all, you might want to say that $F^n$ still has coordinates implied by its components. But, in my opinion, this seems silly, since you cannot do the same in other spaces.

So the short answer is: Yes, there is a difference, because components are part of the objects.

As for your "collection of vectors" notion, they are basically the same. But it is easy to imagine a collection of vectors which is not a vector space: for example the circle in $\mathbb{R}^2$. This is definitely a collection, and the objects in it are definitely vectors, but it is not a vector space.

What I assume you meant by "collection" was what we might call a "meaningfully structured collection", and the meaningful structure is described precisely as an abelian group over which elements can be scaled by objects in a field. In that sense, your notion is correct, though a bit less transparent.

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Initially I started to think of components as fields in object-oriented programming or columns of SQL table whereas coordinates are their values. You simply need N numbers to describe a state -- you define N fields in the object. Programmers simply call (state) vectors the "objects". But, if you say that components may equal coordinates then components are not like fields. They are more like values also. Suppose our basis is half the standard basis. The coordinates will be twice the standard coordinates in this case. What would be the components? –  Val Jun 26 '13 at 8:03
    
The components are still built into the vector, so in that example (and any example) the components are the same as the standard components. As to your other point, I should have been more careful to emphasize this important distinction: coordinates and components are never conceptually the same, but under the standard basis they are numerically equal. As a programmer, you have to deal frequently with memory location vs. return values, so this should be a familiar concept :) But I want to make sure you see that it is happening here. –  Eric Stucky Jun 26 '13 at 17:18
    
@EricStucky describing the act of fixing a basis as an "imposition" is such a fantastic word to use –  citedcorpse Jun 26 '13 at 18:11
    
Wait. The memory address corresponds to component index. Hardly anybody needs such a basis the vector coordinates become <1,2,3> and, for sure, this numeric match does not work for other vectors in the same basis and this meaningless basis only occasionaly can be the standard basis. So, I see no way to identify the components with object fields/memory addresses. –  Val Jun 27 '13 at 8:20
    
I'm afraid I don't understand the objection. Obviously a ($F^n$) vector's components will not match a different vector's components, regardless of basis. It's possible that I have the wrong intuition about the computer metaphor; I have an extremely limited experience with programming. What I meant to say is that components are in some sense "fundamental" in $F^n$ because they are built into the objects. –  Eric Stucky Jun 27 '13 at 16:07
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The vector space $F^n$ is the set of all $n$-tuples of elements of $F$. So given a vector $(a_1, \dots, a_n) \in F^n$, we can define the $i$'th component of $(a_1, \dots, a_n)$ to be $a_i$. To define the coordinate vector of a vector, we need to fix an ordered basis $\beta = \{v_1, \dots, v_n\}$. For any $v \in F^n$, by definition, we can find scalars $c_1, \dots, c_n$ such that

$$v = c_1v_1 + \dots + c_nv_n$$

We define the coordinate vector of $v$ with respect to $\beta$ to be

$$[v]_\beta = (c_1, \dots,c_n)$$

Note that if $\beta$ is the standard basis, we have $[v]_\beta = v$ for any $v \in F^n$. Coordinate vectors are always defined for vectors in a vector space with an ordered basis. However, components are only defined in $F^n$.

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I see no difference between $a^n$ and $c^n$! –  Val Jul 4 '13 at 20:59
    
@Val The two will coincide if and only if $\beta$ is the standard basis. –  Vectk Jul 4 '13 at 21:06
    
@Val Take for instance, $\beta = \{(2, 0), (0, 2) \}$ to be a basis for $\mathbb{R}^2$. Then for $v = (4, 4) \in \mathbb{R}^2$, $[v]_\beta = (2, 2)$, so $a_n$ and $c_n$ are different! –  Vectk Jul 4 '13 at 21:11
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Finally, I see the difference. But, what are (2,0) and (4,4)? Why are they components? I see that coordinates are defined w.r.t specific basis. How do you define (4,4) without the basis? Is standard basis is implied if basis is not specified? What does this differentiation gives to you? You have to stop somewhere and specify your basis vectors in coordinates of some another basis. Do you call vector coordinates in this basic basis "components" or what? What is the problem to define the coordinates the same way: just imply the standard basis if it is not stipulated? –  Val Jul 4 '13 at 22:27
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Components are used to describe general vectors in a given vector space within a basis in terms of linear combinations. Components of a given vector also are coordinates in vector spaces, because univocally determine the position. But talking in more general spaces (like manifolds) I think that nobody says that a given position has certain "components" instead of coordinates.

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try to google up: "components of a vector" versus "coordinates of a vector" and see the difference in quantity of use –  janmarqz Dec 27 '13 at 0:30
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