# Why is there implied an equality between vectors and $n$-tuples?

Are they considered equal in some sense?

For instance, I always write "...for vector $\mathbf{x} \in {\mathbb{R}}^n$ we have ...". I have a small problem with this (not a big one). The problem comes from the fact that the "cartesian power" of a set is defined as: ${\mathbb{R}}^n = \underbrace{ \mathbb{R} \times \mathbb{R} \times \cdots \times \mathbb{R} }_{n}= \{ (x_1,\ldots,x_n) \mid x_i \in \mathbb{R} \ \text{for all} \ 1 \le i \le n \}$. In other words, we are implying that the vector $\mathbf{x}$ is a tuple because we, with the relation "is an element of" ($\in$), are defining the $n$-dimensional vector to be a member of a set ${\mathbb{R}}^n$ in where members are $n$-tuples.

Is there implied a tiny mathematical abuse of notation here? Or have I completely lost it? :)

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I think it is better if you think vectors as attachments to a fixed point, say 0. Denote vectors by $v_0$ and implement them as direction arrows of paths passing though 0. Then the next question will be on transporting this collection of directions to other points of $\mathbb{R}^n$. – niyazi Dec 5 '11 at 13:59
@niyazi: But I thought that we gain most power from seeing vectors as constructs that are exactly not attached to a point? – Ole Thomsen Buus Dec 5 '11 at 15:56

In general, a vector is not a tuple. But in the specific case of $\mathbb{R}^n$, where $n$ is a natural number, the elements of this space are tuples, because that is how the space is defined.

Nevertheless, although these vectors happen to be tuples, it is often more elegant to pretend that they are atomic objects, and work in a "coordinate free" way. This emphasizes the geometry of the vector space rather than its algebraic aspects.

On the other hand, given a finite dimensional vector space $V$ over a field $F$, say of dimension $k$, the original space is isomorphic to the vector space $F^k$ of $k$-tuples of elements of $F$. So in a sense these "tuple spaces" capture all finite dimensional vector spaces up to isomorphism.

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Yes I found what you are talking about in the last paragraph. It makes sense now. I just needed the definition that satisfies me (so the validity comes from the definition of "coordinates spaces"?) – Ole Thomsen Buus Dec 5 '11 at 15:59
Can you show an example of the general case where a vector is not a tuple? Than what is it? – notorious May 15 '15 at 5:22
@notorious: what examples of vector spaces are you familiar with? Matrices? Functions? – Carl Mummert May 15 '15 at 10:52
Can you give a concrete example please? – notorious May 15 '15 at 17:12
The space of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$ is a vector space over $\mathbb{R}$. So the function $f(x) = 3x+1$ is a vector in that space, and is not a tuple. – Carl Mummert May 15 '15 at 19:14

I don't understand the problem. A vector is an element of a vector space. A vector space is any set equipped with appropriate operations satisfying the vector space axioms, and $\mathbb{R}^n$ is an example of such a thing.

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Thanks. Would you say that we are actually, more specifically, talking about a coordinate space (en.wikipedia.org/wiki/Coordinate_space#Definition) which by definition is vector space and a set of $n$-tuples? And, even more specifically, we are talking about the real coordinate space if the underlying field is the set of real numbers $\mathbb{R}$ (i.e., Euclidean space). Because if this is the case, I am satisfied :) – Ole Thomsen Buus Dec 5 '11 at 14:35
@Ole: yes. Is there more to it than that? – Qiaochu Yuan Dec 5 '11 at 15:03
No there is probably not :) – Ole Thomsen Buus Dec 5 '11 at 15:42

Once you choose a basis for your vector space, you get a one-to-one correspondence with $\mathbb R^n$. For most purposes, then, we may say that the vector space is $\mathbb R^n$. But not for all purposes. At a certain point in their education, a math student has to learn to think of an abstract vector space not given as a space of $n$-tuples.

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Thanks. It seems that I had missed a few points with the more fundamental definitions. Se my comment on Qiaochu's answer. If you a are working enterily in $n$-dimensional euclidean space over the field $\mathbb{R}$ (or $\mathbb{C}$) it seems to be a fully valid assumption that ${\mathbb{R}}^n$ is in fact a vector space. hey, I never studied this ... just fooling around with wikipedia and some books :) – Ole Thomsen Buus Dec 5 '11 at 15:52

You can kind of say that a vector is a tuple but there isn't much to gain by doing that. When working with vectors we usually care more about the linear algebraic properties (vectors can be summed, mutiplied by scalars, etc) and less concerned about combinatorial and structural properties (like saying things are a tuple or not).

Also, the tuple analogy falls off a bit when you move to infinite-dimensional vector spaces, like functions or polynomials.

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Well, I am a computer programmar more than a mathematician. I naturally put much emphasis on "type" and the use of that type in certain contexts. I think it might be possible to go into corners of pure math, where such emphasis becomes quite entangled and unecessary. But not in this particular example it seems. – Ole Thomsen Buus Dec 5 '11 at 16:01
Well, computer proof systems, like Coq, also usually give every value a type, and values of different types are usually not interchangeable. It's just that in mathematics, converting between tuples and vectors is often a natural thing to do (so natural that we don't even bother saying we're doing it), just like in computing, converting between, say, characters and numbers is often a natural thing to do (although since our audience is a computer, not a human, we have to explicitly say so). In short, math and programming are not that different in this regard. – Tanner Swett Dec 5 '11 at 16:27

The word "vector" is used to mean different things in different contexts. A point in $\mathbb{R}^n$ is often called a "vector".

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But a point in $\mathbb{R}^n$ is a vector since $\mathbb{R}^n$ is a vector-space - right? – Ole Thomsen Buus Dec 6 '11 at 10:36