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My professor claims that there is no such thing as a vector without a basis, but I claim that there exists "raw" vectors.

For example, let's say you have a basis in $\mathbb{R}$:

$$\beta = \{\langle 1, 0 \rangle, \langle 0, 1 \rangle\}$$

Then the vectors in $\beta$, are they with respect to a basis, or are they "raw" vectors?

Perhaps my terminology is not correct, but I hope my meaning is clear, that vectors can exist without a basis, for example $m \in M_{2 \times 2}(\mathbb{F})$. I think you would write it as a coordinate vector if it were in terms of a basis, and as a matrix in "raw" format.

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    $\begingroup$ Absolutely, the definition of a vector doesn't rely on a basis $\endgroup$ Commented Jun 9, 2023 at 22:38
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    $\begingroup$ Vectors can exist regardless of whether a basis is specified, but how you represent them definitely can be dependent on the basis chosen. Kind of like how you know the number $10$ exists, but you can write it as $9.999\cdots$, or can write it in base $2$ as $1010_2$, or so on. $\endgroup$ Commented Jun 9, 2023 at 22:40
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    $\begingroup$ A vector is an element of a vector space, with properties of closure under addition and scalar multiplication, for example. Considering (1, 0) as some object on its own does not look terribly useful $\endgroup$
    – Paul
    Commented Jun 9, 2023 at 22:45
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    $\begingroup$ Basis is an attribute not of vector, but of vector space. Phrases like “a vector without basis” or “a vector with a basis” are meaningless. $\endgroup$
    – user1551
    Commented Jun 10, 2023 at 1:29
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    $\begingroup$ A basis is made of vectors. So we would be in egg-and-chicken territory if a vector required a basis. $\endgroup$
    – Stef
    Commented Jun 10, 2023 at 14:32

8 Answers 8

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In Euclidean (e.g. plane) geometry: observe pairs of points $(A,B)$ in a plane and define that the pairs $(A,B)$ and $(C,D)$ are equivalent if there exists a translation which maps $A$ to $B$ and $C$ to $D$. (A translation is a composition of two reflections against two parallel, not necessarily distinct, lines.) Now, being equivalent is an equivalence relation on all pairs of points, so now we can observe equivalence classes. Label the equivalence class of the pair $(A,B)$ as $\vec{AB}$ and here you go, you've defined vectors in geometry without ever mentioning coordinates!

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  • $\begingroup$ This feels like cheating because it seems so coordinates, are indirectly baked in when you write it as a pair of points $\endgroup$ Commented Jun 10, 2023 at 9:41
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    $\begingroup$ @HopefulWhitepiller Okay, I should've said what geometry I am talking about. Say, Hilbert's axiomatization, or any other equivalent axiomatization Euclidean style. I am not sure how coordinates would be baked there. (You are probably aware that axiomatizations of geometry have been around since ancient Greek times, while the link to the numbers via coordinates was only established by Descartes.) $\endgroup$
    – user700480
    Commented Jun 10, 2023 at 10:15
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You can certainly present vector spaces, and vectors in them, without any basis.

For instance, take a vector $\newcommand{\x}{\mathbf{x}}\newcommand{\y}{\mathbf{y}} \x \in \newcommand{\R}{\mathbb{R}}\R^n$, and take its orthogonal complement, $V_x := \{ \y \in \R^n \mid \x \cdot \y = 0 \}$. Now $V_x$ is a subspace of $\R^n$, so it’s a vector space, but I haven’t given you a basis for $V_x$, and finding one is non-trivial — but it’s also unnecessary for working in $V_x$ and reasoning about it.

But some basis will always exist — this is an important theorem of classical algebra — and once you choose a basis, every vector can be expressed using it.

So you’re certainly right that you can give a vector space, and vectors in it, without reference to any basis, and you can do lots of work with vector spaces just fine without using a basis. Working this way is crucial for lots of applications in modern algebra, especially in differential geometry/topology. It’s good to learn how to work in a basis-free way.

On the other hand, I guess your professor may have meant “There is no vector space that doesn’t have a basis” — in which case they’re also right. Bases will always exist, even if you’re not using them or looking for them, and vectors will always have basis representations.

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    $\begingroup$ This answer is only true for finite-dimensional vector spaces. If the vector space has infinite dimension then whether it has a basis depends on the axiomatic foundation you choose for doing mathematics. $\endgroup$ Commented Jun 10, 2023 at 16:08
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    $\begingroup$ @DavidA.Craven I work a lot in constructive maths myself, so I appreciate that subtlety and considered mentioning it — but I think it’s a bit of a side point, since taking the axiom of choice as given is pretty much the mainstream standard among mathematicians. $\endgroup$ Commented Jun 10, 2023 at 21:11
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In this answer, for the sake of simplicity I restrict my attention to finite-dimensional real vector spaces. However, all of the ideas generalise straightforwardly to a finite-dimensional vector space over an arbitrary field.

In linear algebra, the term "vector" is an informal term used to refer to an element of a given vector space. For instance, if our vector space is $\mathbb R^2$ together with the usual vector addition and scalar multiplication, then a "vector" is simply an ordered pair of real numbers $(a,b)$. These basic notions do not involve bases at all.

However, given an $n$-dimensional vector space $V$ (where $n\in\mathbb N$), it is possible to represent the elements of $V$ with $n$-tuples of real numbers. For example, consider the set $V$ of real polynomials taking the form $ax^2+bx+c$. There are natural notions of vector addition and scalar multiplication that allow us to regard $V$ as a vector space. Moreover, $V$ has the (ordered) basis $\{x^2,x,1\}$. With this basis, we can represent polynomial $ax^2+bx+c$ with the ordered triple $(a,b,c)$.

In the general case, if $\{v_1,\dots,v_n\}$ is an (ordered) basis of $V$, then every $v\in V$ can be written as $v=a_1v_1+\dots+a_nv_n$ where $a_1,\dots,a_n\in\mathbb R$. The coefficients $a_1,\dots,a_n$ are uniquely determined, meaning that there is a well-defined map $\varphi:V\to \mathbb R^n,v\mapsto (a_1,\dots, a_n)$ (note however that the coefficients $a_1,\dots,a_n$ are dependent on $v$, so we should write something like $a_{v,1},\dots,a_{v,n}$ if we were to use more pedantic notation). It is customary to call $(a_1,\dots,a_n)$ the coordinates of $v$ with respect to the basis $\{v_1,\dots,v_n\}$. For example, if $V$ is the set of polynomials as above, then the coordinates of $ax^2+bx+c$ with respect to the basis $\{2x^2,-x,5\}$ is $(a/2,-b,c/5)$. This example also serves to illustrate that, in general, there is not always a "natural" or "obvious" correspondence between a vector $v$ and the coordinates of $v$ with respect to a certain basis.

If $V=\mathbb R^n$, then we have the slightly confusing situation where the coordinates of a vector $v\in\mathbb R^n$ is another element of $\mathbb R^n$. This means that, in contrast, to the general case, the vectors and the coordinates of vectors "live in the same space". For instance, if $\mathbb R^2$ is our vector space and $\{(1,2),(2,5)\}$ is our basis, then the coordinates of $(4,10)$ is $(0,2)$. However, if our basis is $\{(1,0),(0,1)\}$, then the coordinate of every vector is itself. This is why the basis $\{(1,0),(0,1)\}$ is called the "standard", "canonical", or "natural" basis. A priori, it seems unclear why we would want to represent the representations of vectors in $\mathbb R^n$ in anything other than the standard basis. However, it turns out that in certain situations, changing the basis makes computations like matrix multiplication much easier.

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  • $\begingroup$ Thanks for the very clear response. My only question is you mention ordered and unordered bases. When you say the standard basis $\{(0,1), (1,0)\}$. Is this equivalent to the basis $\{(1,0),(0,1)\}$ (which I am used to calling the standard basis). In class, we learned a basis is an ordered set, so would these be different bases? It seems that in practice, yes, because they create very different coordinate vectors (which I now understand is just another vector in $\mathbb{R}^2$. $\endgroup$ Commented Jun 11, 2023 at 18:55
  • $\begingroup$ @user129393192: The standard basis is $\{(1,0),(0,1)\}$, and in context this should be considered different to $\{(0,1),(1,0)\}$; my answer contained a mistake that I have now edited out. In linear algebra, there tends to be lots of ambiguity about what the formal definition of a basis "is": is it an (unordered) set, an indexed family, a multiset, or a "list" of vectors? Different authors use different conventions. $\endgroup$
    – Joe
    Commented Jun 11, 2023 at 19:56
  • $\begingroup$ The problem with simply saying that a basis is a set of vectors is that, as sets, $\{(1,0),(1,0),(0,1)\}=\{(1,0),(0,1)\}$, so $\{(1,0),(1,0),(0,1)\}$ ought to be a basis of $\mathbb R^2$, but few people want to say that. Note that saying "$\{(1,0),(0,1)\}$ is an ordered basis" is clearly an abuse of notation/language, since the notation $\{\dots\}$ ought to be reserved for (unordered) sets. However, one learns to tolerate such abuses. $\endgroup$
    – Joe
    Commented Jun 11, 2023 at 19:56
  • $\begingroup$ I see. I prefer the pedantic way of things. How would you define a basis pedantically? Ordered, unordered? A set, not a set? $\endgroup$ Commented Jun 11, 2023 at 19:59
  • $\begingroup$ @user129393192: It's actually quite hard to do that in a way that conforms with how people like to think of "linearly independent" and "spanning" collection of vectors. Probably the best way is to use "multisets" – these are like sets in that they are unordered, but unlike sets in that duplicates are accounted for. This allows us to make statements such as $\{(1,0),(1,0),(0,1)\}$ is linearly dependent, which is what we want. However, multisets do not see much usage in mathematics, largely for historical reasons. Now, by default, the term "basis" does not involve any ordering. $\endgroup$
    – Joe
    Commented Jun 11, 2023 at 20:34
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A vector $x$, by definition, is an element of a vector space $V$ (together with a definition of addition and multiplication by scalar), which is a set of elements that satisfy the basic properties that if $x$ and $y$ are elements, then so is $x+y$, as well as commutative and associative properties etc. The trivial vector space $\{0\}$ satisfies all of these properties. So does the real line.

A basis, once chosen, allows us to uniquely represent our vectors in a neat fashion: if $x$ and $y$ have representations $\{1,0\}$ and $\{0,1\}$, then the vector $x+y$ has the representation $\{1,1\}$.

A basis is also useful for linear transformations. A linear transformation maps element $x$ in $V_1$ to element $y$ in $V_2$. But if we have bases for $V_1$ and $V_2$, then we may represent the linear transformation using a matrix, which describes the transformations of the basis elements. So a basis is a convenient tool that allows us to perform computation in a neat and simple fashion.

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    $\begingroup$ How does it answer the question? $\endgroup$
    – Artem
    Commented Jun 10, 2023 at 2:29
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    $\begingroup$ The definition of vector doesn't rely on a basis $\endgroup$ Commented Jun 10, 2023 at 5:41
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The notion of vector arises when the set in subject is a vector space in itself. Each element of a vector space $V$ (on satisfiability of certain axioms) is called as vector. The set $\beta=\{(1,0),(0,1)\}$ being a subset of vector space $\mathbb R^2$ serves as a basis of it so that each vector $v\in\mathbb R^2$ can be expressed as the linear combination of elements of basis $\beta$.

Moreover, contrary to your claim the vector $(1,0)$ too is expressible as $(1,0)={\color{red}1}(1,0)+{\color{red}0}(0,1)$ i.e. the linear combination of the vectors in basis $\beta$.

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Whenever you write down numbers to represent a vector, you are inherently using a basis; you should think of a basis as nothing more than a way of fixing units.

Let's take a simple example in $\mathbb{R}$. After fixing an origin to call the origin on the real line, you next want to describe the relation a point $p$ is from the origin. Colloquially, you might $p$ is either

  1. 5 kilometers
  2. 5000 meters

from the origin. In the first statement is equivalent to picking the basis for $\mathbb{R}$ as the point which is 1 kilometer away from the origin; the second equivalent to picking the basis for $\mathbb{R}$ as the point which is 1 meter from the origin.

Of course, the point $p$ exists outside of the choice of units, so indeed it exists as a "raw vector". However, if you want to assign any numerical value to it, this is equivalent to choosing a basis.

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Yes, but it doesn't matter

Every finite dimensional vector space (what you encounter in basis LA courses) admits a basis. Hence thinking of a vectors through defined by a finite spanning set or thinking of them as their own thing doesn't make much of a difference. Heck, I'd argue that the profs view point maybe be more practical (albeit wrong) since you could do some proofs easier by assuming a basis.

Disregarding that, in the formal definition of a vector space, we have two sets and two binary operations. The two sets are vectors and the scalars, the binary operations are how we can combine vetors, and secondly, how scalars combine with vectors.

So, the vectors in this definition, would be defined by the set itself. For example, you could think of $\mathbb{R^2}$ as everything that could be spanned by the basis $\{ (0,1) , (1,0 ) \}$ , or, by the set of pairs of real numbers.

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@MISC {4716963, TITLE = {Is there such a thing as a vector without a basis?}, AUTHOR = {Mr. Brown (https://math.stackexchange.com/users/448217/mr-brown)}, HOWPUBLISHED = {Mathematics Stack Exchange}, NOTE = {URL:https://math.stackexchange.com/q/4716963 (version: 2023-06-11)}, EPRINT = {https://math.stackexchange.com/q/4716963}, URL = {https://math.stackexchange.com/q/4716963} } I agree. It is like in order to take a measurement, there needs to be a unit to which compare your measurement to.

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  • $\begingroup$ This is a comment to an answer, put it there and delete this answer. $\endgroup$
    – Piita
    Commented Jun 19, 2023 at 0:23

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