In one of Feynman's lectures he mentions that:

Of course it is generally not true that any three numbers form a vector

In order for it to be a vector, not only must there be three numbers, but these must be associated with a coordinate system in such a way that if we turn the coordinate system, the three numbers “revolve” on each other, get “mixed up” in each other, by the precise laws we have already described.

A quick google about vectors and I see things like " a vector is an element of a vector space". Wikipedia (https://en.wikipedia.org/wiki/Vector_space) has a list of 8 axioms that must be satisfied to have a vector space. Reading through these axioms, I can't imagine any 3 numbers that when combined to form a vector, wouldn't satisfy these axioms. I'd love some help in understanding what Feynman is meaning here.

  • 19
    $\begingroup$ I can almost guarantee that you’ll end up with more confusions if you try to understand vectors the way Feynman suggests. Learning linear algebra proper, the way a mathematician would (vector spaces and all) will easily bypass many questions and misconceptions you may come up with/have now and in the future (if you want to learn higher level math and physics…GR, QM…), as I’ve seen repeatedly with the different students I’ve TAed. See for example my answer on MSE What exactly is a "vector" in math (in terms of vector spaces)? for more remarks. $\endgroup$
    – peek-a-boo
    Jun 13 at 5:49
  • 4
    $\begingroup$ @peek-a-boo The central issue is that the definition of vector in maths is much less restrictive than the physical definition of vector. And in this case it should be that the maths that should change—the name "Linear space" is a better fit. So, if you try to look in the maths texts to learn about vectors, it is very likely that the basic stuff would completely miss what Feynman is referring to, i.e. your advice is incredibly dangerous. $\endgroup$ Jun 13 at 6:26
  • 4
    $\begingroup$ Three consecutive house numbers? $\endgroup$
    – jim
    Jun 13 at 9:39
  • 40
    $\begingroup$ I suspect you may have fundamentally misunderstood Feynman's claim. He's not saying "there are some numbers (abstract integers) that don't make up a vector". He's saying "to be a vector those numbers have to be associated in some manner that allows them to interact" - you need numbers and a description of what those numbers mean. So yes, any 3 numbers you can think of form a vector if you interpret them as, say, spacial co-ordinates. $\endgroup$
    – Brondahl
    Jun 13 at 11:02
  • 2
    $\begingroup$ @naturallyInconsistent I agree that the concept of vector Feynman is referring to here is a specialization of the math one, and indeed terminology is somewhat unfortunate, but everyone is guilty here (since math and physics people overload and abuse terminology). Anyway, let’s not make this a math vs Physics thing. My point is that for a student to appreciate the transformation-based definition of a concept (tensor or vector) requires so much more effort on the students’ part, in order to truly separate the idea of the geometric object being referred to from its components. $\endgroup$
    – peek-a-boo
    Jun 13 at 13:14

10 Answers 10


If I click the "share" link under your question, I get the URL https://physics.stackexchange.com/q/768023/44126. If I click on your username, I get the URL https://physics.stackexchange.com/users/295770/zinn. There are three numbers in these two links:

  1. the database index of your question, 768023
  2. the database index of my user account, 44126
  3. the database index of your user account, 295770

You could combine these numbers into a list, where the first number is the post ID, the second number is the user ID of the person who has requested a "share" link, and the third number is the user ID of the person who has made the question. But now you have to ask whether a process of "vector addition," which is a prerequisite to the axioms you list in the Wikipedia article, even makes any sense. If you just use ordinary addition on the different "components" of the "vector," then the results of "adding up two posts" don't really mean anything. In particular, the space of posts is not closed under addition: if we construct one of these "vectors" for somebody else's post, and add up the database IDs, we will almost certainly not get a "vector" that points to another Physics Stack Exchange post. The additive identity isn't a member of the vector space, because there isn't any post or user with database identifier zero. There is no additive inverse in the vector space, because there aren't any database identifiers with negative numbers.

Note the Wikipedia page you link to defines a mathematical vector space. When physicists talk about vectors, including in your Feynman quote, we are typically not talking about any possible abstract vector space, but specifically about vectors which can describe the locations of objects relative to one another in the world where we live. Those physical vectors are subject to additional transformation rules, which allow you to get from the three numbers you would choose as the coordinates of your vector to the three numbers which would be chosen by someone who is looking at the same objects from a different seat at your table. If this other observer is sitting across from you, or is suspended from the ceiling above you, their coordinates will be related to yours by a "rotation." It is the rotation properties which define physics vectors.

  • 12
    $\begingroup$ I find it quite dangerous and not scientifically based to maintain a difference between vectors in Physics and in Mathematics. $\endgroup$ Jun 13 at 7:16
  • 2
    $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 I agree. But in the mathematical sense, what physicists call spinors are also "vectors", though they differ from physicists' vectors in the way they transform under rotation. Perhaps giving another name to physicists' vectors would solve the problem? Or perhaps it'll become even more confusing... $\endgroup$
    – Jono94
    Jun 13 at 9:15
  • 9
    $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 Keeping the distinction between physics and mathematics is essential. The objects mathematics studies do not exist outside of human minds, but the objects that physics studies exist in nature. The disciplines are different: mathematics is constrained by consistency while physics is constrained by real phenomena. Mathematical objects like vectors are thus only useful to physics to the extent that they accurately model the phenomena in question. That requires meaning, not consistency. Indeed, we get to be sloppy about consistency, because nature rules. $\endgroup$
    – John Doty
    Jun 13 at 12:41
  • 22
    $\begingroup$ @JohnDoty, that's an opinion not shared by all the physicists. Usually, after studying some more math, one discovers that sloppy usage of math is not necessary to do good Physics. $\endgroup$ Jun 13 at 13:07
  • 2
    $\begingroup$ The distinction between vectors in mathematics and physics is a pragmatic one. Yes, to mathematicians any three (or two, or whatever) numbers can form a vector. A physicist additionally simply demands the vector to be a useful sensible entity for physics purposes, or they are just not interested. $\endgroup$ Jun 13 at 22:59

Temperature, density, and water vapor concentration at one point in the atmosphere are three numbers. However, any triple of such values is unaffected by a change in the spatial reference frame. More or less, this is the idea behind Feynman's words.

However, Feynman's argument was not quite accurate from a mathematical point of view. He made a good point, stressing that a triple of numbers does not necessarily represent a vector, but he made a wrong example and missed the real problem.

Let me show a counterexample where the weakness of Feynman's argument can be easily seen, showing at the same time why not every triple of number can be interpreted as the set of components of a $3D$ vector.

Let's associate an arrow to a rotation in the space, such that the arrow orientation encodes the axis of rotation and the rotation direction, the arrow length is the size of the rotation angle.

Of course, we can associate three components to such arrows, and clearly, any change of the cartesian axes would result in the "correct" transformation: if we turn the coordinate system, the three numbers “revolve” on each other, get “mixed up” in each other, by precise laws as discussed by Feynman.

Still, we cannot say that the set of all possible rotations in $3D$ is a vector space, and consequently, our well-transforming three numbers are not the components of a vector representing a rotation. The reason is that in order to have a vector we must say something about the meaning of the operations defined on the set of objects. In particular, to make contact with the mathematical definition of vector space, we must define the multiplication by a scalar and the sum of two rotations. No problem emerges from the multiplication by a scalar. The sum of two rotations instead, is not the same as the usual parallelogram rule for two arrows. It can be easily seen by applying two successive $90^{\circ}$ rotations around two orthogonal directions, say around the $z$ and $x$ axis. The final result of applying before the $z$, and after the $x$ axis rotations is not the same if the order of the two rotations is inverted (some nice images can be found in the answers to this question on Mathematics Stack Exchange ). This implies that the sum of two rotations, defined as the final rotation, is not commutative, and this implies the failure of associating a vector space to the $3D$ rotations. Actually, it is well known that only infinitesimal rotations can be associated with vectors.

In summary, the main claim in Feynman's lecture is correct: it is not enough to associate three real numbers to a physical quantity to have a vector. However, the reason is not what Feynman's said, but the fact that the mathematical definition of a vector space contains, as a fundamental ingredient, the association of combination rules to the set of elements, and such rules have some constraints. In particular, the sum must be commutative. Notice that the composition rules are at the core of the possibility of introducing the vector components and their transformation rules. Therefore they should be considered the most basic requirement to define what a vector is.

  • $\begingroup$ I like your answer the best but I think it still misses the fundamental difference between vectors and representations of vectors as linear combinations of elements of a basis of that vector space. Components only appear once we fix a basis and now we have the problem that there are many possible choices for the basis. A combination of coefficients without a choice of basis is still meaningless, isn't it? Of course at this point you might as well give a full course on linear algebra and vector spaces, which the OP will probably not appreciate. $\endgroup$ Jun 13 at 5:12
  • 2
    $\begingroup$ @FlatterMann The difference between vectors and their representation through their component in a basis exists, but from the algebraic point of view, it is not that fundamental, due to the isomorphism existing between the two sets. Anyway, you hit the point: I had to stop before starting a full course on linear algebra :-) $\endgroup$ Jun 13 at 5:40
  • $\begingroup$ I am not denying that in much more advanced maths, they would sometimes cover vectors with a definition that includes this extra "combination rules" or "constraints". However, most basic undergrad texts in maths on vector spaces are actually only covering linear spaces, and so the students would have completely missed this important distinction. Feynman most likely predated the popularisation of what you are talking about, so you should cut him some slack. $\endgroup$ Jun 13 at 6:29
  • $\begingroup$ @naturallyInconsistent Certainly, I put Feynman's words in the context of the time they were expressed. I do not blame him The problem is that nowadays, some people think that everything in the Lectures remains the ultimate word. $\endgroup$ Jun 13 at 7:13
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 That is rather unfortunate. After all, aren't we supposed to think that there would be progress? looks around, despairs $\endgroup$ Jun 13 at 7:30

Every mathematical object both is and is not a vector, depending on context. To ask whether (3,5,9) is a "vector" is like asking whether Barack Obama is a "member". He is in fact a member of some clubs and not of others. To meaningfully ask "is Barack Obama a member?", you have to have some particular club in mind. To meaningfully ask whether (3,5,9) is a vector, you have to have some particular vector space in mind.

A (real) vector space is a collection of mathematical objects together with some rule for adding two vectors together and some rule for multiplying a scalar (that is, a real number) times a vector. (There are also some axioms these operations must satisfy, which others have listed.) When you have a vector space in mind, the elements of that vector space are called vectors. When you don't have a vector space in mind, those same elements are not called vectors.

Example 1: There is a vector space consisting of all triples of real numbers. You can add two of these, and you can multiply by scalars, in the usual way. When you have that vector space in mind, any triple of real numbers a vector, and nothing else is a vector.

Example 2: There is a vector space consisting of all quadratic polynomials with real coefficients. You can add two of these, and you can multiply by scalars, in the usual way. When you have that vector space in mind, any quadratic polynomial with real coefficients is a vector, and nothing else is a vector.

Example 3 (this is bizarre one, but perfectly legitimate): There is a vector space consisting of all triples of real numbers except for (0,0,0), along with the number 8. You've decided to add these and to multiply by scalars according to the usual rules for triples, together with the rule that v+8=v for any vector v, and r x 8 = 8 for any scalar r. When you have that vector space in mind, (3,7,1), (2,-1,0) and 8 are all vectors, but (0,0,0) is not a vector.

Example 4: Just like example 4, but instead of the number 8, use the unit circle (not the various elements of the unit circle, but the unit circle itself). With this vector space in mind, the unit circle is a vector. With any of Examples 1,2,3 in mind, it is not.

Finally: Take the example from @rob's answer. He is talking about triples for which you have no good definition of addition. (You could try adding two of these triples in the usual way, but you'd get a third triple that isn't one of the allowable triples). If you have this collection of triples in mind, then you do not have any vector space in mind at all. Then (768023, 44126, 295770) is not a vector (and neither is anything else). But if you have example 1 in mind, then the same triple is a vector.

  • $\begingroup$ I think (but am not sure) that Feynman's argument also contains the following point: if V is a real vector space with dimension 3, then for every basis B of V, a triple of real numbers represents a vector in V; which means that each triple of real numbers corresponds to different vectors for different bases, and conversely each vector of V is represented by different triples under different bases; so that "vector = triple of real numbers" is not a good simplification, even when vectors are represented by triples of real numbers. $\endgroup$
    – Stef
    Jun 14 at 17:53
  • $\begingroup$ So, the reason that any three numbers don't form a vector is due to the fact that the vector space definitions need to be taken into consideration. Three arbitrary numbers won't necessarily form a vector if the vector space is defined as only consisting of a limited set of number combinations. Is that right? $\endgroup$
    – Zinn
    Jun 29 at 23:38

If you write down any three $\mathbb R$-numbers, then they constitute an element of the vector space $\mathbb R^3$. This is obviously true, which is leads to (understandable) confusion about what physicists mean when we say that not all sets of numbers constitute vectors.

What we mean is that if you consider three $\mathbb R$-valued quantities which have their own transformation rules and then compile them into a 3-tuple, then that list does not generally transform in the same way that an honest vector does. I'll give you an example, in 2D for simplicity.

Let $\mathbf u = (1,0)$ and $\mathbf v = (2,1)$ be elements of $\mathbb R^2$. Under the action of a rotation matrix $R_\theta$, we have

$$\mathbf u \mapsto \mathbf u' = R_\theta \mathbf u = \pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}\pmatrix{1\\0} = \pmatrix{\cos\theta\\\sin\theta}$$ $$\mathbf v \mapsto \mathbf v' = R_\theta\mathbf v = \pmatrix{2\cos\theta -\sin\theta\\ 2\sin\theta + \cos\theta}$$

More generally, a vector $\mathbf w$ is sent to $\mathbf w' = R_\theta\mathbf w$ which has components $w'^i = (R_{\theta})^i_{\ \ j} w^j$.

Now I will define a list $\mathbf x = \big(\Vert \mathbf u \Vert , \mathbf u \cdot \mathbf v\big) = (1,2)$. Its components are $\mathbb R$-valued, but they are defined in terms of quantities which already have their own prescribed transformation behaviors. That is, under the action of the rotation we would have that

$$\mathbf x \mapsto \mathbf x' = \big(\Vert \mathbf u' \Vert , \mathbf u' \cdot \mathbf v'\big) = (1,2)$$

However, it's easy to see that because the norm and dot product are invariant under rotations, $\mathbf x' = \mathbf x$. If $\mathbf x$ were a vector, then we would have that $$\mathbf x' = R_\theta \mathbf x = \pmatrix{\cos\theta - 2 \sin\theta\\ \sin\theta + 2\cos\theta} \neq \mathbf x$$

So to summarize, given a vector space $V\simeq \mathbb R^n$, vectors transform in a particular way under the action of a linear transformation $T$. Any arbitrary list of $n$ $\mathbb R$-numbers defines a vector (in some basis, at least). However, not any arbitrary list of $n$ $\mathbb R$-valued quantities - which have their own independent existences - transforms in the same way.


For a mathematician, a vector is a member of a vector space. A vector space is a set of vectors that form a group under $+$. Multiplication by any real (or complex) number gives another vector in the space.

You have to define the set and the operations to make a vector space. You can't define a single vector in isolation. Unless you are defining a trivial vector space containing only the vector $\vec 0$.

There are many ways to define a set and two operations that do not make a vector space. Take a finite set. Take a + operation where $\vec x + \vec y$ always adds to the same vector.

Perhaps you are thinking that given any three numbers, you can always define a vector space that holds them. Perhaps.

Take something off the wall like the x coordinate of position and the y and z coordinates of velocity. You can make a vector space out of this. Take addition to add coordinates as usual. Take multiplication to multiply each coordinate as usual. It is indeed a vector space.

Physicists have one more requirement. It must transform correctly when you change the basis.

The coordinates must be physically meaningful. And when you change the basis by taking a linear combination of them, they must still be physically meaningful.

In practice, this means they must all be the same kind of quantity. E.G. All must be distance, or all must be velocity.

So a phase space from thermodynamics is a real example of a mathematical vector space that is not a physics vector space. For a system with $n$ atoms, it is a $6n$ dimensional space. It has the $3$ spatial coordinates of each particle, and the $3$ momentum coordinates. A single very complex point represents the position and momentum of every atom.

Physicists use such a space, but not as a vector space. As the system evolves in time, they track the state of the system by the position of the point in the space. They never add two points together, or multiply the point by a number. They certainly never change the basis to see the system from the point of view of another inertial frame of reference.

Another instructive example is space time. This combines $3$ spatial coordinates and $1$ time coordinate into a single $4$D vector space. In this space, you do make linear combinations out of space and time. They are physically meaningful. They are needed to correctly represent the system as seen from another inertial frame of reference. This works because space and time are much more alike than you would guess from everyday experience.

One such similarity is the failure of simultaneity. Alice stands by the side of the road, while Bob drives by. At $t_0$, they are both in the same place. At $t_1$, Alice says she is in the same place as before, but Bob has moved. Bob disagrees. He is still in the place he was at. Nobody is surprised by this disagreement.

They are surprised by the equivalent disagreement about time intervals. Alice$_1$ and Alice$_2$ are at opposite ends of a train station. Bob$_1$ and Bob$_2$ ride by in the caboose and engine of a train. The Bobs have synchronized their watches. They agree that Bob$_1$ passes Alice$_1$ at the same time that Bob$_2$ passes Alice$_2$. The Alices have also synchronized their watches. They say that when Bob$_1$ passes Alice$_1$, Bob$_2$ has not yet reached Alice$_2$. Bob$_1$ passes Alice$_1$ at a later time.

  • 4
    $\begingroup$ There is a good reason why the 6N dimensional phase space is not usually treated as a vector space. In general, it isn't, due to the nature of the generalized coordinates used in the Hamiltonian formalisms. $\endgroup$ Jun 13 at 5:44
  • 1
    $\begingroup$ Physicists have one more requirement. It must transform correctly when you change the basis. Actually, this is not an additional requirement. It is a consequence of the vector space axioms. $\endgroup$ Jun 14 at 6:17

Every triple of (real) numbers is an element of the vector space $\mathbb{R}^3$ and hence by definition a vector as you have already found out.

In physics we often deal with the euclidean space $E$ (the Riemannian manifold of all points in 3D-space). We describe this space by picking an orthogonal coordinate system on $E$, that is an isometry $i: E \to \mathbb{R}^3$. Now what Feynman really means is that the numbers produced by $i$ and the numbers produced by any other orthogonal coordinate system on $E$ must be related to each other in a specific way. I will describe this relation below:

Coordinate isomorphism and change of basis

First of there is a natural isometry between $E$ and the tangent space of $E$ at any point $o$, where a point $p\in E$ corresponds to the vector pointing from $o$ to $p$. So we get a full description of $E$ by considering only the tangent space of $E$ for a point $o \in E$ (the origin).

We choose an origin $o$ and a basis $A= (a_1, a_2 ,a_3)$ of the tangent space $T_o E$ of $E$ at $o$ and then we have an isomorphism (an invertible linear map) \begin{align} J: E &\longrightarrow \mathbb{R}^3 \\ \alpha_1 a_1 + \alpha_2 a_2 + \alpha_3 a_3 &\longmapsto \begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} \end{align} that sends an euclidean vector to its components with respect to the basis $A$. Now if we take a different basis $B =(b_1, b_2, b_3)$ of the tangent space to $o$ of $E$, then we obtain another isomorphism $I: E \to \mathbb{R}^3 $ of the same kind. We can ask the question what the relation between the components of a vector $v \in E$ with respect to $A$ and $B$ is. We can do this as follows: For $i \in \{1,2,3\}$ we have that $b_i = \sum_{j=1}^{3} c_{ji}a_j$ for some coefficients $c_{ji} \in \mathbb{R}$ since $A$ is a basis. We can form a matrix $C=(c_{ji})_{j,i} \in \mathbb{R}^{3 \times 3}$ from these coefficients.

Let $v = \sum_{j=1}^3 \alpha_j a_j \in E$ and also $v =\sum_{i=1}^3 \beta_i b_i $ . Then we have $$v = \sum_{i=1}^3 \beta_i b_i = \sum_{i=1}^3 \sum_{j=1}^3 \beta_i c_{ji} a_j $$ and therefore since the coordinates are unique we have using matrix multiplication $$\begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} = C \cdot \begin{pmatrix} \beta_1 \\ \beta_2 \\ \beta_3 \end{pmatrix} .$$ And so we found the relationship between the old coordinates and the new ones.

Now in the case of euclidean space $E$, we are only interested in the case where $A$ and $B$ are both an orthonormal basis so that $J$ and $I$ are isometric isomorphisms when $\mathbb{R}^3$ carries the usual scalar product. This further restricts $C$ to be a member of the orthognal group $O(3)$. Note that we can also choose different origins $o$ and look at how the coordinate system transforms in that case.


I can't imagine any 3 numbers that when combined to form a vector, wouldn't satisfy these axioms

I assume that when you say "combined to form a vector" you actually mean "combined to form a tuple" (i.e. an ordered list) - otherwise you would be asking "are there vectors which are not vectors". So your question becomes "are there tuples which are not vectors". And the answer is definitely yes.

For example, suppose we define the sum of two 3-tuples $A = (a_1,a_2,a_3)$ and $B = (b_1,b_2,b_3)$ as $A+B = (\max(a_1,b_1), \max(a_2,b_2), \max(a_3,b_3))$. and the multiplication of $A$ by a scalar $t$ as $tA = (a_1, ta_2, t^2a_3)$. Then we have made some sort of algebraic space from our collection of 3-tuples, but it is certainly not a vector space.

A vector is not an ordered list of numbers. It is a mathematical object that can be added to other vectors and multiplied by scalars in a way that conforms with the axioms of a vector space. A vector may be represented by an ordered list of numbers, which are its co-ordinates relative to a given co-ordinates system. And once we introduce the co-ordinate representation then we can define vector addition and multiplication in terms of operations on these co-ordinates. It is a mistake to identify a vector with its co-ordinate representation because (a) a vector has many different co-ordinate representations, depending on the chosen co-ordinates system and (b) just being handed a tuple of numbers does not guarantee that that tuple is part of a vector space, as the example above shows.


Of course it is generally not true that any three numbers form a vector

In order for it to be a vector, not only must there be three numbers, but these must be associated with a coordinate system in such a way that if we turn the coordinate system, the three numbers “revolve” on each other, get “mixed up” in each other, by the precise laws we have already described.

He is saying that a vector in a $3$-dim. vector space can be identified with a function that assigns three numbers to each basis (coordinate system) and that satisfies a certain transformation rule. Here is how to put this into math language:

Let $V$ be an $n$-dim. $F$-vector space. A basis corresponds to a vector space isomorphism $B\in L(V,F^n)$. Let $A\subset L(V,F^n)$ be the set of all bases and $$W:=\{f:A\to F^n:\forall B,\tilde B\in A:f(\tilde B)=\tilde B B^{-1}f(B)\}$$ then we can define a bijection $$\Phi:V\to W$$ by setting $$\forall v\in V:\forall B\in A:\Phi(v)(B)=B(v).$$


Consider the list \begin{align} (N_f,\, N_c,\, N_s) = (10, 1, 0) \end{align} where $N_f$ is the number of fingers on my hands, $N_c$ is the number of cats on my bed, and $N_s$ is the number of shoes on my feet. These numbers don't change under any Euclidean coordinate transformation, so this list of numbers is not a Euclidean vector.

  • 4
    $\begingroup$ Of course these change under coordinate transformations. Let P be the number of figures on your hand minus the number of cats on your bed, Q the number of cats on your bed minus the number of shoes on your feet, and R twice the number of shoes on your feet. Then your triple transforms in the new coordinates to (P=9,Q=1,R=0). $\endgroup$
    – WillO
    Jun 13 at 3:34
  • $\begingroup$ @WillO I've edited my answer to conform to your pedantry. $\endgroup$
    – d_b
    Jun 13 at 3:50
  • 3
    $\begingroup$ d_b: The transformation here is an element of SO(3). If that doesn't count as euclidean, I can't imagine what does. And for novices struggling to understand basic concepts, pedantry is very often exactly what's needed. $\endgroup$
    – WillO
    Jun 13 at 4:23
  • $\begingroup$ @WillO I don't see how that can be true... It's not an orthogonal transformation. Anyway, that's beside the point. Of course you can take any list of three numbers, treat it as an element of $\mathbf{R}^3$, and look at transformations on that space. The point is that the relevant transformations act on physical space, not some abstract vector space, and I'm sure you know exactly what Feynman means in the quoted passage. I'm not sure what point you're trying to make, and if you have substantive feedback I'm happy to try to incorporate it in my answer, but, respectfully, your comment reads... $\endgroup$
    – d_b
    Jun 13 at 5:17
  • 2
    $\begingroup$ Like @WillO, I also honestly never understood the "physics" definition of vector. The coordinates of any vector transform in a certain way because that's how the math works out when you work it out. I can't imagine a vector whose coordinates transform in a "wrong" way. I've tried but I just don't get it. $\endgroup$
    – Thierry
    Jun 14 at 17:27

The problem is not wether three numbers form a vector; the issue is whether three variables form a vector. Calling a variable "a number" is likely a simplification and your question exposes a shortcoming of that simplification.

There's nothing intrinsecally vector-like or vector-unlike in the values $(1,10,\pi)$ that make them "a vector" or not. It depends on what those numbers represent in your model of... something.

If the three numbers are, say, geometrical coordinates, then the mathematical structure of a vector space (scalar multiplication, addition etc.) will be meaningful in how your model describes space: vector addition describes a traslation, scalar multiplication an enlargement/reduction and several geometrical transformation can be expressed in the language of linear algebra.

If you pick unrelated variables (i.e. temperature, userID, number of students enrolled in physycs 101) or variables that are related but for which the vector space operations do not make sense (such as temperature, humidity and air pressure) then you have what most people would call "not-a-vector" because even if you could do linear algebra with those numbers, those operations do not correspond to anything meaningful about what those numbers are describing, so they are not "a vector" in that sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.