I'm really confused about vector spaces. We're learning about them in Linear Algebra, and my book doesn't give good examples of what a vector space is. I understand sets and vectors, but I don't understand vector spaces. From the definitions they've provided so far, it seems anything can be a vector space.

Can someone provide a simple example of what isn't a vector space so I can make a distinction?

Gerry Myerson
  • 168,500
  • 12
  • 196
  • 359
  • 2,795
  • 4
  • 22
  • 47
  • 1
    From the answers so far, it seems the existence of the zero vector is an important identity of a vector space. Why is this? – AleksandrH Oct 02 '17 at 13:45
  • 1
    this is pretty much by definition... Vector space must contain a zero element. – Yanko Oct 02 '17 at 13:45
  • 4
    this is because of two points 1)closure due addition and 2)existence of additive inverse ,so if $a \in V$ then $a + (-a) \in V$ – BAYMAX Oct 02 '17 at 13:48
  • 119
    I know that's not appropriate and this will be probably be moderated away, but my inner 12yo urges me to point out that your mom is not a vector space, and I'm not one to ignore my inner 12yo. – Nico Oct 02 '17 at 14:58
  • 23
    @Nico: Technically you're right, but the OP's mom is the underlying set of a vector space. This kind of set theoretic abstraction is possibly relevant to the OP's question. – Pete L. Clark Oct 02 '17 at 17:40
  • Anything which is not the image of a linear transformation. – Kimball Oct 02 '17 at 21:39
  • 8
    @PeteL.Clark Are you calling OP's mom a zero? – MartianInvader Oct 02 '17 at 23:26
  • 9
    @PeteL.Clark Even if we take the "everything is a set" philosophical view, if OP's mom were a set with six elements it would not admit a vector space structure, if I am not mistaken. The set $\{\text{OP's mom}\}$ is isomorphic to the zero vector space, but that's a different thing. (Obligatory joke: your mom is so fat, that if she was a vector space she'd have infinite dimension.) – Federico Poloni Oct 03 '17 at 09:18
  • 2
    @Nico Lmao. Never hurts to have a little humor, even in a serious setting :) Thanks for the laughs. – AleksandrH Oct 03 '17 at 11:37
  • A blackbird. Love. Monotheism. The empty set. The smell of cloves. The sound of a moonbeam. – PJTraill Oct 03 '17 at 14:12
  • 1
    @Nico - "Your mom" was also my immediate thought. Kudos for having the courage to say it. – MackTuesday Oct 03 '17 at 14:56
  • 1
    @Nico: Maybe not a vector space, but unfortunately people can easily be vectors... – user541686 Oct 03 '17 at 23:31
  • 4
    @Federico: Since we're going to "your mom" jokes here, it is obvious that we are talking about infinite sets. – Asaf Karagila Oct 04 '17 at 05:04
  • @Nico anything's a vector space if you're brave enough – Tom Anderson Oct 04 '17 at 10:33
  • 2
    I can't help but point out that you cannot truthfully claim to “understand vectors” if you have not understood vector spaces. A vector is not a list of numbers, a vector is an element of a vector space. A list of numbers just happens to be a popular example of an element of the vector space $\mathbb{R}^n$. – Christopher Creutzig Oct 04 '17 at 10:56
  • @TomAnderson How is $\{0,1,2,3,4,5,6\}$ or $\mathbb{N}$ a vector space? – Christopher Creutzig Oct 04 '17 at 10:57
  • @ChristopherCreutzig easy just biject $\mathbb{N}$ with $\mathbb{Q}$ and define vector addition and scalar multiplication in the obvious way :P – Kai Rüsch Oct 04 '17 at 22:38
  • Most things are not vector spaces. For example Strings (which are free monoids), with the zero element "", addition defined as concatenation, and scalar multiplication defined as repetition, are a bit like vector spaces (it has the two operations required and they are closed under both, and meet 3/8 of the axioms). – Frames Catherine White Oct 05 '17 at 06:32

14 Answers14


First off, a vector space needs to be over a field (in practice it's often the real numbers $\Bbb R$ or the complex numbers $\Bbb C$, although the rational numbers $\Bbb Q$ are also allowed, as are many others), by definition. Thus, for instance, the set of pairs of integers with the standard componentwise addition is not a vector space, even though it has an addition and a scalar multiplication (by integers) that fulfills all of the properties we ask of a vector space.

A vector space needs to contain $\vec 0$. Thus any subset of a vector space that doesn't, like $\Bbb R^2 \setminus \{\vec 0\}\subseteq \Bbb R^2$ with the standard vector operations is not a vector space. Similarily, a vector space needs to allow any scalar multiplication, including negative scalings, so the first quadrant of the plane (even including the coordinate axes and the origin) is not a vector space.

A more subtle example is the circle (with some chosen zero) where addition is done by adding distances along the circle from the chosen zero (equivalently by adding angles), and scalar multiplication is done by multiplying distances (angles). Here we get into trouble with scalar multiplication again, because the zero vector is simultaneously representing $360^\circ$, so what should $0.5$ multiplied by that vector be? $0^\circ$? $180^\circ$? It would be both at the same time, which is not good.

  • 187,016
  • 14
  • 158
  • 288
  • 3
    Thank you :) Unfortunately, my book omitted the point you made in your first paragraph—the fact that a vector space must be defined over a field. Also, it placed way too much emphasis on examples of vector spaces instead of distinguishing between what is and what isn't a vector space. – AleksandrH Oct 02 '17 at 14:23
  • 28
    I don't like that this answer identifies a vector space as a *set* and does not explicitly mention the addition and scalar multiplication operations. In particular $\mathbb{R}^2 \setminus \{0\}$ is not any more or less a vector space than $\mathbb{R}^2$: you have to say what the operations are. – Pete L. Clark Oct 02 '17 at 17:43
  • 1
    @PeteL.Clark You're right. I have clarified that example. – Arthur Oct 02 '17 at 18:15
  • 1
    So if I'm getting this right, pairs of integers wouldn't form a vector field, as they only form a commutative ring rather than a proper field due to their lack of multiplicative inverses. Correct? – LegionMammal978 Oct 03 '17 at 02:57
  • @LegionMammal978 Yes, that's right. – Arthur Oct 03 '17 at 05:01
  • 1
    To the proposer: We usually use the same symbol $+$ for addition of two members of the field as for addition of two vectors . And the same symbol $\cdot$ or $\times$ for multiplication of two members of the field as for multiplication of a vector by a field-member. And when it is clear from the context we often use the same symbol $0$ for the additive identity of the field and for the additive identity of the vectors (the $0$-vector). This is called "abuse of notation" and is acceptable. – DanielWainfleet Oct 04 '17 at 06:44
  • 1
    The last example with the circle fails because the (attempted) scalar multiplication is too vaguely defined. Suppose we tried to fix that like this: The circle points are identified with $\left[ 0^\circ, 360^\circ \right)$, so we always prefer the "principal representative" in this half-open interval. Now define both operation as before, with "reduction" to $\left[ 0^\circ, 360^\circ \right)$ included in them. Now the operations are well-defined. For example, $0.5(0^\circ)$ is $0^\circ$. To see why it is not a vector space now, find an axiom that is violated (exercise to the asker!). – Jeppe Stig Nielsen Oct 04 '17 at 12:42
  • Pairs of integers is a fine example, but so is $\mathbb{Z}$ itself. – G Tony Jacobs Oct 05 '17 at 04:52

Vector spaces are not just a set! They are an abstract concept, involving a set $V$, a field $\mathbb{F}$, and operations \begin{align*} + &: V \times V \rightarrow V \\ \cdot &: \mathbb{F} \times V \rightarrow V, \end{align*} addition and scalar multiplication respectively, satisfying a bunch of axioms. There's a lot more at play here than the set $V$ itself. Sets that can be made into vector spaces with the right field and operations are extremely common, but it's much rarer to be a vector space if the set already comes with the field and operations.

For example, the set of positive numbers $(0, \infty)$ doesn't seem like it's a vector space, but with scalar field $\mathbb{R}$ and with the (non-standard) operations,

\begin{align*} \oplus &: (0, \infty) \times (0, \infty) \rightarrow (0, \infty) : (x, y) \mapsto xy \\ \odot &: \mathbb{R} \times (0, \infty) \rightarrow (0, \infty), (\lambda, x) \mapsto x^\lambda \end{align*}

it forms a vector space. Even the natural numbers could be defined to be a vector space over a finite field, or a countable field like $\mathbb{Q}$ (although the operations would look a little funky).

So, to answer your question, can I come up with a set that is definitely not a vector space? Yes. As it turns out, all finite fields have a cardinality that takes the form $q = p^m$, where $p$ is prime and $m \in \mathbb{N}$. As such, finite vector spaces over such a finite field, which must have some finite dimension $n \ge 0$, must have cardinality $q^n$. Therefore, a set with a number of elements not equal to a prime power $p^{mn}$ must not be a finite vector space under any operations. For example, a set with $6$ elements is definitely not a vector space!

Theo Bendit
  • 42,945
  • 2
  • 42
  • 92
  • If $(0, \infty)$ is real vector space under $\oplus$ and $\odot$ then $\oplus$ had to satisfy inverse axiom. That is for all $x \in (0, \infty)$ there exist $y \in (0, \infty)$ such that $x \oplus y = 0$ but if that is true then either $x = 0$ or $y = 0$, both is impossible because $0 \in (0, \infty)$ is false. – user8277998 Oct 02 '17 at 15:29
  • 3
    @123: In this group the identity is $1$. – Matthew Leingang Oct 02 '17 at 15:57
  • @MatthewLeingang I am very sorry. I always think that the identity is $0$. – user8277998 Oct 02 '17 at 16:56
  • @123: No worries. This is a pretty obscure example of a vector space anyway. – Matthew Leingang Oct 02 '17 at 18:13
  • 5
    @MatthewLeingang: Well, depends on what you mean by "obscure". It's really just $\mathbb R$ (viewed as a vector space over itself) relabed by the map $x \mapsto e^x$. – Ilmari Karonen Oct 02 '17 at 20:27
  • 1
    @Ilmari By *obscure* I meant “hidden” or “concealed.” The answerer took an obvious vector space and applied (or hid it behind) the exponential map. I don't dispute that it's a vector space. – Matthew Leingang Oct 03 '17 at 08:33
  • 1
    OP is just learning introductory linear algebra. How can you possibly think that this answer is even remotely comprehensible to them? – J... Oct 04 '17 at 11:35
  • 3
    @J... For one, it's not just about the OP. Why do you think this site encourages closing duplicate questions, or allows questions to be answered after an answer is accepted? This site is a mathematics resource, motivated by a question and answer format. Other people with further knowledge may look at these answers and get more out of these answers. – Theo Bendit Oct 04 '17 at 11:58
  • 1
    @J... For two, I can say from my experience of tutoring a second year introductory Linear Algebra that this isn't above and beyond my students. I don't expect them to know how to prove $\mathbb{Q}$ is countable (or to know exactly what it means), or to be able to prove anything about finite fields, but everything else they would generally be able to comprehend. The only advanced bit of jargon is "dimension", which they cover in reasonably short order anyway. The OP may find value in looking back at these answers as they progress. – Theo Bendit Oct 04 '17 at 11:58
  • 3
    @J... The actual essence of my answer is to point out the importance of specifying operations, and even if the OP can't fill in the details, provide an example of a set that cannot be made into a vector space. – Theo Bendit Oct 04 '17 at 11:59
  • @J... By reading it carefully, possibly with the formal definition of a vector space close at hand. There should not be any issue. And if there is, they can always ask. – Evpok Oct 04 '17 at 20:29

I. the set of points $(x,y,z)\in \mathbb R^3$ satisfying $x+y+z=1$ is not a vector space, because $(0,0,0)$ isn't in it. However if you change the condition to $x+y+z=0$ then it is a vector space.

II. The set of all functions from $\mathbb R$ to $\mathbb R$ is a vector space but the subset consisting of those functions which only take positive values is not.

  • 62,572
  • 4
  • 66
  • 113
  • More simply: the set of points $(x,y)\in \mathbb R^2$ satisfying $x+y=0$ – user Oct 03 '17 at 00:53
  • Sets of points aren't vector spaces, because a vector space is a set of points **and** *a field, two operations, and a distinguished element from the set* (that satisfy certain properties). – Yakk Oct 04 '17 at 20:22

Examples of subsets of $\mathbb{R}^n$ which are not vector spaces with respect to the usual operations (and assuming that the scalars are the real numbers).

  • Any subset which does not contain the origin.
  • Any bounded set which is different from $\{\bf{0}\}$.
Robert Z
  • 136,560
  • 12
  • 93
  • 176
  • 1
    I think this should be edited to specify these aren't vector spaces if we have the usual operations. $(0,+\infty) \subseteq \Bbb R^1$ does not contain the origin but it is a vector space over $\Bbb R$ with "vector addition" being standard multiplication and "scalar multiplication" being exponentiation. –  Oct 03 '17 at 14:22
  • @tilper Is it better now? – Robert Z Oct 03 '17 at 14:34
  • This receives the official and entirely useless tilper seal of approval. Also +1. –  Oct 03 '17 at 14:35

This might be an instructive example...

Define the set $S$ as being the set $\mathbb{R}^2$ of all ordered pairs of real numbers, considered only as a set. Then $S$ is not a vector space.

Or, put more tersely: The set $\mathbb{R}^2$ is not a vector space.

It's useful to realize that "the set $\mathbb{R}^2$" and "the vector space $\mathbb{R}^2$" are different, distinct and separate mathematical objects. The set $\mathbb{R}^2$ is a set, not a vector space, and the vector space $\mathbb{R}^2$ is a vector space, not a set.

It happens that there's exactly one standard way to convert the set $\mathbb{R}^2$ into a vector space. (There are other ways to do it, but those other ways are not standard or conventional.) Since this vector space is the only "standard" vector space whose underlying set is $\mathbb{R}^2$, mathematicians haven't bothered to give this vector space its own name; we just call it $\mathbb{R}^2$ too.

Tanner Swett
  • 8,885
  • 28
  • 51
  • 1
    strictly speaking, all vector spaces are sets (of course, the vectors), but not all sets admit vector space structure, i.e. not all sets _can_ be a vector space. – Michal Dvořák Jul 31 '20 at 10:55

Lots of good non-examples already, but let me add $$ C = \left\{(x,y) \in \mathbb{R}^2 \mid x\geq 0,\ y \geq 0 \right\} $$ In words, $C$ is the set of all ordered pairs of real numbers with both coordinates nonnegative. Addition and scalar multiplication are defined as they usually are in $\mathbb{R}^2$.

The set $C$ is closed under addition, because if two pairs have nonnegative coordinates, their sum has nonnegative coordinates. It also contains the zero element $(0,0)$.

But $C$ is not closed under scalar multiplication. For instance $(1,1) \in C$, but $(-1)(1,1) = (-1,-1) \notin C$. It is closed under positive scalar multiplication, though. This is an example of what is called a convex cone in linear algebra.

Matthew Leingang
  • 23,120
  • 1
  • 33
  • 57

An example of a concept which is "a vector space, but less so" is a module. A module is "a vector space, but over a ring instead of a field": it's a set with binary operations $+: M \times M \to M$ and $\cdot: R \times M \to M$ where $R$ is a ring, such that

  • $(M, +)$ is an abelian group,
  • $\cdot$ distributes over $+$,
  • $\cdot$ distributes over the ring multiplication,
  • $1 \cdot x = x$.

(Same axioms as a vector space.)

Modules are much more general than vector spaces. For example, if you let $R$ be the ring $\mathbb{Z}$, and $M$ be any abelian group, then you get a "$\mathbb{Z}$-module" which turns out just to be that same abelian group, by means of $$n \cdot g = \underbrace{g + g + \dots + g}_{\text{$n$ times}}$$

Also if $R$ is a ring, then let $M$ be any ideal of $R$, and let $\cdot$ just be the ring multiplication; then we get another kind of $R$-module.

Patrick Stevens
  • 34,379
  • 5
  • 38
  • 88

Let $S=\{\vec{a}\}$, where $\vec{a}\neq\vec{0}.$

Hence, $\vec{a}+\vec{a}\not\in S$.

Thus, $S$ is not a vector space.

Michael Rozenberg
  • 185,235
  • 30
  • 150
  • 262

Geometrically consider the positive $x$ and $y$ axis(including origin) and the $3$rd quadrant as your set, then it is not a vector space since any linear combination of the vectors from say,positive x and y axis (vector addition(applying paralleogram law))lie in first quadrant which is not in your space.Hence this set is not a Vectorspace.

If you see,you can remove the $3rd$ quadrant from the example above,still it is not a vector space!you see why!

  • 5,494
  • 4
  • 27
  • 54

Here's one from linear algebra. $GL_n(F)$ - the set of non-singular $n\times n$ matricies over $F$ with matrix multiplication as a binary operation, matrix inverse as an inverse to matrix multiplication, and the identity matrix as unit for multiplication is not a vector space over $F$. The reason is that for some matrix $A$,the matrix $0\cdot A$ is not in $GL_n(F)$. However, the structure I gave on $GL_n(F)$ does give it a structure of a (non-abelian if $n \ge 1$) group.

  • 2,357
  • 2
  • 21
  • 35

The following sets and associated operations are not vector spaces: (1) The set of $n \times n$ magic squares (with real entries) whose row, column, and two diagonal sums equal $s \neq 0$, with the usual matrix addition and scalar multiplication; (2) the set of all elements $u$ of $\mathbb{R}^3$ such that $||u|| = 1$, where $|| \cdot ||$ denotes the usual Euclidean norm (and with the usual $n$-tuple addition and scalar multiplication); (3) the set of all $n \times n$ nonsymmetric matrices, with the usual matrix operations.

Each of these examples violates the closure requirement. (Example (1) works if $s = 0$.)

  • 816
  • 5
  • 9

An abstract finite set $\{a_1,\dots, a_n\}$ is not a vector space in general, but could be the basis for a vector space of formal sums $\sum x_ia_i$, where $x_i\in\mathbb R$ (or any other field).

  • 13,268
  • 4
  • 23
  • 72

A finite set with 0, 6, 10 elements can by no means made to a vector space.

Edit: If a vector space $V$ contains a finite number of vectors, it is either the zero space or contains a 1-dimensional subspace that has the same cardinality as the underlying Field $F$. So $F$ is finite and has cardinality $q = p^n$ where $p$ is a prime number. The cardinality of $V$ is then $q^m = p^{nm}$.

  • 69
  • 3

Here is also some nice list of vector spaces examples: https://en.wikipedia.org/wiki/Examples_of_vector_spaces

  • 291
  • 1
  • 10