# What isn't a vector space?

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?

• 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 2 '17 at 13:48
• 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 2 '17 at 14:58
• @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 2 '17 at 17:40
• @PeteL.Clark Are you calling OP's mom a zero? – MartianInvader Oct 2 '17 at 23:26
• @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 3 '17 at 9:18

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.

• 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 2 '17 at 14:23
• 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 2 '17 at 17:43
• @PeteL.Clark You're right. I have clarified that example. – Arthur Oct 2 '17 at 18:15
• 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 3 '17 at 2:57
• 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 4 '17 at 6:44

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!

• @123: In this group the identity is $1$. – Matthew Leingang Oct 2 '17 at 15:57
• @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 2 '17 at 20:27
• @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 3 '17 at 8:33
• @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 4 '17 at 11:58
• @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 4 '17 at 11:59

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.

• More simply: the set of points $(x,y)\in \mathbb R^2$ satisfying $x+y=0$ – user Oct 3 '17 at 0:53
• 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 4 '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 $\{\bf{0}\}$.
• 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. – user307169 Oct 3 '17 at 14:22
• @tilper Is it better now? – Robert Z Oct 3 '17 at 14:34
• This receives the official and entirely useless tilper seal of approval. Also +1. – user307169 Oct 3 '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.

• 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.

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.

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.

• Why isn't $\vec{a} + \vec{a}$ in the set $S$? – AleksandrH Oct 2 '17 at 13:43
• @AleksandrH Because if $\vec{a}+\vec{a}\in S$ then $\vec{a}+\vec{a}=\vec{a}$, which gives $\vec{a}=\vec{0}$, which is contradiction. – Michael Rozenberg Oct 2 '17 at 13:45
• I guess I'm a bit confused. What if the set S contains the vectors 2 and 4? Then isn't 2 + 2 = 4? – AleksandrH Oct 2 '17 at 13:47
• @AleksandrH, In Michael's example, $S$ contains exactly one element. – Yanko Oct 2 '17 at 13:48
• @yanko Oh, gotcha. I thought it was supposed to represent the set of all possible "a". – AleksandrH Oct 2 '17 at 13:55

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!

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.

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 $s = 0$.)

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).

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 $q^m = p^{nm}$.

• Would you mind elaborating why? – Frenzy Li Oct 5 '17 at 15:29