# What are examples of vectors that are not usually called vectors?

In algebra, a vector is an element of a vector space. An example of such an element is a matrix.

In linear algebra, a vector is a shorthand name for a $1 \times m$ or a $n \times 1$ matrix. (Whereas a matrix itself is also a vector, by definition, but rarely referred to as such.)

In (analytic) geometry, a (euclidean) vector is a geometric object with a magnitude and direction. These can be represented by tuples.

Both matrices and tuples are clearly vector elements in some spaces, but are for some reason not called vectors by name, unlike euclidean vectors and row vectors.

Are there any other examples of vectors that are not called vectors, like matrices and tuples?

• If I understand you correctly I have 2 examples: in quantum mechanics, "states", and in Fourier analysis & SL theory, "functions". – Ofek Gillon Feb 27 '17 at 18:13
• Complex numbers are vectors as $\Bbb{C}$ is a $\Bbb{R}$-vector space. – Crostul Feb 27 '17 at 18:26
• Some other examples: Every vector space homomorphism (linear map) is a vector. Every field is a vector space over itself. – Henricus V. Feb 27 '17 at 19:34
• The set of strictly positive real numbers is a one-dimensional real vector space in which the 'sum' of the 'vectors' $x$ and $y$ is $xy$, the 'product' of the 'vector' $x$ by the 'scalar' $\lambda$ is $x^\lambda$, and the 'coordinate' of the 'vector' $x$ with respect to the basis $a$ is the 'scalar' $\log_a{x}$. – Calum Gilhooley Feb 28 '17 at 1:03
• Isn't everything really a vectorspace if you just define it together with some operations that fullfill the criterias ? You could probably construct those operations for everything. – HopefullyHelpful Feb 28 '17 at 17:16

Of course there are: A few examples that come to mind:

1) The vector space of all continuous functions with the usual function addition and scalar multiplication

2) The vector space of all sequences $u_n: \mathbb{N} \rightarrow \mathbb{R}$

3) Polynomials

The examples above are vectors in specific vector spaces, but still, we prefer to call them functions (1), sequences (2), polynomials (3).

You can think of more exotic examples, but I thought these were good examples to answer your question as these are things you have already encountered.

• +1 - My first reaction was "polynomials." They often come up in introductory linear algebra as a classic example of a vector space that most students wouldn't initially think of. – NoseKnowsAll Feb 28 '17 at 3:32
• #2 is such a good one that I always forget about. bravo. – The Count Mar 1 '17 at 3:50

Anything.

Take any set $X$. Consider the set $A:=\{f:\{X\} \to \mathbb{R}\}$. $A$ has an obvious vector space structure. Now, let $\mathcal{C}:=(A-\{f_1\}) \cup \{X\}$, where $f_1$ maps $X$ to $1$. Then $\mathcal{C}$ also has an obvious vector space structure (the one which makes the "identity" an isomorphism), and $X$ is an element of $\mathcal{C}$, hence a vector.

This answer is just to emphasize that being a vector is not a quality in itself. The importance is the whole structure (the vector space, the sum, the multiplication by scalar etc) and its relevance to the situation which is being applied to, which is why we don't refer to a lot of things as vectors even though they are (because being a vector is not relevant to the situation at hand).

• I can see where some users might no like this answer, but I think it is glorious. +1 – The Count Feb 27 '17 at 21:25
• I meant to write a comment somewhat along these lines. I do not fully agree with "being a vector is not a quality in itself" though. It is true for the 'definition' of vector in OP, but I feel it should be stressed that the fault is with that definition. "In algebra, a vector is an element of a vector space." is just not true exactly like that. What is true is that elements of a (implicitly) given vector space are sometimes are referred to as vectors in that particular context (to distinguish them from scalars, operators on the space, etc). – quid Feb 28 '17 at 10:40
• Why not just define $\mathcal{C} = (\mathbb{R}\setminus \{1\}) \cup \{X\}$? – leftaroundabout Feb 28 '17 at 16:29
• @AloizioMacedo Is that the free vector space? Shouldn't be then $f : X\to\mathbb{R}$? Otherwise, it looks as if $f$ is only defined on a single point, namely $X$. – chi Feb 28 '17 at 18:13
• I see. So, $A \simeq \mathbb{R}$ as a vector space, and the example is, up to iso, the one given by @leftaroundabout above, but with different objects used as vectors, just for the sake of showing we can do that. I think I got it now, thanks. – chi Feb 28 '17 at 18:17

The quality of a "vector" you are describing is the ability to write it as a set of coordinates with respect to some basis of the space in which they reside. In finite dimensions, provided that you maintain the ordering of the basis elements, this takes the form of a "tuple", or a row or column matrix. This "picture" of a vector space gets somewhat complicated once you move past finite dimensions, since our basis now contains infinitely many elements.

Sometimes we can still write the vectors in our space as some kind of ordered list (like the space of all real sequences) even though the vector space is infinite dimensional. However we cannot write down a basis for these spaces, but we know that one must exist (assuming the axiom of choice). For such vector spaces, the basis becomes fairly useless, and we have to adopt other tools to study them.

There are many examples of vector spaces which do not "look" like vectors in the traditional sense. The set of functions from a given set into the real or complex numbers will do, where we take the "pointwise operations" of addition and scalar multiplication.

You have mentioned matrix space as a vector space. The answer by Math_QED has given three frequent examples. One other often-mentioned example that immediately comes to mind is a field $\mathbb K$ as a vector space over a subfield $\mathbb F$.

For instance, $\mathbb R$ is a vector space over $\mathbb Q$, where each real number is a "vector" and each rational number is a "scalar". Vector addition is defined by the usual addition of real numbers, and scalar multiplication is also the usual multiplication of real numbers, but only the multiplication of a "vector" (a real number) by a "scalar" (a rational number) is considered a legitimate scalar multiplication. This vector space is infinite dimensional, as explained by a popular posting on this site.

Another very important example, which is somewhat similar to but not exactly the same as what Aloizio Macedo describes in his answer, is the concept of a free vector space.

Let $X$ be any set, and $\mathbb F$ be any field. For any $x\in X$, let $f_x:X\to\mathbb F$ denotes the membership function defined by $f_x(x)=1$ (the $1$ here is the identity element in $\mathbb F$) and $f_x(y)=0$ (the zero element in $\mathbb F$) if $y\ne x$. Then the linear span of $\{f_x: x\in X\}$, with the usual addition of functions and multiplication of a function by a scalar, is a vector space over $\mathbb F$. In other words, every "vector" here is a function $f:X\to \mathbb F$ whose preimage $f^{-1}(\mathbb F)$ is a finite set.

The moral here is that every (yes, every) set $X$ can be made a vector space over any field $\mathbb F$, even when $X$ and $\mathbb F$ are completely unrelated. Whether we can define a vector addition and a scalar multiplication that makes the vector space interesting or useful is another matter.

The concept of free vector space is useful to our understanding of the concepts of universal mapping and tensor product, but I shall not go into details here, as this is covered in many books on multilinear algebra.

• It looks like, if $X$ is finite with cardinality $|X|=n$, this "free vector space" is simply (isomorphic to) $\mathbb F^n$. If $X$ is infinite, can we also say that the free vector space is $\mathbb F^{|X|}$? – mr_e_man Jun 22 '18 at 4:05

# Everything.

Given any set $S$, you can consider the space of (formal) finite linear combinations of its elements $\{a_1\mathbf{x}_1 + a_2\mathbf{x}_2 + \dots + a_k\mathbf{x}_k: k\in\mathbb{N}, \mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_k \in S \}$. This has a vector space structure.

In the "linear algebra" sense, scalars (numbers) are "vectors" of dimension 0 and all tensors are "vectors".

• I think you mean scalars are vectors with dimension 1. – mb7744 Feb 27 '17 at 19:37

The set of operators between vector spaces is itself a vector space.

The set of functions mapping vectors to the field on which they live is a vector space, as well, but this is just a special case of the first statement.

An example that is often hard to get for students on first encounter is given by vectors and spinors in particle physics: The representations of the Lorentz group $SO(1,3)$ fall into two classes, having integer or and half-integer spin, called vectors and spinors, respectively. Hence, you find statements such as "$\psi$ is a spinor, not a vector!" This can be particularly confusing as both are elements of a vector space, both are written as rows or columns of (commonly four) numbers (well, possibly anticommuting numbers), and both appear together in typical particle physics applications.

This is actually a general feature of the orthogonal groups, related to universal covering groups and projective representations. In particle physics, these classes of representations are used to desribe very different classes of particles, fermions vs. bosons.

Any ordered set of numbers (or other values) can be considered a vector; it's just an abstraction.

For example, in the field of data processing, it's common to consider graphics/audio files as vectors (or sets vectors) with thousands or millions of elements each, then use the tools of Linear Algebra to manipulate them.

An exotic example with an interesting application: $\mod 2$ homology.

The formal sums of faces/edges/vertices of a polyhedron with coefficents in $\Bbb F_2$ are called $k$-chains $\mod 2$ ($k =$ dimension of objects) and form a vector space over $\Bbb F_2$.

See Proofs and Refutations by Imre Lakatos for a proof of the Euler's polyhedron formula using $\mod 2$ homology.

Quaternion's coefficients: are just 4 real numbers, so just a vector in $\Bbb R^4$. Note, I explicitly told "coefficients" because in reality a quaternion is a extension of the complex numbers.

Sorry for the short answer, but it cannot be longer than that without repeating something that is already in other answers :).

A vector space $V$ over the field $F$ can be more formally described as a set with some properties. These properties include closure under addition, and closure under multiplication with a scalar in the field $F$.

Many concepts can be thought of as vector spaces. Whether it is useful to think of them as vector spaces is another question.

Here are some examples:

Real numbers

The set of real numbers is an infinite dimensional vector space over the field of rational numbers.

The set of real numbers is a 1 dimensional vector space over the field of real numbers.

Rational Numbers

The set of rational numbers is a 1 dimensional vector space over the field of rational numbers. It is not a vector space over the field of real numbers because it is not closed under scalar multiplication. It is not a vector space over integers because integers does not make a field.

Functions

If we take the co-domain to be the set of real numbers or the set of complex numbers:

The set of functions over a set domain is a vector space over both the real and rational numbers.

The set of bounded functions over a set domain is a vector space ...

The set of [insert description here] functions is a vector space over the real numbers/rational numbers that you can add any two such functions, or multiply one of these functions with a real/rational and still get a function that fits the description. For example: continuous, uniformly continuous, integrable, $C_n$, polynomials, rational functions, etc.

Magic squares

The set of magic squares is a vector space (I believe 4 dimensional) over the real numbers. It is also a vector space over rational numbers

Equations

The set of equations is a vector space.

Regarding fields

Ultimately, you can come up with some really silly example of vector spaces if you use some really unfamiliar/obscure fields. For example, this field.

• A vector space must have a zero, so the empty set is not a vector space. Similarly, the empty set is not a field (must have an identity). For the functions, you need to require a particular domain and range from them to be a real or rational vector space. – TokenToucan Feb 28 '17 at 5:36
• Ah, you are right about the zero thing. I forgot about it. I think I did say that functions over a specific domain makes a vector space. – Fluidized Pigeon Reactor Feb 28 '17 at 6:22
• Why is the field with 4 elements silly? While this is a rather small field, finite fields in general have many concrete practical applications in coding theory and cryptography that crucially depend on vector spaces over them. =) – user21820 Feb 28 '17 at 15:46
• No, the set of functions over a set domain is not always a real or rational vector space. What if I gave you functions from {0,1,2} to {cow,moose}. How are scalar multiplication and addition defined, what's the zero vector and so on. – TokenToucan Feb 28 '17 at 16:51
• user21820: You are correct. I should change the wording to "unfamiliar/obscure". TokenToucan: Ah, I see. You are correct. I guess it should be functions from a subset of complex numbers to the set of complex numbers/real numbers. – Fluidized Pigeon Reactor Feb 28 '17 at 21:02

A physical vector space is the one in which quantum states lives, namely a Hilbert space, which is a vector space of square integrable functions. When the space dimension is infinite, then the vectors are just functions square-integrable.