Why do we use the word “scalar” and not “number” in Linear Algebra?

During a year and half of studying Linear Algebra in academy, I have never questioned why we use the word "scalar" and not "number". When I started the course our professor said we would use "scalar" but he never said why.

So, why do we use the word "scalar" and not "number" in Linear Algebra?

• Actually the better question is "What is a 'number'?" – Deepak Jun 21 '16 at 16:45
• One can do, for example, linear algebra on the booleans, where the scalars are only $\{0, 1\}$ and not all (natural, integer, real, complex) numbers. – Rahul Jun 21 '16 at 16:52
• Part of the reason is surely tradition. See en.wikipedia.org/wiki/Scalar_(mathematics)#Etymology – Barry Cipra Jun 21 '16 at 18:02
• ...to scare away laymen? – noɥʇʎԀʎzɐɹƆ Jun 21 '16 at 22:25
• Independent of the historical reasons or Ian's conceptual ones, I think the best reason may be @Rahul's point that we often define vector spaces over fields that bear little to no resemblance with the subfields of $\mathbb{C}$ which we typically call "numbers" seems the strongest case to me. – AJY Jun 22 '16 at 2:11

So first of all, "integer" would not be adequate; vector spaces have fields of scalars and the integers are not a field. "Number" would be adequate in the common cases (where the field is $\mathbb{R}$ or $\mathbb{C}$ or some other subfield of $\mathbb{C}$), but even in those cases, "scalar" is better for the following reason. We can identify $c$ in the base field with the function $*_c : V \to V,*_c(v)=cv$. Especially when the field is $\mathbb{R}$, you can see that geometrically, this function acts on the space by "scaling" a vector (stretching or contracting it and possibly reflecting it). Thus the role of the scalars is to scale the vectors, and the word "scalar" hints us toward this way of thinking about it.

• I like this answer because it generalizes: Why do we have numbers called "lengths"? Why do we have numbers called "areas"? Et cetera. The general answer is: we name numbers in different contexts according to what the numbers measure. – Lee Mosher Jun 21 '16 at 16:49
• The set of integers, $\mathbb{Z}$, is not a field, but there are finite fields whose members are all integers. en.wikipedia.org/wiki/Finite_field – Solomon Slow Jun 21 '16 at 22:42
• @JimLarge: the elements of finite fields can be put into bijection with an initial segment of the naturals, sure, but that doesn't mean they are integers. You could just as well say that they 'are' residue classes of polynomials with coefficients in some field of prime order. – yatima2975 Jun 22 '16 at 8:54
• @JimLarge In fact you could say the same thing about $\Bbb Z$, because there are countable fields (to wit, $\Bbb Q$). Just pick a bijection to $\Bbb Z$ and voila, $\Bbb Z$ is a field. – Mario Carneiro Jun 23 '16 at 2:33
• @yatima2975, or strawberries – Mariano Suárez-Álvarez Jun 23 '16 at 19:12

Not all fields are fields of numbers. For instance, it makes sense to talk about vector spaces over the field of rational functions $\mathbb R(X)$ but the scalars in this case are definitely not numbers.

• Yep. To make things worse, since $\mathbb{R}^3$ is a commutative ring, we can therefore speak of $\mathbb{R}^3$-modules. So $\mathbb{R}^3$ can act as our (ring) of scalars. – goblin Jun 21 '16 at 17:13
• @goblin What ring structure on $\mathbb{R}^3$ do you have in mind? – Jeppe Stig Nielsen Jun 21 '16 at 22:13
• @JeppeStigNielsen There's $(x,y,z) \cdot (x',y',z') = (xx', yy', zz')$ for example (I don't know which one goblin had in mind but it's certainly a natural one). It's not a field, but you can speak of modules over a ring... – Najib Idrissi Jun 22 '16 at 10:43
• Conversely, sometimes the vectors are numbers. Ian's answer has a trivial example. An interesting case is that of field extensions. For instance, $\mathbb{R}$ can be seen as a vector space over $\mathbb{Q}$. – filipos Jun 22 '16 at 12:29
• @JeppeStigNielsen, of course, if I call $\mathbb{R}^3$ a ring, I mean the object $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$, where $\times$ is interpreted as the Cartesian product bifunctor $\mathbf{Ring} \times \mathbf{Ring} \rightarrow \mathbf{Ring}$. – goblin Jun 22 '16 at 14:50

Scalar gives you a sense of what the "number" does. A scalar scales a vector, stretching or contracting each of its coordinates by the same amount.

While yes it is a number in common parlance (as long as you are working over a field of numbers, which you probably are), in the context of linear algebra, numbers really just serve this purpose (unless you are in one dimension, in which case vectors $\textit{are}$ numbers and there is some ambiguity).

In mathematics, as well as elsewhere in life, the same thing can be called by different names depending on which aspect of it is of interest to us at the time. For example, I have a roommate who is also a friend; when talking about him I can either call him "my roommate" or "a friend of mine", depending on what is relevant to the conversation.

Similarly, even though when doing linear algebra the scalars are often numbers, when we talk about them we often don't care about the fact that they're numbers, we care about their connection to the vectors and the linear space (which is defined in the linear space axioms). The term for an entity which has this particular connection is "scalar", so that is what we use to discuss them.

Other points:

1. The term "number", in general, isn't defined. The scalar field underlying a linear space needn't be anything that would be normally called a set of numbers.

2. The same number can be either a scalar or a vector, depending on how you look at it. If you consider $\mathbb{C}$ to be a two-dimensional vector space over $\mathbb{R}$, then $1+i$ is a vector. If $\mathbb{C}$ is the underlying field for a vector space, such as $\mathbb{C}^4$, then $1+i$ is a scalar. So saying that $1+i$ is a "number" is completely useless in the context of linear algebra. Saying that it is a scalar, or that it is a vector, as the case may be, is useful.

This is a great question!

The point is that we can look at vector spaces over more general objects called fields. The scalars in a vector space are exactly the elements from this field. If the field is the real, rational, or complex numbers, then the scalars are numbers. But we usually don't refer to elements from all fields as numbers.

Another point is that a vector space can be a field at the same time, for example every field $K$ is trivially a $K$-vector space, and $\mathbb{R}$ is a $\mathbb{Q}$-vector space.

Saying number would be ambiguous, so we distinguish between scalars and vectors.

You can find this mentioned briefly in the link Barry Cipra provides in a comment on the OP, but the usage of "scalar" in this context comes from quaternions, as does the term "vector". If you are not familiar with quaternions, you can think of them as complex numbers on a bad acid trip. If you are willing to sacrifice commutivity of multiplication, you can have more than one imaginary unit. Instead of the complex $i$, quaternions have three: $i, j, k$. So where a complex number has the form $a + bi$, a quaternion has form $a + bi + cj + dk$. In studying them, their inventor, W.R. Hamilton, referred to $a$ as the scalar part, and to $bi + cj + dk$ as the vector part.

When they were first introduced, quaternions revolutionized the study of analytic geometry and physics, providing a much more concise and easier to use notation than the free-floating sets of coordinate variables in use before. But it didn't take too long for those early developers (Heaviside in particular) to realize that all that was really needed for these applications was the vector part of the quaternion, and thus the more general concept of a vector space was born. For the reasons already given in the other answers, the term "scalar" was a convenient name to use for the real multipliers of the vector elements, and so that terminology remains, expanding to now mean any field element, not just reals.

Start with etymology: scalar derives from a Latin adjective (scalaris), akin to the word "ladder" (scala). Its first known usage is from François Viète, one of the fathers of "modern algebra". In his In artem analyticem isagoge (Analytic art, 1591), one finds:

Magnitudes that ascend or descend proportionally in keeping with their nature from one kind to another may be called scalar terms (Magnitudines quae ex genere ad genus sua vi proportionaliter adscendunt vel descendunt, vocentur scalares)

What is in germ here was notably noted by @Ian and @Lee Mosher: "keeping their nature" denotes a "class of objects". Nature can denote areas, lengths, objects of different dimensions. Objects, within the same nature class, can vary in magnitude. And this does not affect their nature. If I multiply a segment of magnitude $3$ meters, by another quantity of $2$ meters, I get $6$ square meters, and have affected the nature of the initial segment. If I multiply it by a unitless quantity, meters remain meters.

So multiplications are binary operations that may operate on quantities of seemingly different nature:

• integer $\times$ fraction,
• complex $\times$ function,
• matrix $\times$ vector.

Possibly, these multiplies are somewhat different in their properties. In computer programming, binary operations can be cast to different nature.

Scalar multiplication denotes the simplest form (yet very generic) of a linear transformation of object $B\in \mathcal{B}$ by scalar $a\in\mathcal{A}$, such that $a\times B\in \mathcal{B}$ is of the same nature as $B$. It is an external binary operation, from $\mathcal{A}\times \mathcal{B}$ to $\mathcal{B}$. Objects in $\mathcal{A}$ are often called scalars, and those in $\mathcal{B}$ vectors.

For practical usefulness, additional requirements are imposed for consistency, such as $a'\times (a\times B)=(a'\times a)\times B$. Scalar multiplication is at play in the definition of a vector space $V$ over field $K$, or in a module $M$ ("vectors") over commutative ring $R$ ("scalars").

So you could have a module defined over the ring of integers, here your scalars would be integers: for instance a $\mathbb{Z}$-module, which agrees with the definition of Abelian groups.

Apparently, Hamilton used "scalar" in English for the first time to denote the real part of quaternions.

Outside algebra, scalar processors handle only one datum at the time, as opposed to vector processors.

There are several reasons I can identify:

1. History. As described by Paul Sinclair, the nomenclature comes from quaternions, where distinguishing between a number and a component of a quaternion is a good idea. This distinction is not so directly relevant to vector spaces, but the name stuck anyhow.
2. Fields other than $\mathbb{R}$ or $\mathbb{C}$. Vector spaces can be constructed over any field you want, and the name "scalar" can be interpreted as a general way of referring to elements of whatever base field you're using.
3. The most important reason, in my opinion, for using "scalar" instead of "number" comes from tensor analysis. A tensor is (roughly speaking) an array of numbers for which the exact value of the numbers depends on a coordinate system. For example, the velocity of a particle in $\mathbb{R}^3$ is represented by an array of three numbers (also known as a vector), and the values of those numbers will change if you, say, rotate your coordinate axes. The precise way in which a tensor transforms when you change coordinate systems determines whether it is a scalar, vector, pseudoscalar, pseudovector, or something else (all special cases of tensors). A scalar is something whose value does not change at all when you change coordinate system. For example, the temperature in a room doesn't depend on any coordinates you have laid down in that room, so we say temperature is a scalar. This is to be contrasted with the previous example of the velocity of a particle, which does depend on what coordinates we use. Because of the precise way velocity transforms (which I won't explain in detail here; see this link for details), we say that velocity is a vector. The other terms I mentioned above---pseudoscalar and pseudovector---are similarly defined in terms of their transformation properties under changes of coordinate system.
In the context of linear algebra, a change of coordinate system is essentially a change of basis, which causes the components of the base field (the scalars), the elements of the vector space (the vectors), and linear maps on the vector space (higher order tensors) to change in different ways. In particular, the elements of the base field DO NOT change at all. This is to be contrasted with, say, a single component of a vector in some basis, which DOES change when you change basis. Both the scalar and the component of a vector are numbers, but they are distinguished by how they are related to other numbers in your vector space: The component of the vector is "attached inseparably" to the other components of the vector; the scalar is on its own. The fact that numbers can have different behavior depending on whether they are a part of a vector or scalar (or a component of a matrix representing a linear transformation, etc.) is why it is good to have a name, scalar, to indicate that a particular number does not transform under changes of coordinate system. This elaborate nomenclature may seem like overkill for pure linear algebra, and I would agree that it is. The distinction really only becomes useful in geometry (or physics).

We don't use integer because not all are integers, I can use $\pi$ as a scalar and it'll be just fine. The reason for scalar vs number is that they are distinct in the sense that a number is an individual entity in a ring while a scalar is an operation on a vector space that changes the vector itself in some sense.

In abstract algebra, where linear algebra belongs, there is a distinct difference between a module, where vector spaces are a subset, and rings. Objects in a so called ring, which is in vector spaces a field, are just individual elements but they ACT UPON the elements of a module, or in your case the vector space, to produce a new vector.

So the distinction lies in that a scalar acts upon a vector to produce a new vector while a number is just an element in the field in question.

• What do you mean by saying "individual entity in a ring"? – LiziPizi Jun 21 '16 at 16:32
• It's a point a single element in a ring. Are you familiar with abstract algebra? – Zelos Malum Jun 21 '16 at 16:33
• I feel like if LiziPizi is just learning linear algebra, there's no way they'd be familiar with abstract algebra. – Samuel Yusim Jun 21 '16 at 16:35
• Unfortunately, no – LiziPizi Jun 21 '16 at 16:37
• @LiziPizi That explains it, i'll give a better explination then – Zelos Malum Jun 21 '16 at 16:40

To me, the reason has to do with levels of abstraction. Namely, when we say "number" we instantly carry with it all the baggage of arithmetic and elementary algebra, be it an integer, a rational number or even a complex number.

Now, when we say "scalar" we indicate what this thing tends to do with the object (usually, an abstract thing called "vector") without actually concerning ourselves what that thing is (people say that it "comes from some scalar field, F" or something along those lines). Specifically, it acts as if it was scaling the vector.

Yes, the term "scaling" still retains some reference to multiplying by a number; however, it is also more open and general.

Using the term "scalar" in the context of Linear Algebra we emphasize that it is an atomic element rather than a collection of elements like "vector" or "matrix". This the main point. In this context we do not care about a nature of such element. Particularly in this context we do not care whether it is a number or not. In this context we do not care if the set of all the scalars is a field.

• I have no idea what an "atomic element" is or how "scalar" emphasizes this. – user223391 Jun 25 '16 at 20:27
• @ZacharySelk: He means it's not a set or tensor – jmoreno Jun 25 '16 at 22:18
• @jmoreno And how does the term "scalar" emphasize this? – user223391 Jun 25 '16 at 22:47

protected by user223391 Jun 25 '16 at 0:05

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