Is zero a scalar?

Is zero considered a scalar?

In other words, is $\begin{bmatrix}0\\0\\\end{bmatrix}$ a scalar multiple of $\begin{bmatrix}a\\b\\\end{bmatrix}$ where $a$ and $b$ are real numbers?

• $0$ is a scalar. – Mike Earnest Sep 18 '15 at 4:36
• Perhaps the confusion is in the term 'scalar multiple', which doesn't mean a scalar that is a multiple but rather a vector that is a multiple (where the multiplier is a scalar) of another, in a similar fashion to how a complex vector space isn't usually made of complex numbers. Not too likely the reason here, but just offering up an opinion. – Vandermonde Sep 19 '15 at 1:44

Yes, zero is a scalar.

So that people reading this might learn something, it turns out that the term "scalar" was first used by François Viète in his book Analytic Art (translated from In artem analyticem isagoge). He called two things scalar if they were proportionally related to each other (roughly speaking).

The first usage of "scalar" in English was apparently by Hamilton, although he used it to mean the "real" part of a quaternion, i.e. the $t$ in standard notation $t + xi + yj + zij$.

Yes, $0$ is indeed a scalar. But, to be clear, $\begin{bmatrix}0\\0\\\end{bmatrix}$ is not a scalar; it's a vector, the result of multiplying the scalar, $0$ by the vector, $\begin{bmatrix}a\\b\\\end{bmatrix}$, where $a$ and $b$ are real numbers.

• Funny, I never thought of it as "multiplying the scalar by the vector", always as "multiplying the vector by the scalar". – G. Bach Sep 18 '15 at 12:01
• @G.Bach isn't the multiplication of a scalar and a vector commutative? – Mindwin Sep 18 '15 at 14:39
• Depends on whether you define it as a function on $K\times V$, on $V\times K$, or on the union, $K$ being the field and $V$ the vector space. In the last case, yes, it is commutative. :) – MickG Sep 18 '15 at 15:52
• @Mindwin: usually commutative refers to binary operations on a set with itself (I disagree with the venerable hotmath here). Scalars and vectors come from different sets. But you're right that left- and right-multiplication of a vector by a scalar coincide. – Matthew Leingang Sep 18 '15 at 16:34

As the other answers say, $0$ is indeed a scalar, and $0 {\begin{bmatrix}a \\ b \end{bmatrix}} = {\begin{bmatrix}0 \\ 0 \end{bmatrix}}$ is indeed a scalar multiple of ${\begin{bmatrix}a \\ b \end{bmatrix}}$. More precisely, whatever field or ring of scalars your matrices are defined over, it surely has a zero element, which we conventionally denote by "$0$".

However, it may be worth noting that the symbol $0$ is also often used to denote a zero matrix, i.e. a matrix whose elements are all zero. Thus, one might e.g. write $AB = 0$, where $A$ and $B$ are matrices, to indicate that the matrix product of $A$ and $B$ contains all zeros. This does not usually cause any ambiguity, since it's clear from context what type of entity the symbol $0$ represents in each case. (Similarly, we often use $I$ to denote "the" identity matrix, without explicitly specifying what the dimensions of this particular identity matrix are.) Another justification for this notation is that, in the matrix ring of $n \times n$ square matrices, the $n \times n$ zero matrix is indeed the zero element of the ring.

Some authors might use e.g. boldface "$\mathbf 0$" or an underlined "$\underline{\underline 0}$" instead of a plain "$0$" to represent a zero matrix, especially if they follow a convention to highlight all matrix symbols in the same manner. Similarly, a zero vector might (in contexts where one makes a distinction between vectors and row/column matrices) be denoted e.g. by "$\vec 0$" instead of a plain "$0$". Many authors, however, especially in more advanced texts, will not bother with such conventions and will just expect you to figure out what type of object "$0$" represents from context.

Yes. In the context of vector spaces, a scalar is a member of the underlying field. That especially includes the neutral element of field addition, known as $0$.

• To the down-voter: please state the reason for the downvote. – celtschk Sep 21 '15 at 18:49