The following was taken from Thomas A. Garrity's textbook "All the Mathematics You Missed."

Definition 1.3.1 A set $V$ is a vector space over the real numbers $\mathbb{R}$ if there are maps:

$\hspace{0.5cm}$1.$\hspace{.3cm} \mathbb{R} \times V \rightarrow V$, denoted by $a \cdot v$ or $av$ for all real numbers $a$ and elements $v$ in $V$,

$\hspace{0.5cm}$2.$\hspace{.3cm} V \times V \rightarrow V$, denoted by $v + w$ for all elements $v$ and $w$ in the vector space $V$,

with the following properties:

$\hspace{0.5cm}$a)$\hspace{.3cm}$There is an element $0$, in $V$ such that $0 + v = v$ for all $v \in V$.

$\hspace{0.5cm}$b)$\hspace{.3cm}$For each $v \in V$, there is an element $(-v) \in V$ with $v + (-v) = 0$.

$\hspace{0.5cm}$c)$\hspace{.3cm}$For all $v, w \in V$, $v + w = w + v$.

$\hspace{0.5cm}$d)$\hspace{.3cm}$For all $a \in \mathbb{R}$ and for all $v, w \in V$, we have that $a(v + w) = av + aw$.

$\hspace{0.5cm}$e)$\hspace{.3cm}$For all $a,b \in \mathbb{R}$ and all $v \in V$, $a(bv) =(a \cdot b)v$.

$\hspace{0.5cm}$f)$\hspace{.3cm}$For all $a, b \in \mathbb{R}$ and all $v \in V$, $(a + b)v = av + bv$.

$\hspace{0.5cm}$g)$\hspace{.3cm}$For all $v \in V$, $1 \cdot v = v$.

For the sake of absolute clarity, could someone explain to me what the first two statements are saying in exhaustive detail?


As noted by Andreas Blass, the litany for vector space criteria should—rather than the above definition—be read as follows:

$\hspace{0.5cm}$ Definition. A vector space (or linear space) consists of the following:

$\hspace{0.5cm}$ 1. $\hspace{0.3cm}$ a field $F$ of scalars;

$\hspace{0.5cm}$ 2. $\hspace{0.3cm}$ a set $V$ of objects, called vectors;

$\hspace{0.5cm}$ 3. $\hspace{0.3cm}$ a rule (or operation), called vector addition, which associates with each pair of vectors $\alpha$, $\beta$ in $V$ a vector $\alpha + \beta$ in $V$, called the sum of a $\alpha$ and $\beta$, in such a way that

$\hspace{.8cm}$ (a) $\hspace{0.3cm}$ addition is commutative, $\alpha + \beta = \beta + \alpha$;

$\hspace{.8cm}$ (b) $\hspace{0.3cm}$ addition is associative, $\alpha + (\beta + \gamma) = (\alpha + \beta) + \gamma$;

$\hspace{.8cm}$ (c) $\hspace{0.3cm}$ there is a unique vector $0$ in $V$, called the zero vector, such that $\alpha + 0 = \alpha$ for all $\alpha$ in $V$.

$\hspace{.8cm}$ (d) $\hspace{0.3cm}$ for each vector $\alpha$ in $V$ there is a unique vector $-\alpha$ in $V$ such that $\alpha + (-\alpha) = 0$;

$\hspace{0.5cm}$ 4. $\hspace{0.3cm}$ a rule (or operation), called scalar multiplication, which associates with each scalar $c$ in $F$ and vector $\alpha$ in $V$ a vector $c\alpha$ in $V$, called the product of $c$ and $\alpha$, in such a way that

$\hspace{.8cm}$ (a) $\hspace{0.3cm}$ $1\alpha = \alpha$ for every $\alpha$ in $V$;

$\hspace{.8cm}$ (b) $\hspace{0.3cm}$ $(c_1c_2)\alpha = c_1(c_2\alpha)$;

$\hspace{.8cm}$ (c) $\hspace{0.3cm}$ $c(\alpha + \beta) = c\alpha + c\beta$;

$\hspace{.8cm}$ (d) $\hspace{0.3cm}$ $(c_1 + c_2)\alpha = c_1\alpha + c_2\alpha$.

A definition which can be found in Hoffman & Kunze's textbook "Linear Algebra."

  • 2
    $\begingroup$ It might be easier if you mentioned what you find troublesome. From your post it seems as though it might be the word "map" or the concept of a function. $\endgroup$ – user2055 Feb 25 '13 at 0:58
  • $\begingroup$ I have my own interpretations of things-as does everyone, but I'd like to know how anyone else interprets "$A \times B \rightarrow C$". $\endgroup$ – Trancot Feb 25 '13 at 1:03
  • 1
    $\begingroup$ What is $A\times B$? It is the set of all ordered pairs $\langle a,b\rangle$. So, a function $A\times B\to C$ is like a machine that inputs an element from $A$ and an element from $B$ and outputs an element of $C$. $\endgroup$ – Berci Feb 25 '13 at 1:04
  • $\begingroup$ So, "$\times$" should be seen as a set of operators then, correct? $\endgroup$ – Trancot Feb 25 '13 at 1:08
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    $\begingroup$ Almost. $\times$ is a single operator (not a set of operators). You input two sets, $A$ and $B$, and the result is $A \times B$, which is shorthand for $\{(a,b): a \in A, b \in B\}$. $\endgroup$ – goblin Feb 25 '13 at 1:18

The 1st one says there's a way to multiply a vector by a number and get a vector as the answer.

The 2nd one says there's a way to add two vectors and get a vector as the answer.


The first line means that for every tuple $(a,v)$ where $a$ is a real number and $v$ is an element from the vector space $V$ there is an associated element in the vector space, denoted $av$.

The second line says that for every tuple $(u,v)$ where $u$ and $v$ are elements from the vector space $V$ there is an associated vector in $V$, denoted $u+v$.

  • 1
    $\begingroup$ New spelling touple ... $\endgroup$ – GEdgar Feb 25 '13 at 4:12
  • $\begingroup$ @GEdgar: thanks 8)! $\endgroup$ – harlekin Feb 25 '13 at 12:46
  • $\begingroup$ Understandable, since we have the similar pronunciation of 'couple' $\endgroup$ – user50229 Feb 25 '13 at 12:52

A binary operation on a set $X$ is defined as a function $X\times X\to X$.

The addition of vectors is a binary operation on the set of all vectors which behaves very similarly to the addition of numbers (in fact, if we use coordinates for vectors, addition is done coordinatewise), and thus it is not a coincidence that addition of vectors is also denoted by $+$.

Among vectors, there is another characterisitic operation, it is that we can multiply a vector by a scalar number. Now $V$ denotes the set of all vectors and $\Bbb R$ the set of the real numbers, and this scalar multiplication itself is thus a function $\Bbb R\times V\to V$.

The notion of vector space is an abstraction that uses exactly these operations and their characteristic properties.

For example, the set of all vectors on the plane, is a vector space, which -once we fixed a coordinate system- becomes isomorphic to the set of all ordered pairs of real numbers, that is, $\Bbb R^2:=\Bbb R\times\Bbb R$.

If we take $V=\Bbb R^2$, then the operation of addition is the function $V\times V\to V$ which maps $$\langle\pmatrix{a\\b},\pmatrix{c\\d}\rangle \mapsto \pmatrix{a+c\\b+d}$$ where elements of $V=\Bbb R^2$ are written as culomn vectors.

  • $\begingroup$ Could you please explain your usage of angle brackets? $\endgroup$ – Trancot Feb 25 '13 at 4:04
  • $\begingroup$ I used it for ordered pairs. $A\times B=\{\langle a,b\rangle : a\in A,\,b\in B\}$, and $f:x \mapsto y$ denotes the function which satisfy $f(x)=y$ for all $x$ in its domain. $\endgroup$ – Berci Feb 25 '13 at 11:50

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