
## Section20.1Definitions and Examples

A vector space $V$ over a field $F$ is an abelian group with a scalar product $\alpha \cdot v$ or $\alpha v$ defined for all $\alpha \in F$ and all $v \in V$ satisfying the following axioms.

• $\alpha(\beta v) =(\alpha \beta)v\text{;}$

• $(\alpha + \beta)v =\alpha v + \beta v\text{;}$

• $\alpha(u + v) = \alpha u + \alpha v\text{;}$

• $1v=v\text{;}$

where $\alpha, \beta \in F$ and $u, v \in V\text{.}$

The elements of $V$ are called vectors; the elements of $F$ are called scalars. It is important to notice that in most cases two vectors cannot be multiplied. In general, it is only possible to multiply a vector with a scalar. To differentiate between the scalar zero and the vector zero, we will write them as 0 and ${\mathbf 0}\text{,}$ respectively.

Let us examine several examples of vector spaces. Some of them will be quite familiar; others will seem less so.

###### Example20.1

The $n$-tuples of real numbers, denoted by ${\mathbb R}^n\text{,}$ form a vector space over ${\mathbb R}\text{.}$ Given vectors $u = (u_1, \ldots, u_n)$ and $v = (v_1, \ldots, v_n)$ in ${\mathbb R}^n$ and $\alpha$ in ${\mathbb R}\text{,}$ we can define vector addition by

\begin{equation*} u + v = (u_1, \ldots, u_n) + (v_1, \ldots, v_n) = (u_1 + v_1, \ldots, u_n + v_n) \end{equation*}

and scalar multiplication by

\begin{equation*} \alpha u = \alpha(u_1, \ldots, u_n)= (\alpha u_1, \ldots, \alpha u_n). \end{equation*}
###### Example20.2

If $F$ is a field, then $F[x]$ is a vector space over $F\text{.}$ The vectors in $F[x]$ are simply polynomials, and vector addition is just polynomial addition. If $\alpha \in F$ and $p(x) \in F[x]\text{,}$ then scalar multiplication is defined by $\alpha p(x)\text{.}$

###### Example20.3

The set of all continuous real-valued functions on a closed interval $[a,b]$ is a vector space over ${\mathbb R}\text{.}$ If $f(x)$ and $g(x)$ are continuous on $[a, b]\text{,}$ then $(f+g)(x)$ is defined to be $f(x) + g(x)\text{.}$ Scalar multiplication is defined by $(\alpha f)(x) = \alpha f(x)$ for $\alpha \in {\mathbb R}\text{.}$ For example, if $f(x) = \sin x$ and $g(x)= x^2\text{,}$ then $(2f + 5g)(x) =2 \sin x + 5 x^2\text{.}$

###### Example20.4

Let $V = {\mathbb Q}(\sqrt{2}\, ) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q } \}\text{.}$ Then $V$ is a vector space over ${\mathbb Q}\text{.}$ If $u = a + b \sqrt{2}$ and $v = c + d \sqrt{2}\text{,}$ then $u + v = (a + c) + (b + d ) \sqrt{2}$ is again in $V\text{.}$ Also, for $\alpha \in {\mathbb Q}\text{,}$ $\alpha v$ is in $V\text{.}$ We will leave it as an exercise to verify that all of the vector space axioms hold for $V\text{.}$

To prove (1), observe that

\begin{equation*} 0 v = (0 + 0)v = 0v + 0v; \end{equation*}

consequently, ${\mathbf 0} + 0 v = 0v + 0v\text{.}$ Since $V$ is an abelian group, ${\mathbf 0} = 0v\text{.}$

The proof of (2) is almost identical to the proof of (1). For (3), we are done if $\alpha = 0\text{.}$ Suppose that $\alpha \neq 0\text{.}$ Multiplying both sides of $\alpha v = {\mathbf 0}$ by $1/ \alpha\text{,}$ we have $v = {\mathbf 0}\text{.}$

To show (4), observe that

\begin{equation*} v + (-1)v = 1v + (-1)v = (1-1)v = 0v = {\mathbf 0}, \end{equation*}

and so $-v = (-1)v\text{.}$ We will leave the proof of (5) as an exercise.