## 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)\text{.} \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}\text{,} \end{equation*}

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