## Section 20.1 Definitions 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.

### Example 20.1.

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

and scalar multiplication by

### Example 20.2.

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 \(\alpha p(x)\text{.}\)

### Example 20.3.

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 \((2f + 5g)(x) =2 \sin x + 5 x^2\text{.}\)

### Example 20.4.

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 \(V\text{.}\)

### Proposition 20.5.

Let \(V\) be a vector space over \(F\text{.}\) Then each of the following statements is true.

\(0v ={\mathbf 0}\) for all \(v \in V\text{.}\)

\(\alpha {\mathbf 0} = {\mathbf 0}\) for all \(\alpha \in F\text{.}\)

If \(\alpha v = {\mathbf 0}\text{,}\) then either \(\alpha = 0\) or \(v = {\mathbf 0}\text{.}\)

\((-1) v = -v\) for all \(v \in V\text{.}\)

\(-(\alpha v) = (-\alpha)v = \alpha(-v)\) for all \(\alpha \in F\) and all \(v \in V\text{.}\)

### Proof.

To prove (1), observe that

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

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