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.

\(\alpha(\beta v) =(\alpha \beta)v\);

\((\alpha + \beta)v =\alpha v + \beta v\);

\(\alpha(u + v) = \alpha u + \alpha v\);

\(1v=v\);

where \(\alpha, \beta \in F\) and \(u, v \in V\).

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 \({\mathbf 0}\), respectively.

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

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Example20.1

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

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Example20.2

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

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Example20.3

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

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Example20.4

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

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Proposition20.5

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

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

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

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

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

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

To prove (1), observe that
\begin{equation*}0 v = (0 + 0)v = 0v + 0v;\end{equation*}
consequently, \({\mathbf 0} + 0 v = 0v + 0v\). Since \(V\) is an abelian group, \({\mathbf 0} = 0v\).

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

To show (4), observe that
\begin{equation*}v + (-1)v = 1v + (-1)v = (1-1)v = 0v = {\mathbf 0},\end{equation*}
and so \(-v = (-1)v\). We will leave the proof of (5) as an exercise.