20.1: Definitions and Examples
- Page ID
- 81199
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
Solution
vector addition by
\[ u + v = (u_1, \ldots, u_n) + (v_1, \ldots, v_n) = (u_1 + v_1, \ldots, u_n + v_n) \nonumber \]
and scalar multiplication by
\[ \alpha u = \alpha(u_1, \ldots, u_n)= (\alpha u_1, \ldots, \alpha u_n)\text{.} \nonumber \]
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
Solution
\(\alpha p(x)\text{.}\)
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
Solution
\((f+g)(x)\) 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
Solution
\(u + v = (a + c) + (b + d ) \sqrt{2}\) 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
\[ 0 v = (0 + 0)v = 0v + 0v; \nonumber \]
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
\[ v + (-1)v = 1v + (-1)v = (1-1)v = 0v = {\mathbf 0}\text{,} \nonumber \]
and so \(-v = (-1)v\text{.}\) We will leave the proof of (5) as an exercise.