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Section20.1Definitions and Examples

ΒΆ

A VV over a field FF is an abelian group with a Ξ±β‹…v\alpha \cdot v or Ξ±v\alpha v defined for all α∈F\alpha \in F and all v∈Vv \in V satisfying the following axioms.

  • Ξ±(Ξ²v)=(Ξ±Ξ²)v;\alpha(\beta v) =(\alpha \beta)v\text{;}

  • (Ξ±+Ξ²)v=Ξ±v+Ξ²v;(\alpha + \beta)v =\alpha v + \beta v\text{;}

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

  • 1v=v;1v=v\text{;}

where α,β∈F\alpha, \beta \in F and u,v∈V.u, v \in V\text{.}

The elements of VV are called ; the elements of FF are called . 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 0,{\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 nn-tuples of real numbers, denoted by Rn,{\mathbb R}^n\text{,} form a vector space over R.{\mathbb R}\text{.} Given vectors u=(u1,…,un)u = (u_1, \ldots, u_n) and v=(v1,…,vn)v = (v_1, \ldots, v_n) in Rn{\mathbb R}^n and Ξ±\alpha in R,{\mathbb R}\text{,} we can define vector addition by

u+v=(u1,…,un)+(v1,…,vn)=(u1+v1,…,un+vn)\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

Ξ±u=Ξ±(u1,…,un)=(Ξ±u1,…,Ξ±un).\begin{equation*} \alpha u = \alpha(u_1, \ldots, u_n)= (\alpha u_1, \ldots, \alpha u_n). \end{equation*}
Example20.2

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

Example20.3

The set of all continuous real-valued functions on a closed interval [a,b][a,b] is a vector space over R.{\mathbb R}\text{.} If f(x)f(x) and g(x)g(x) are continuous on [a,b],[a, b]\text{,} then (f+g)(x)(f+g)(x) is defined to be f(x)+g(x).f(x) + g(x)\text{.} Scalar multiplication is defined by (αf)(x)=αf(x)(\alpha f)(x) = \alpha f(x) for α∈R.\alpha \in {\mathbb R}\text{.} For example, if f(x)=sin⁑xf(x) = \sin x and g(x)=x2,g(x)= x^2\text{,} then (2f+5g)(x)=2sin⁑x+5x2.(2f + 5g)(x) =2 \sin x + 5 x^2\text{.}

Example20.4

Let V=Q(2 )={a+b2:a,b∈Q}.V = {\mathbb Q}(\sqrt{2}\, ) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q } \}\text{.} Then VV is a vector space over Q.{\mathbb Q}\text{.} If u=a+b2u = a + b \sqrt{2} and v=c+d2,v = c + d \sqrt{2}\text{,} then u+v=(a+c)+(b+d)2u + v = (a + c) + (b + d ) \sqrt{2} is again in V.V\text{.} Also, for α∈Q,\alpha \in {\mathbb Q}\text{,} Ξ±v\alpha v is in V.V\text{.} We will leave it as an exercise to verify that all of the vector space axioms hold for V.V\text{.}

Proposition20.5

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

  1. 0v=00v ={\mathbf 0} for all v∈V.v \in V\text{.}

  2. α0=0\alpha {\mathbf 0} = {\mathbf 0} for all α∈F.\alpha \in F\text{.}

  3. If Ξ±v=0,\alpha v = {\mathbf 0}\text{,} then either Ξ±=0\alpha = 0 or v=0.v = {\mathbf 0}\text{.}

  4. (βˆ’1)v=βˆ’v(-1) v = -v for all v∈V.v \in V\text{.}

  5. βˆ’(Ξ±v)=(βˆ’Ξ±)v=Ξ±(βˆ’v)-(\alpha v) = (-\alpha)v = \alpha(-v) for all α∈F\alpha \in F and all v∈V.v \in V\text{.}

Proof

To prove (1), observe that

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

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

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

To show (4), observe that

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

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