Skip to main content
Logo image

Section 20.1 Definitions and Examples

A vector space V over a field F is an abelian group with a scalar product αv or αv defined for all αF and all vV satisfying the following axioms.
  • α(βv)=(αβ)v;
  • (α+β)v=αv+βv;
  • α(u+v)=αu+αv;
  • 1v=v;
where α,βF and u,vV.
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 0, 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 Rn, form a vector space over R. Given vectors u=(u1,,un) and v=(v1,,vn) in Rn and α in R, we can define vector addition by
u+v=(u1,,un)+(v1,,vn)=(u1+v1,,un+vn)
and scalar multiplication by
αu=α(u1,,un)=(αu1,,αun).

Example 20.2.

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 αF and p(x)F[x], then scalar multiplication is defined by αp(x).

Example 20.3.

The set of all continuous real-valued functions on a closed interval [a,b] is a vector space over 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 (αf)(x)=αf(x) for αR. For example, if f(x)=sinx and g(x)=x2, then (2f+5g)(x)=2sinx+5x2.

Example 20.4.

Let V=Q(2)={a+b2:a,bQ}. Then V is a vector space over Q. If u=a+b2 and v=c+d2, then u+v=(a+c)+(b+d)2 is again in V. Also, for αQ, αv is in V. We will leave it as an exercise to verify that all of the vector space axioms hold for V.

Proof.

To prove (1), observe that
0v=(0+0)v=0v+0v;
consequently, 0+0v=0v+0v. Since V is an abelian group, 0=0v.
The proof of (2) is almost identical to the proof of (1). For (3), we are done if α=0. Suppose that α0. Multiplying both sides of αv=0 by 1/α, we have v=0.
To show (4), observe that
v+(1)v=1v+(1)v=(11)v=0v=0,
and so v=(1)v. We will leave the proof of (5) as an exercise.