Vector_Space
One of the fundamental concepts in Linear Algebra is that of a Vector Space. For simplicity, we define a Vector Space over a FielD.
Definition: A SeT V is a vector space over a FielD F, if given the operation vector addition defined in V, denoted v+w for all v,w in V, and the operation scalar multiplication in V, denoted a*(v) for all v in V and a in F, the following 10 properties hold:
*1. For all v,w in V, v+w belongs to V.
V is closed under vector addition.
*2. For all u,v,w in V, u+(v+w)= (u+v)+w.
Associativity of vector addition in V.
*3 . For all v in V, there exists an element 0 in V, such that 0+v=v.
Existence of an additive identity element in V.
*4. For all v in V, there exists an element –v in V, such that v+(-v)=0.
Existence of an additive inverse in V.
*5. For all v,w in V, v+w=w+v.
Commutativity of vector addition in V.
*6. For all a in F and v in V, a*v belongs to V.
V is closed under scalar multiplication.
*7. For all a,b in F and v in V, a*(b*v)=(a*b)*v.
Associativity of scalar multiplication in V.
*8. For all v in V, there exists and element 1 in F, such that 1*v=v.
Existence of an identity element for scalar multiplication in V.
*9. For all v,w in V and a in F, a*(v+w)=a*v+a*w.
Distributivity with respect ot vector addition.
*10. For all a,b in F and v in V, (a+b)*v=a*v+a*w.
Distributivity with respect to field addition.
Properties 1 through 5 indicate that V is a commutative group under vector addition. Properties 6 through 10 apply to scalar multiplication of a vector v in V by a scalar a in F.----
The members of a Vector Space are called vectors.
The concept of a Vector Space is entirely abstract like the concepts of a Group, Ring, and Field. To determine if a set V is a vector space one must specify the set V, a field F and define vector addition and scalar multiplication in V. Then if V satisfies the above 10 properties it is a Vector Space over the field F.----
'Terminology:
If the field F=R={real numbers}, then a vector space over R is called a Real Vector Space.
If the field F=C={complex numbers}, then a vector space over C is called a Complex Vector Space.
Example I:
Let V be the set of all n-tuples, [v1,v2,v3,…,vn] where vi, for i={1,2,3,...n} is a member of R={real numbers}. Let the field be R, as well.
Define Vector Addition:
For all v, w, in V, define v+w=[v1+w1,v2+w2,v3+w3,…vn+wn].
Define Scalar Multiplication:
For all a in F and v in V, a*v=[a*v1,a*v2,a*v3,…a*vn].
Then V is a Vector Space over R.
Proof:
*1. If v,w, in V then v+w=[v1+w1,v2+w2,v3+w3,…vn+wn]. But for all vi, wi, where
i={1,2,3,…n}, vi+wi is in R, since R is a field. Therefore, for all u,v in V, v+w is
in V.
*2. If u,v,w in V then u+(v+w)= [u1,u2,u3,…,un]+ [v1+w1,v2+w2,v3+w3,…vn+wn]=
[u1+(v1+w1),u2+(v2+w2),u3+(v3+w3),…un+(vn+wn)]. But for all ui,vi.wi, where
i={1,2,3,…n}, ui+(vi+wi)=(ui+vi)+wi, since ui,vi,wi in R and R is a field.
Therefore, u+(v+w)=(u+v)+w, for all u,v,w in V.
*3. Since R is a field there exists an additive identity in R, say, 0. Consider [0,0,0,…0].
Then 0 is in V. But then for all v in V, 0+v=[v1+0,v2+0,v3+0,…,vn+0]=
[v1,v2,v3,…vn] since vi in R for all i={1,2,3…n}, and 0+vi=vi for all vi ,
where i={1,2,3…n}, since R is a field.
*4. Since R is a field, there exists for every a in R and element –a in R such that
a+(-a)=0. For v in V=[v1,v2,v3,…,vn], Consider –v=[-v1,-v2,-v3,…,-vn].
-v is in R and v+(-v)=[v1+(-v1),v2+(v2),v3+v3+(-v3),…v+(-vn)]=0, since
vi+(-vi)=0 for all I={1,2,3,..n} since R is a field.
*5. Since R is a field, for a,b in R a+b=b+a. Then
v+w=[v1+w1,v2+w2,v3+w3,…,vn+wn]= [w1+v1,w2+v2,w3+v3,…,wn+vn]
=w+v, since for each i={1,2,3,…,n} vi+wi=wi+vi, since R is a field.
*6. Since R is a field, if a,b in R a*b in R. Then a*v=[a*v1,a*v2,a*v3,…,a*vn].
Then a*vi for I={1,2,3…n} is in R. Therefore, a*v in V.
*7. Since R is a field, R has a multiplicative identity 1, such that 1*a=a for all
a in R. Then for v in V, 1*v=[1*v1,1*v2,1*v3,…1*vn]=
[v1,v2,v3,…,vn]=v, since vi, for I={1,2,3,…,n}, a*vi=a*vi.
*8. Since R is a field for a,b,c in R a*(b+c)=a*b+a*c. Then for v in V
a*(v+w)=a*[v1+w1,v2+w2,v3+w3,…,vn+wn]=
[a*(v1+w1),a*(v2+w2),a*(v3+w3),…a*(vn+wn)]=
[a*v1+aw1,a*v2+a*w2,a*v3+a*w3,…a*vn+a*wn]=
a*[v1,v2,v3,…,vn]+a*[w1,w2,w3,…,wn]=a*v+a*w.
*9. Since R is a filed, for a,b,c in R a*(b*c)=(a*b)*c.
Then a*(b*v)=a*[b*v1,b*v2,b*v3,…b*vn]=
[(a*b)v1,(a*b)v2,(a*b)v3,…,(a*b)vn]=(a*b)*v.
*10. Since R is a filed, for a,b,c in R, (a+b)*c=a*b+a*c.
Then (a+b)v=(a+b)[v1,v2,v3,…vn]=[(a+b)v1,(a+b)v2,(a+b)v3,…(a+b)vn]=
[a*v+b*v1,a*v2+a*v2,a*v3+b*v3,…a*vn+b*n]=[a*v1,a*v2,a*v3,…,a*vn]+
[b*v1,b*v2,b*3,…,b*vn]=a*v+b*v.
This vector space is denoted Rn.
Example II:
Let M be the set of all (mxn) matrices, with complex elements. Let C be the field of complex numbers. Then if
P is in M, P= |p11 p12 p13...p1n|
|p21 p22 p23...p2n|
|p31 p32 p33...p3n|
|.......................|
|.......................|
|pm1 pm2 pm3...pmn|
- where pij is in C.
Define vector addition in M:
P+Q= |p11 p12 p13...p1n| |q11 q12 q13...q1n|
|p21 p22 p23...p2n| |q21 q22 q23...q2n|
|p31 p32 p33...p3n| |q31 q32 q33...q3n| =
| . | + | . |
| . | | . |
|pm1 pm2 pm23 pmn| |qm1 qm2 qm3...qmn|
|p11+q11 p12+q12 p13+q13...p1n+q1n|
|p21+q21 p22+q22 p23+q23...p2n+q2n|
|p31+q31 p32+q32 p33+q33...p3n+q3n|
|. |
|. |
|pm1+qm1 pm2+qm2 pm3+qm3...pmn+qmn|
Define scalar multiplication:
|p11 p12 p13...p1n| |c*p11 c*p12 c*p13...c*p1n|
|p21 p22 p23...p2n| |c*p21 c*p22 c*p23...c*p2n|
c* |p31 p32 p33...p3n| |c*p31 c*p32 c*p33...c*p3n|
| . | = | |
| . | | |
|pm1 pm2 pm3...pmn| |c*pm1 c*pm2 c*pm3...c*pmn|
Then M is a vector space over C and we denote this as Cmxn.
So Example I would be denoted R1xn, or more simply, Rn.
In Analysis, many function sets have the structure of a Vector Space. In Analysis, a Vector Space is called a Linear Space.
Example III:
Let the set F[a,b]={all functions f defined on the closed interval [a,b]->R}.
Define vector addition:
(f+g)(x)=f(x)+g(x).
Define scalar multiplication: If a in R={real numbers} and F in F, then
(a*f)(x)=a*f(x).
Then F is a vector space over the field R.
Proof
*1. Since R is a field, if r,s, in R, then r+s in R.
Then for f,g in F and cin [a,b], f(c)+g(c) in R.
*2. Since R is a field, if r,s,t in R, then r+(s+t)=(r+s)+t.
Then for f,g,h, in F and c in [a,b], f(c)+(g(c)+h(c))=((f(c)+g(c))+h(c).
*3. Since R is a field, consider the function 0, where for c in [a,b], 0(c)=0.
0 is in F, since for all c in [a,b], 0(c) is in R. But, for f in F and c in [a,b],
0(c)+(f(c)=0+f(c)=f(c).
*4. Since F is the set of all functions from [a,b] to R, for f in F, consider –f in F,
defined by –f(c)=-(f(c)). –f is in F since it is defined from [a,b] to R.
since if f(c) in R then –(f(c)) in R since R is a field.
*5. Since R is a field, for r,s in R, r+s=s+r.
Then for f,g in F and c in [a,b], f(c)+g(c)=g(c)+f(c).
*6. Since R is a field, if r,s in R r,s,t in R, r*(s+t)=r*s+r*t. Then for f,g, inF,
r in R, and c in [a,b], r*(f(c)+g(c)))=r*(f(c)+r*(g(c).
*7. Since R is a field, consider the function 1, defined by, for all c in [a,b], 1(c)=c.
Then 1 is in F since for all c in [a,b], 1(c)=c in R.
But (1*f)(c)= 1*f(c)=f(c).
*8. Since R is a field, if r,s,t in R, r*(s+t)=r*s+r*t.
Then if f,g in F and r in R, for c in [a,b], (r*(f(c)+g(c)))=r*f(c)+r*g(c).
*9. Since R is a field, if r,s,t in R then r*(s*t)=(r*s)*t.
Then for r,s in R, f in F, and c in [a,b], r*(s*f(c))= (r*s)(f(c)).
*10. Since R is a field, is r,s,t in R, then (r+s)*t=r*t+s*t.
Then for r,s in R, f in F and c in [a,b], then (r+s)f(c)=r*f(c)+s*f(c).