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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).