Transcription of Vectors and Vector Spaces
1 Chapter 1 Vectors and Vector Vector SpacesUnderlying every Vector space (to be defined shortly) is a of scalarfields are the real and the complex numbersR:= real numbersC:= complex are the onlyfields we use spaceVis a collection of objects with a ( Vector )addition and scalar multiplication defined that closed under both operationsand which in addition satisfies the following axioms:(i) ( + )x= x+ xfor allx Vand , F(ii) ( x)=( )x(iii)x+y=y+xfor allx, y V(iv)x+(y+z)=(x+y)+zfor allx, y, z V(v) (x+y)= x+ y(vi) O Vz0+x=x; 0 is usually called theorigin(vii) 0x=0(viii)ex=xwhereeis the multiplicative unit 1. Vectors AND Vector SPACESThe closed property mentioned above means that for all , Fandx, y V x+ y V( you can t leaveVusing Vector addition and scalar multiplication). Also,when we write for , Fandx V( + )xthe + is in thefield, whereas when we writex+yforx, y V,the + isin the Vector space.
2 There is a multiple usage of this (1)R2={(a1,a2)|a1,a2 R}two dimensional space.(2)Rn={(a1,a2,.. ,an)|a1,a2,.. ,an R},ndimensional space.(a1,a2,.. ,an)iscalledann-tuple.(3)C2andCnrespecti vely toR2andRnwhere the underlyingfield isC,the complex numbers.(4)Pn=ln j=0ajxj|a0,a1,.. ,an RMis called thepolynomial spaceofall polynomials of degreen. Note this includes not just the polynomialsof exactly degreenbut also those of lesser degree.(5)fp={(ai,..)|ai R, |ai|p< }. This space is comprised ofvectors in the form of infinite-tuples of numbers. Properly we wouldwritefp(R)orfp(C)to designate thefield.(6)TN=FN n=1ansinn x|a1,.. ,an Rk, trigonometric Vectors inRne1=(1,0,.. ,0)e2=(0,1,0,.. ,0)e3=(0,0,1,0,.. ,0)..en=(0,0,.. ,0,1)These are theunit vec-torswhichpoint in thenorthogonal Vector SPACES9 Precise definitions will be given , the standard Vectors are10 CHAPTER 1.
3 Vectors AND Vector SPACESe1=(1,0)e2=(0,1)(1,0)(0,1)(0,0)12e Graphical representa-tion ofe1ande2in theusual two the usual Vector addition in the plane uses the parallelogram ruleyx+yForR3, the standard Vectors aree1=(1,0,0)e2=(0,1,0)e3=(0,0,1)(0,0,1) (1,0,0)eee(0,1,0)231 Graphical representa-tion ofe1,e2,ande3inthe usualLinear algebra is the mathematics of Vector Spaces and their subspaces. Wewill see that many questions about Vector Spaces can be reformulated asquestions about arrays of SubspacesLetVbe a Vector space andU closed under Vector addition, scalar multiplication and satisfies all of thevector space axioms. We also use the termlinear Vector will be given laterletV=R3={(a, b, c)|a, b, c R}( )U={(a, b,0)|a, b R}.ClearlyU Vand alsoUis a subspace ,v2 R3( )W={av1+bv2|a, b R}Wis a subspace this case we sayWis spanned by{v1,v2}.
4 In general, letS V,avector space, have the formS={v1,v2,.. ,vk}.ThespanofSis the setU= k3j=1ajvj|a1,.. ,ak R .We will use the notionS(v1,v2,.. ,vk)for thespanof a set of say thatu=a1v1+ +akvkis alinear combinationof the vectorsv1,v2,.. , a Vector space andU closed undervector addition and scalar multiplication, thenUis a subspace remark that this result provides a short cut to proving that aparticular subset of a Vector space is in fact a subspace. The actual proofof this result is simple. To show (i), note that ifx Uthenx Vand so(ab)x=ax+ , bx, ax+bxand (a+b)xareallinUby the closure hypothesis. Theequality is due to Vector space properties (i) the other axioms is proved 1. Vectors AND Vector SPACESA very important corollary follows about a Vector space andS={v1,v2.}
5 ,vk} (v1,.. ,vk)is a linear subspace merely observe thatS(v1,.. ,vk)=lk31ajvj|a1,.. ,ak means that the closure is built right into the definition of span. Thus,ifv=a1v1+ +akvkw=b1v1+ +bkvkthen bothv+w=(a1+b1)v1+ +(ak+bk)vkandcv=ca1v+ca2v+ +cakvare closed under both operations; thereforeUis a (Product Spaces .) LetVandWbe Vector Spaces definedover the samefield. We define the new Vector spaceZ=V WbyZ={(v, w)|u V, w W}We define Vector addition as (v1,w1)+(v2,w2)=(v1+v2,w1+w2)andscalar multiplication by (v, w)=( v, w). With these operations,Zis avector space, sometimes called set-builder notation, defineV13={(a,0,b)|a, b, R}.ThenUis a subspace can also be realized as the subspace ofthe standard vectorse1=(1,0,0) ande3=(0,0,1), that is to sayV13=S(e1,e3). LINEAR INDEPENDENCE AND LINEAR DEPENDENCE13 Example subspaces are two other importantmethods to construct subspaces ofR3.
6 Besides the set builder notationused above, we have just considered the method of spanning sets. Forexample, letS={v1,v2} (S) is a subspace , ifT={v1} (T) is a subspace construct subspaces is by using inner products. Letx, w (x1,x2,x3)andw=(w1,w2,w3).Definethe inrner product ofxandwbyx w=x1w1+x2w2+ {x R3|x w=0}is a subpace ofR3. To prove this it is neces-sary to prove closure under Vector addition and scalar multiplication. Thelatter is easy to see because the inner product is homogeneous in ,that is,( x) w= x1w1+ x2w2+ x3w3= (x w).Therefore ifx w=0soalso is ( x) additivity is also straightforward. Letx, y sum(x+y) w=(x1+y1)w1+(x2+y2)w2+(x3+y3)w3=(x1w1+x2 w2+x3w3)+(y1w1+y2w2+y3w3)=0+0=0 However, by choosing two vectorsv, w, R3we can defineUv,w={x R3|x y=0andx w=0}.EstablishingUv,wis a subspace ofR3is provedsimilarly.
7 In fact, what is that both these sets of subspaces, those formedby spanning sets and those formed from the inner products are the same setof subspaces. For example, referring to the previous example, it follows thatV13=S(e1,e3)=Ue2. Can you see how to correspond the others? Linear independence and linear dependenceOne of the most important problems in Vector Spaces is to determine ifa given subspace is the span of a collection of Vectors and if so, to deter-mine a spanning set. Given the importance of spanning sets, we intend toexamine the notion in more detail. In particular, we consider the conceptof uniqueness of {v1,.. ,vk} V, a Vector space, and letU=S(v1,.. ,vk)(orS(S) for simpler notation). Certainly we know that any vectorv Uhasthe representationv=a1v1+ +akvkfor some set of scalarsa1.
8 ,ak. Is this representation unique?Or,canwe14 CHAPTER 1. Vectors AND Vector Spaces find another set of scalarsb1,.. ,bknotallthesameasa1,.. ,akrespec-tively for whichv=b1v1+ + need more information aboutSto answer this question either {v1,.. ,vk} V, a Vector space. We say thatSislinearly dependent( ) if there are scalarsa1,.. ,aknot all zero forwhicha1v1+a2v2+ +akvk=0.(T)OtherwisewesaySislinearly independent( ). we allow all the scalars to be zero we can always arrange for (T)to hold, making the concept {v1,.. ,vk} V, a Vector space, is linearlydependent, then one member of this set can be expressed as alinear combi-nationof the know that there are scalarsa1,.. ,aksuch thata1v1+a2v2+ +akvk=0 Since not all of the coefficients are zero, we can solve for one of the vectorsas a linear combination of the other we have shown that there isnovector with aunique representation inS(S).
9 Corollary S={v1,.. ,vk},thenSis linearly {v1,.. ,vk}is linearly independent then everysubset ofSis linearly BasesThe idea of a basis is that offinding a minimal generating set for a vectorspace. Through basis, unicity of representation and a number of other usefulproperties, both theoretical and computational, can be concluded. Thinkingof the concept in operations research ideas, a basis will be a redundancy freeand complete generating set for a Vector spaceDefinition a Vector space andS={v1,.. ,vk} setfor the subspaceU=S(S).Suppose thatVis a Vector space, andS={v1,.. ,vk}is a linearlyindependent spanning set called correspondingly for a basis ofV, then every Vector has a {v1,.. ,vk}andv + +akvkfor some choice of scalars. If there is a second choice of scalarsb1.
10 ,bknot all the same, respectively, asa1,.. ,ak,wehavev=b1v1+ +bkvkand0=(a1 b1)v1+ +(ak bk) not all of the differencesa1 b1,.. ,ak bkare zero we must havethatSis linearly dependent. This is a contradiction to our hypothesis, andthe result is {e1,e2,e3}.ThenSis a basis spanned byS. Now suppose that0=a1e1+a2e2+a3e3or(0,0,0) =a1(1,0,0) +a2(0,1,0) +a3(0,0,1)=(a1,a2,a3).Hencea1=a2=a3=0. Thustheset{e1,e2,e3}is linearly 1. Vectors AND Vector SPACESR emark how we resolved the linearly dependent/linearly in-dependent issue by converting a Vector problem to a numbers problem. Thisis at the heart of linear {v1,v2}={(1,0,1),(1, 1,0)} R3. Show thatSislinearly independent and therefore a basis ofS(S). Extension to a basisIn this section, we show that given a linearly independent set of vectorsfrom a Vector space with afinite spanning set, it is possible add to this setmore Vectors until it becomes a basis.
