Contents Notes on Elementary Linear Algebra

NotesonElementaryLinearAlgebraAdamCoffmanJuly30,2007Contents1Realvectorspaces22Subspaces63AdditiveFunctionsandLinearFunctions74Distancefunctions95Bilinearformsandsesquilinearforms116InnerProducts177Orthogonalandunitarytransformationsfornon-degenerateinnerproducts248Orthogonalandunitarytransformationsforpositivedefiniteinnerproducts28TheseNotesarecompiledfromclassroomhandoutsforMath351and511atIPFW.Theyarenotself-contained,butsupplementtherequiredtexts,[A]and[FIS].11RealvectorspacesDefinition1.1.GivenasetV,andtwooperations+(addition)and·(scalarmultiplication),Viscalleda“realvectorspace”iftheoperationshaveallofthefollowingproperties:1.ClosureunderAddition:Foranyu∈Vandv∈V,u+v∈V.2.AssociativeLawforAddition:Foranyu∈Vandv∈Vandw∈V,(u+v)+w=u+(v+w).3.ExistenceofaZeroElement:Thereexistsanelement0∈Vsuchthatforanyv∈V,v+0=v.4.ExistenceofanOpposite:Foreachv∈V,thereexistsanelementofV,called−v∈V,suchthatv+(−v)=0.5.ClosureunderScalarMultiplication:Foranyr∈Randv∈V,r·v∈V.6.AssociativeLawforScalarMultiplication:Foranyr,s∈Randv∈V,(rs)·v=r·(s·v).7.ScalarMultiplicationIdentity:Foranyv∈V,1·v=v.8.DistributiveLaw:Forallr,s∈Randv∈V,(r+s)·v=(r·v)+(s·v).9.DistributiveLaw:Forallr∈Randu,v∈V,r·(u+v)=(r·u)+(r·v).ThefollowingtheoremsrefertoarealvectorspaceV.Theorems1.2through1.11useonlythefirstfouraxiomsaboutaddition.Theorem1.2(RightCancellation).Givenu,v,w∈V,ifu+w=v+w,thenu=v.Proof.u+wandv+wareelementsofVbyAxiom1.Sincew∈V,thereexistsanopposite,alsocalledan“additiveinverse,”−w∈V.Addingthistobothsidesofu+w=v+wontherightgives(u+w)+(−w)=(v+w)+(−w),andtheassociativelawgivesu+(w+(−w))=v+(w+(−w)),sou+0=v+0.ByAxiom3,u=v.Theorem1.3.Givenu,w∈V,ifu+w=w,thenu=0.Proof.Sincew∈V,thereexistsanadditiveinverse−w∈V.Addingthistobothsidesofu+w=wontherightgives(u+w)+(−w)=w+(−w),andtheassociativelawgivesu+(w+(−w))=w+(−w),sou+0=0.ByAxiom3,u=0.Theorem1.4.Foranyv∈V,(−v)+v=0.Proof.(−v)+v∈VbyAxiom1.ThefollowingstepsuseAxioms2,3,4.((−v)+v)+((−v)+v)=(((−v)+v)+(−v))+v=((−v)+(v+(−v)))+v=((−v)+0)+v=(−v)+v,sothepreviousTheoremapplieswithuandwbothequalto(−v)+v,toshow(−v)+v=0.2Theorem1.5.Foranyv∈V,0+v=v.Proof.Weusethefactthatvhasanadditiveinverse,theassociativelaw,andthepreviousTheorem.0+v=(v+(−v))+v=v+((−v)+v)=v+0=v.Theorem1.6(LeftCancellation).Givenu,v,w∈V,ifw+u=w+v,thenu=v.Proof.w+uandw+vareinVbyAxiom1.Sincew∈V,thereexistsanadditiveinverse−w∈V.Addingthistobothsidesofw+u=w+vontheleftgives(−w)+(w+u)=(−w)+(w+v),andtheassociativelawgives((−w)+w)+u=(−w)+w)+v.ByTheorem1.4,0+u=0+v,andbythepreviousTheorem,u=v.Theorem1.7(UniquenessofAdditiveIdentity).Givenu,w∈V,ifw+u=w,thenu=0.Proof.w=w+0byAxiom3,soifw+u=w,thenw+u=w+0,andthepreviousTheoremgivesu=0.Theorem1.8(UniquenessofAdditiveInverse).Givenv,w∈V,ifv+w=0thenv=−wandw=−v.Proof.v+(−v)=0byAxiom4,soifv+w=0,thenv+w=v+(−v),andtheleftcancellationtheoremgivesw=−v.(−w)+w=0byTheorem1.4,soifv+w=0,thenv+w=(−w)+w,andtherightcancellationtheoremgivesv=−w.Theorem1.9.−0=0.Proof.0+0=0byAxiom3,sothepreviousTheoremapplies,withv=0andw=0,toshowthat0=−0.Theorem1.10.Foranyv∈V,−(−v)=v.Proof.Since−v∈Vandv+(−v)=0byAxiom4,Theorem1.8applies,withw=−v,toshowv=−w=−(−v).Theorem1.11.Givenu,x∈V,−(u+x)=(−x)+(−u).Proof.Note−xand−uareinVbyAxiom4,andu+xand(−x)+(−u)areinVbyAxiom1.Considerthesum(u+x)+((−x)+(−u)).Usingtheassociativelaw,itsimplifies:u+(x+((−x)+(−u)))=u+((x+(−x))+(−u))=u+(0+(−u))=(u+0)+(−u)=u+(−u)=0.So,Theorem1.8applies,withv=u+xandw=(−x)+(−u),toshoww=−v,and(−x)+(−u)=−(u+x).ThepreviousresultsonlyusedAxioms1–4,about“+,”butthenextresult,eventhoughitsstatementrefersonlyto+,usesascalarmultiplicationtrick,togetherwiththedistributiveaxioms,whichrelatescalarmultiplicationtoaddition.3Theorem1.12(CommutativePropertyofAddition).Foranyv,w∈V,v+w=w+v.Proof.Westartwith(1+1)·(v+w),whichisinVbybothclosureaxioms,setLHS=RHS,andusebothdistributivelaws:(1+1)·(v+w)=(1+1)·(v+w)((1+1)·v)+((1+1)·w)=(1·(v+w))+(1·(v+w))((1·v)+(1·v))+((1·w)+(1·w))=(v+w)+(v+w)(v+v)+(w+w)=(v+w)+(v+w).Then,theassociativelawgivesv+(v+(w+w))=v+(w+(v+w)),andleftcancellationleavesv+(w+w)=w+(v+w).Usingtheassociativelawagain,(v+w)+w=(w+v)+w,andrightcancellationgivestheresultv+w=w+v.Theorem1.13.Foranyv∈V,0·v=0.Proof.0·v∈VbyAxiom5.Thedistributivelawisneeded.0·v=(0+0)·v=(0·v)+(0·v).Theorem1.3applies,withuandwbothequalto0·v,toshow0·v=0.Theorem1.14.Foranyv∈V,(−1)·v=−v.Proof.(−1)·v∈VbyAxiom5.UsingAxiom7,thedistributivelaw,andthepreviousTheorem,v+((−1)·v)=(1·v)+((−1)·v)=(1+(−1))·v=0·v=0.Theorem1.8applies,withw=(−1)·v,toshow−v=w=(−1)·v.Theorem1.15.Foranyr∈R,r·0=0.Proof.r·0∈VbyAxiom5.Usingthedistributivelaw,r·0=r·(0+0)=(r·0)+(r·0).Theorem1.3applieswithu=w=r·0,toshowr·0=0.Theorem1.16.Foranyr∈Randu∈V,(−r)·u=−(r·u).Proof.(−r)·uandr·uareinVbyAxiom5.Usingthedistributivelaw,andTheorem1.13,(r·u)+((−r)·u)=(r+(−r))·u=0·u=0.Theorem1.8applies,withv=r·uandw=(−r)·u,toshoww=−v,so(−r)·v=−(r·v).Theorem1.17.Givenr∈Randu∈V,ifr·u=0,thenr=0oru=0.Proof.Therearetwocases:givenr∈R,eitherr=0,inwhichcasetheTheoremisprovedalready,orr=0.So,supposingr=0,multiplybothsidesofr·u=0by1r,toget1r·(r·u)=1r·0.ByAxioms6and7,theLHSsimplifiesto(1rr)·u=1·u=u,andby