Transcription of Introduction - math.uconn.edu
1 ISOMETRIES OF RnKEITH a functionh:Rn Rnthat preserves the distance between vectors:||h(v) h(w)||=||v w||for allvandwinRn, where||(x1,..,xn)||= x21+ + identity transformation: id(v) =vfor allv : id(v) = vfor allv : fixingu Rn, lettu(v) =v+u. Easily||tu(v) tu(w)||=||v w||.Example around points and reflections across lines in the plane are isome-tries ofR2. Formulas for these isometries will be given in Example and Section effects of a translation, rotation (around the origin) and reflection across a line inR2are pictured below on sample line composition of two isometries ofRnis an isometry. Is every isometry invertible? Itis clear that the three kinds of isometries pictured above (translations, rotations, reflections)are each invertible (translate by the negative vector, rotate by the opposite angle, reflect asecond time across the same line).In Section 2, we show the close link between isometries and the dot product onRn,which is more convenient to use than distances due to its algebraic properties.
2 Section 3is about the matrices that act as isometries on onRn, called orthogonal matrices. Section4 describes the isometries ofRandR2geometrically. In Appendix A, we will look moreclosely at reflections , dot products, and linearityUsing translations, we can reduce the study of isometries ofRnto the case of isometry ofRncan be uniquely written as the compositiont kwheretis a translation andkis an isometry fixing the :Rn Rnbe an isometry. Ifh=tw k, wheretwis translation by a vectorwandkis an isometry fixing0, then for allvinRnwe haveh(v) =tw(k(v)) =k(v) + getw=h(0), sowis determined byh. Thenk(v) =h(v) w=h(v) h(0),sokis determined byh. Turning this around, if we definet(v) =v+h(0) andk(v) =h(v) h(0), thentis a translation,kis an isometry fixing0, andh(v) =k(v)+h(0) =tw k,wherew=h(0). Theorem a functionh:Rn Rn, the following are equivalent:(1)his an isometry andh(0) =0,(2)hpreserves dot products:h(v) h(w) =v wfor allv,w link between length and dot product is the formula||v||2=v (1).
3 Then for any vectorsvandwinRn,( )||h(v) h(w)||=||v w||.As a special case, whenw=0in ( ) we get||h(v)||=||v||for allv Rn. Squaring bothsides of ( ) and writing the result in terms of dot products makes it(h(v) h(w)) (h(v) h(w)) = (v w) (v w).Carrying out the multiplication,( )h(v) h(v) 2h(v) h(w) +h(w) h(w) =v v 2v w+w first term on the left side of ( ) equals||h(v)||2=||v||2=v vand the last term onthe left side of ( ) equals||h(w)||2=||w||2=w w. Canceling equal terms on both sidesof ( ), we obtain 2h(v) h(w) = 2v w, soh(v) h(w) =v assumehsatisfies (2), so( )h(v) h(w) =v wfor allvandwinRn. Therefore||h(v) h(w)||2= (h(v) h(w)) (h(v) h(w))=h(v) h(v) 2h(v) h(w) +h(w) h(w)=v v 2v w+w wby ( )= (v w) (v w)=||v w||2,so||h(v) h(w)||=||v w||. Thushis an isometry. Settingv=w=0in ( ), we get||h(0)||2= 0, soh(0) =0. Corollary only isometry ofRnfixing0and the standard basis is the :Rn Rnbe an isometry that satisfiesh(0) =0, h(e1) =e1.
4 , h(en) = saysh(v) h(w) =v wfor allvandwinRn. Fixv Rnand letwrun over the standard basis vectorse1,e2,..,en,so we seeh(v) h(ei) =v eachei,h(v) ei=v + +cnen, we geth(v) ei=cifor alli, soh(v) =c1e1+ +cnen=v. Asvwas arbitrary,his the identity onRn. It is essential in Corollary that the isometry fixes0. An isometry ofRnfixing thestandard basiswithoutfixing0need not be the identity! For example, reflection across thelinex+y= 1 inR2is an isometry ofR2fixing (1,0) and (0,1) but not0= (0,0). a functionh:Rn Rn, the following are equivalent:(1)his an isometry andh(0) =0,(2)his linear, and the matrixAsuch thath(v) =Avfor allv RnsatisfiesAA>= an isometry andh(0) =0. We want to prove linearity:h(v+w) =h(v)+h(w) andh(cv) =ch(v) for allvandwinRnand allc R. The mappinghpreservesdot products by Theorem :h(v) h(w) =v wfor allvandwinRn. For the standard basise1,..,enofRnthis saysh(ei) h(ej) =ei ej= ij, soh(e1),..,h(en) is an orthonormal basis ofRn.
5 Thus two vectors inRnareequal if they have the same dot product with each ofh(e1),..,h(en).For alluinRnwe haveh(v+w) h(u) = (v+w) uand(h(v) +h(w)) h(u) =h(v) h(u) +h(w) h(u) =v u+w u= (v+w) u,4 KEITH CONRADsoh(v+w) h(u) = (h(v) +h(w)) h(u) for allu. Lettingu=e1,..,enshowsh(v+w) =h(v) +h(w). Similarly,h(cv) h(u) = (cv) u=c(v u) =c(h(v) h(u)) = (ch(v)) h(u),so again lettingurun throughe1,..,entells ush(cv) =ch(v). Thushis the matrix forh:h(v) =Avfor allv Rn, whereAhasjth columnh(ej). Wewant to showAA>=In. Sincehpreserves dot products, the conditionh(v) h(w) =v wfor allv,w RnsaysAv Aw=v w. The fundamental link between the dot product andmatrix transposes, which you should check, is that we can move a matrix to the other sideof a dot product by using its transpose:( )v Mw=M>v wfor everyn nmatrixMandv,w Rn. UsingM=AandAvin place ofvin ( ),Av Aw=A>(Av) w= (A>A)v is equal tov wfor allvandw, so (A>A)v w=v wfor allvandwinRn. Sincethe (i,j) entry of a matrixMisMej ei, lettingvandwrun through the standard basis ofRntells usA>A=In, soAis invertible.
6 An invertible matrix commutes with its inverse,soA>A=In AA>= the converse, assumeh(v) =Avforv RnwhereAA>=In. Triviallyhfixes0. Toshowhis an isometry, by Theorem it suffices to show( )Av Aw=v wfor allv,w Rn. SinceAand its inverseA>commute, we haveA>A=In, soAv Aw=A>(Av) w= (A>A)v w=v w. Corollary ofRnare invertible, the inverse of an isometry is an isometry,and two isometries onRnthat have the same values at0and any basis ofRnare gives a second proof of Corollary as a special :Rn Rnbe an isometry. By Theorem ,h=k+h(0) wherekis anisometry ofRnfixing0. Theorem tells us there is an invertible matrixAsuch thatk(v) =Avfor allv Rn, soh(v) =Av+h(0).This has inverseh 1(v) =A 1(v h(0)). In particular,his isometry condition||h(v) h(w)||=||v w||for allvandwinRnimplies||v w||=||h 1(v) h 1(w)||for allvandwinRnby replacingvandwin the isometry conditionwithh 1(v) andh 1(w). Thush 1is an isometry isometries ofRnthat are equal on0and a basis then the functionsk1(v) =h1(v) h1(0) andk2(v) =h2(v) h2(0) are linear and are equal on that basis, soby linearityk1=k2onRn.
7 That is,h1(v) h1(0) =h2(v) h2(0) for allvinRn. Sinceh1(0) =h2(0) we geth1=h2onRn. Remark isometries ofRnfixing0are linear and invertible is a special case ofthe following more general result: for a finite-dimensional vector spaceVover an arbitraryfield and a nondegenerate bilinear formBonV, a functionA:V Vfor whichB(v,w) =B(A(v),A(w)) for allvandwinVmust be linear and invertible. A more general versionof this is due to A. Vogt [3, Lemma , Theorem ], and a proof can be found thereor in my answer A physicallyISOMETRIES OFRn5interesting example of this overRbesidesRnwith its usual dot product is 4-dimensionalspace (x,y,z,ct) with the indefinite bilinear form associated tox2+y2+z2 c2t2in specialrelativity (Minkowski spacetime).Definition , a set ofn+ 1 pointsP0,P1,..,Pnis said to be ingeneral positionif they don t all lie in a concept abstracts the idea of 3 points inR2not being collinear. In the definition,the hyperplanes inRnare translated subspaces of dimensionn 1, so they need not passthrough the origin.
8 For example, a line inR2need not be a linear subspace ofR2since aline doesn t have to contain the origin. Three points inR2are in general position if no linepasses through all of them and four points inR3are in general position if no plane passesthrough all of them. SayingP0,P1,..,Pnare in general position inRndoes not mean thesen+ 1 points are linearly independent as vectors inRn, but rather that thendifferencesP1 P0,..,Pn P0are linearly independent vectors: a nontrivial linear relation wouldplace thesendifferences, along with0, in a common subspace of dimensionn 1, so addingP0to all of the differences and to0would putP0,P1,..,Pnin a common a common vector to points in general position keeps them in general positionsince the added vector cancels out when taking ,P1,..,Pnben+ 1points inRnin general position . Two isome-tries ofRnthat are equal atP0,..,Pnare the know isometries ofRnare invertible. Ifh1andh2are isometries ofRnwith thesame values at eachPithenh 12 h1is an isometry that fixes eachPi.
9 Therefore to proveh1=h2it suffices to show an isometry ofRnthat fixesP0,..,Pnis the an isometry ofRnsuch thath(Pi) =Pifor 0 i n. Sett(v) =v P0, whichis a translation. Thentht 1is an isometry with formula(tht 1)(v) =h(v+P0) (tht 1)(0) =h(P0) P0=0, sotht 1is linear by Theorem Also (tht 1)(Pi P0) =h(Pi) P0=Pi ,..,Pnare in general position, the differencesP1 P0,..,Pn P0form a basisofRn. Therefore by Corollary ,tht 1is the identity, sohis the identity. matricesWe have seen that the isometries ofRnthat fix0come from matricesAsuch thatAA>=In. These matrices have a nmatrixAis calledorthogonalifAA>=In, or equivalently ifA>A= matrix is orthogonal when its transpose is its inverse. Since det(A>) = detA, anorthogonal matrixAsatisfies (detA)2= 1, so detA= 1. (Forn 2 not all matrices withdeterminant 1 are orthogonal, such as (3 15 2). The orthogonal 1 1 matrices are 1.)Example onRn(Example ) is an isometry that is described by thematrix In, which is orthogonal: ( In)( In)>= ( In)( In) = 2.
10 By algebra,AA>=I2if and only ifA= (a bb a), wherea2+b2= 1 and = 1. Writinga= cos andb= sin , we get the matrices (cos sin sin cos ) and6 KEITH CONRAD(cos sin sin cos ). Algebraically, these types of matrices are distinguished by their determinants:the first type has determinant 1 and the second type has determinant geometric effects of these two types of matrices differ. Below on the left, (cos sin sin cos )is a counterclockwise rotation by angle around the origin. Below on the right, (cos sin sin cos )is a reflection across the line through the origin at angle /2 with respect to the positivex-axis. (Check (cos sin sin cos ) squares to the identity, as any reflection should.)vwA(v)A(w)A= (cos sin sin cos )vwA(v)A(w)A= (cos sin sin cos )Let s explain why (cos sin sin cos ) is a reflection at angle /2. See the figure below. Pick alineLthrough the origin, say at an angle with respect to the positivex-axis. To find aformula for reflection acrossL, we ll use a basis ofR2with one vectoronLand the othervectorperpendiculartoL.
