really ultimately take the view that this geometric Liveartrausfomatiousthesaelofliuearalgebrwe doing function that of what A is will
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1 vecr However 14 Feb wht is mtrix wnt more geometric picture doing product I I in Livertrusfomtiousselliuerlgebrwe we will ultimtely tke view tht this geometric ction Is wht mtrices re bout relly Let's consider 2 2 cse first we cn think s function tht lkes vecr IERZ nd returns Foti nor vecr IR? we write tee R2 We cn't grph this becuse tht would require ll 224 sptil dimensions we cn mke movie where ech vecr ( which we 'll think s
2 just point its skirts destintion i Picture t loction its I nd moves ( 0 1 y o Flt Its # ' Solution Lets consider few points me fiefdom 9 ][8] [ [ minsty Fct p ] Coted ] loin lid fit 183 Exmple µwgggjy mps tcxey function just for fun \Netuedul'motii # ustueeudin poiutu only ctully tells rottion he refer We like looks it so gives us find liner is Proved mgpwtge following s orem Intuition liner for e?fe0fdw R2 R2 FFE # isbijedive Imi function s 2 Let heorem D regrded geometric some trnsformtions not trnsformtion liner in your re is if some be mening 2 2 mtrix ' so Furr lines function n 5 for tht lines nd iij for ll I I 8 ll 8 ' # '
![Exmples & Nonexruples mtrix domin > rnge rottion [ YD scle food sher fj3] projection [122] quiches down ( soueline rection](/docs-images/95/124421812/images/3-0.jpg "reflecting ( scle direction inx nonliner squring complex ( ifyoitw#rested [209] none nonliner conformlmdp none ify")
3 Exmples & Nonexruples mtrix domin > rnge rottion [ YD scle food sher fj3] projection [122] quiches down ( soueline rection reflecting ( scle direction inx nonliner squring complex ( ifyoitw#rested [209] none nonliner conformlmdp none ify onyheskd
4 Find sme stuff pplies mxn mtrices but is bit trickier visulize n mxn mtrix tkes vecr I e R nd returns vecr t e Rm liner mp from so it represents Rntjjtmifgmngd this mupnud o Let be liner trnsformtion from R2 R2 which mps [ f ] [ F ll x nd [ ] [ ng ] Exerts for y x 8] # solution webuow (xd xltftyff ] E±xI6DEtd# Since is liner n (x[ 8 y[9h xfst y[ Yd
5 stndrd bsis vecrs t gllirds 2it3j Visully knowing tht else ge ujht nd j [ 9 J tells us where mps everything mpstfdu Xtdfsoelbd rte * E Furr vecrs tht [ 83 nd G mp cn be red f y 're its columns! Exercise Locte tlw solutionwuq wwwvg#tgpsoteitutztw*ydf& 7
6 liner 1 So liner Exercise Show tht lists np liner linerly dependent linerly dependent lists under trnsformtions solution Suppose Cn not ll zero wke C V t e Cntn D tking both sides ( t cutu IE cexercise 9H cut (e so { CI ( Eu } Is linerly dependent µ chtuug y mxtb is in sense term we're using here unless b 0 ; if b FO we cll it ffine lgebr ffine u#liner Linttg lgt with 0inlpt
7 JUS FON FOR ppetite & ( line for Pro so If L CM nd intersected L in nd R2 R2 nd gontinuous CL tx Is bijectvie L line K cu C 8 n re for lines exist ll I c R? n distinct point y ( intersection for ssume e if mp Hei is becuse let's if 5 point sme 8 o would t prllel n prllel re re nd 1122 in M nd M nd tht tht now pwegfdstwmpilyoituftfippmed exminble every B nd LG If heory non ftpifeng f L ( [ h ] n on lines nd we continue nd xe C 4 ] refore similrly hs lie y # [ 13 in reson this gend HYE cn tee tht wy ( [ I ] fill out ft ]
8 But whole Integer grid e e is we know sends ech nd FI ; t se this mens tht lso fixes points itself se o! ny intersection point 8 two line segments with endpoints on grid this includes ll points with rtionl coordintes Heft s not tricky exercise ] Since ext I for ll E with rtionl coordintes nd is continuous t I for ll E o hndle generl D cse Define tent I nd he 2 C 8 Suppose mp
9 b b S (e I 5 where columns re tee nds ( so s sends 88 e sme plces does 51 z n Is Ez G o 5 ' fixes 8 e Ez We E lines lines imdmps hve lredy shown tht tins implies o 5 ' is identity mp tht is S #
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