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-1 -1 "new century schoolboo k" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 3 336 1 {CSTYLE "" -1 -1 "new century schoolbook" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 3 337 1 {CSTYLE "" -1 -1 "new century schoolbook" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 338 1 {CSTYLE " " -1 -1 "new century schoolbook" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 339 1 {CSTYLE "" -1 -1 "new ce ntury schoolbook" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 258 "" 0 "" {TEXT 261 34 "Matrices as linear tran sformations" }}{PARA 258 "" 0 "" {TEXT 257 16 "by Jerry Kazdan\n" }} {PARA 260 "" 0 "" {TEXT -1 123 "This demonstration illustrates how mat rices can be viewed as (linear) transformations of vector spaces. As u sual, we begin:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{SECT 1 {PARA 263 "" 0 "" {TEXT -1 13 "Preliminaries" }}{EXCHG {PARA 261 "" 0 "" {TEXT -1 67 "Because the y are easiest to visualize, we will use 2-by-2 matrices:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "A:=matrix([[ 2,-1],[1,1]]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 " " 0 "" {TEXT -1 123 "Since A is a 2-by-2 matrix, we can multiply it by a 2-element vector and the result will be a 2-element vector, as foll ows:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "v:=vector([2,6]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalm(A&*v);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 258 323 "Both v and A*v may be viewed as column \+ matrices. In any case, the operation \"multiplication by A\" transform s 2-component vectors into other two-component vectors. If we think of these 2-component vectors as the coordinates of points in the 2-dimen sional plane, then multiplication by A is a mapping from the plane to \+ itself" }{TEXT -1 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {SECT 1 {PARA 264 "" 0 "" {TEXT -1 26 "Our map A and the letter F" }} {EXCHG {PARA 265 "" 0 "" {TEXT -1 116 "To understand the nature of the mapping of the plane defined by A, we consider what A does to a whole set of points:" }}{PARA 266 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 267 "> " 0 "" {MPLTEXT 1 0 73 "pts:=[[0,0],[1,0],[1,2],[2,2],[2, 3],[1,3],[1,4],[3,4],[3,5],[0,5],[0,0]];" }}}{EXCHG {PARA 268 "> " 0 " " {MPLTEXT 1 0 37 "plot(pts,x=-8..8,y=-8..8,style=LINE);" }}}{EXCHG {PARA 269 "" 0 "" {TEXT -1 113 "Because A is a linear map, the lines b etween the points get mapped to the lines between the images of the po ints." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 270 "" 0 "" {TEXT -1 341 "The following is the definition of a program called \"linmap\" -- the program takes a 2-by-2 matrix Q and applies it to all of the poin ts in the \"F\" given by the list \"pts\" above ... it then draws the \+ original \"F\", and also the shape to which the F gets mapped (you're \+ a real Maple aficionado if you can figure out what all the statements \+ do):" }}{PARA 271 "> " 0 "" {MPLTEXT 1 0 15 "linmap:=proc(Q)" }}{PARA 272 "> " 0 "" {MPLTEXT 1 0 15 "local B,cv,F,G;" }}{PARA 273 "> " 0 "" {MPLTEXT 1 0 31 "B:=(x,Q)->evalm(evalm(Q) &* x);" }}{PARA 274 "> " 0 " " {MPLTEXT 1 0 23 "cv:=x->convert(x,list);" }}{PARA 275 "> " 0 "" {MPLTEXT 1 0 35 "F:=plot(pts,style=LINE,color=blue):" }}{PARA 276 "> \+ " 0 "" {MPLTEXT 1 0 63 "G:=plot(map(cv,map(B,pts,Q)),style=LINE,color= red,thickness=3):" }}{PARA 277 "> " 0 "" {MPLTEXT 1 0 65 "plots[displa y](\{F,G\},view=[-11..11,-11..11],scaling=constrained);" }}{PARA 278 " > " 0 "" {MPLTEXT 1 0 5 "end;\n" }{XPPMATH 20 "6#>%'linmapG:6#%\"QG6&% \"BG%#cvG%\"FG%\"GG6\"F-C'>8$:6$%\"xGF'F-6$%)operatorG%&arrowGF--%&eva lmG6#-%#&*G6$-F86#9%9$F-F->8%:6#F3F-F4F--%(convertG6$F@%%listGF-F->8&- %%plotG6%%$ptsG/%&styleG%%LINEG/%&colorG%%blueG>8'-FL6%-%$mapG6$FB-FZ6 %F0FNF@FO/FS%$redG-&%&plotsG6#%(displayG6%<$FJFV/%%viewG7$;!#5\"#5Fdo/ %(scalingG%,constrainedGF-F-" }}}{EXCHG {PARA 279 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 280 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 14 "To illu strate:" }{MPLTEXT 1 0 12 "\nlinmap(A);\n" }}}{EXCHG {PARA 281 "" 0 " " {TEXT -1 57 "Thus A stretches a bit and rotates a bit. What does th e " }}{PARA 282 "" 0 "" {TEXT -1 16 "inverse of A do?" }}{PARA 283 "> \+ " 0 "" {MPLTEXT 1 0 19 "linmap(inverse(A));" }}}}{SECT 1 {PARA 284 "" 0 "" {TEXT -1 49 "Now we'll be more systematic. Simplest examples:" } }{EXCHG {PARA 285 "> " 0 "" {MPLTEXT 1 0 39 "ID:=matrix([[1,0],[0,1]]) ;\nlinmap(ID);\n" }}}{EXCHG {PARA 286 "> " 0 "" {MPLTEXT 1 0 30 "B:=2* ID;\nevalm(B);\nlinmap(B);\n" }}}{EXCHG {PARA 287 "" 0 "" {TEXT -1 110 "So B stretches by 2 in both directions. How does the area of the \+ transformed F compare with the original area?" }}{PARA 288 "> " 0 "" {MPLTEXT 1 0 8 "det(B);\n" }}}{EXCHG {PARA 289 "> " 0 "" {MPLTEXT 1 0 18 "C:=-ID;\nlinmap(C);" }}}}{SECT 1 {PARA 290 "" 0 "" {TEXT -1 25 "Mo re complicated examples" }}{EXCHG {PARA 291 "> " 0 "" {MPLTEXT 1 0 28 "S:= matrix([[1,0],[0,2]]); " }{TEXT -1 29 "stretch in vertical direc tion" }}{PARA 292 "> " 0 "" {MPLTEXT 1 0 11 "linmap(S);\n" }}}{EXCHG {PARA 293 "> " 0 "" {MPLTEXT 1 0 28 "R:= matrix([[-1,0],[0,1]]); " } {TEXT -1 28 "reflect across vertical axis" }}{PARA 294 "> " 0 "" {MPLTEXT 1 0 11 "linmap(R);\n" }}}{EXCHG {PARA 295 "> " 0 "" {MPLTEXT 1 0 28 "det(R); " }{TEXT -1 20 "what does det(R) < 0" }{MPLTEXT 1 0 1 " " }{TEXT -1 9 "tell us?\n" }}}{EXCHG {PARA 296 " > " 0 "" {MPLTEXT 1 0 27 "Ri:= inverse(R); " }{TEXT -1 29 "p redict what will happen to F" }{MPLTEXT 1 0 1 " " }}{PARA 297 "> " 0 " " {MPLTEXT 1 0 12 "linmap(Ri);\n" }}}{EXCHG {PARA 298 "> " 0 "" {MPLTEXT 1 0 26 "H:= evalm(R&*S); " }{TEXT -1 17 "stretch & r eflect" }}{PARA 299 "> " 0 "" {MPLTEXT 1 0 11 "linmap(H);\n" }}} {EXCHG {PARA 300 "> " 0 "" {MPLTEXT 1 0 28 "J:= matrix([[0,-1],[1,0]]) ; " }{TEXT -1 21 "rotate by +90 degrees" }}{PARA 301 "> " 0 "" {MPLTEXT 1 0 11 "linmap(J);\n" }}}{EXCHG {PARA 302 "> " 0 "" {MPLTEXT 1 0 25 "Jj:=J^2; " }{TEXT -1 29 "predict what will hap pen to F" }}{PARA 303 "> " 0 "" {MPLTEXT 1 0 25 "linmap(Jj); \+ " }{TEXT -1 41 "Note J^2 = - I so J is like sqrt(-1)." } {MPLTEXT 1 0 3 " \n" }}}{EXCHG {PARA 304 "" 0 "" {TEXT -1 36 "what ab out J^3 and J^4?\n " }}}{EXCHG {PARA 305 "> " 0 "" {MPLTEXT 1 0 24 "Ji:= inverse(J); " }{TEXT -1 28 "predict what \+ will happen to " }{TEXT 260 1 "F" }}{PARA 306 "> " 0 "" {MPLTEXT 1 0 12 "linmap(Ji);\n" }}}{EXCHG {PARA 307 "> " 0 "" {MPLTEXT 1 0 28 "K:= \+ matrix([[1,1],[0,1]]); " }{TEXT -1 5 "shear" }}{PARA 308 "> " 0 "" {MPLTEXT 1 0 11 "linmap(K);\n" }}}{EXCHG {PARA 309 "> " 0 "" {MPLTEXT 1 0 26 "L:= evalm(J&*K); " }{TEXT -1 16 "shear and rotate" } {MPLTEXT 1 0 2 " " }}{PARA 310 "> " 0 "" {MPLTEXT 1 0 11 "linmap(L); \n" }}}{EXCHG {PARA 311 "> " 0 "" {MPLTEXT 1 0 26 "M:= evalm(K&*J); \+ " }{TEXT -1 21 "rotate and then shear" }}{PARA 312 "> " 0 "" {MPLTEXT 1 0 2 " " }{TEXT -1 54 "Reversing the order matters! Matrix \+ multiplication is " }{TEXT 256 3 "not" }{TEXT -1 12 " commutative" } {MPLTEXT 1 0 23 " " }}{PARA 313 "> " 0 "" {MPLTEXT 1 0 10 "linmap(M);" }}}{EXCHG {PARA 314 "> " 0 "" {MPLTEXT 1 0 26 "N:= evalm(S&*J); " }{TEXT -1 23 "rotate and then stretc h" }}{PARA 315 "> " 0 "" {MPLTEXT 1 0 11 "linmap(N);\n" }}}{EXCHG {PARA 316 "> " 0 "" {MPLTEXT 1 0 27 "Q:= evalm(J&*S); " } {TEXT -1 23 "stretch and then rotate" }}{PARA 317 "> " 0 "" {MPLTEXT 1 0 11 "linmap(Q);\n" }}}{EXCHG {PARA 318 "> " 0 "" {MPLTEXT 1 0 28 "P := matrix([[1,0], [0,0]]); " }{TEXT -1 16 "projection onto " }{TEXT 259 1 "x" }{TEXT -1 5 " axis" }}{PARA 319 "> " 0 "" {MPLTEXT 1 0 28 "l inmap(P); " }{TEXT -1 23 "can you see the image?\n" } }{PARA 320 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 321 "> " 0 "" {MPLTEXT 1 0 26 "P2:= P^2; " }{TEXT -1 8 "predict!" } {MPLTEXT 1 0 26 "\nlinmap(P^2);\nevalm(P^2);\n" }{TEXT -1 5 "Thus " } {MPLTEXT 1 0 0 "" }{TEXT -1 72 " P^2=P. This is an example other than the obvious cases when P=0 or I. " }{MPLTEXT 1 0 1 "\n" }}}{EXCHG {PARA 322 "> " 0 "" {MPLTEXT 1 0 25 "T:= evalm(J&*P); " } {TEXT -1 23 "project and then rotate" }}{PARA 323 "> " 0 "" {MPLTEXT 1 0 11 "linmap(T);\n" }}}{EXCHG {PARA 324 "> " 0 "" {MPLTEXT 1 0 19 "T t:=evalm(T^2); " }{TEXT -1 35 "project and then rotate -- repeated " }}{PARA 325 "> " 0 "" {MPLTEXT 1 0 12 "linmap(Tt);\n" }{TEXT -1 80 " Thus T^2 = 0, even though T is not zero itself. This example may sur prise you." }{MPLTEXT 1 0 2 "\n\n" }}}}{SECT 1 {PARA 326 "" 0 "" {TEXT -1 50 "Find a linear transformation with given properties" }} {EXCHG {PARA 327 "" 0 "" {TEXT -1 1 " " }{TEXT 263 33 "Reflection acro ss the line x = y\n" }{TEXT -1 213 " We use the observation that t he first column in the matrix M representing this linear transformatio n is the image of the basis vector e1 = [1.0], while the second column is the image of the vector e2 = [0,1]." }}{PARA 328 "" 0 "" {TEXT -1 65 " It is clear that Me1 = e2 and Me2 = e1. Thus the matrix is " }}{PARA 329 "> " 0 "" {MPLTEXT 1 0 39 "R1:=matrix([[0,1],[1,0]]);\nl inmap(R1);\n" }}}{EXCHG {PARA 331 "" 0 "" {TEXT 262 20 "Rotation by an angle" }{TEXT 277 1 " " }{XPPEDIT 278 0 "theta" "I&thetaG6\"" }{TEXT -1 146 "\n By the above reasoning, to find the matrix we need only \+ see where the vectors [1,0] and [0,1] go. By elementary geometry [1,0 ] goes to [cos " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 6 ", s in " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 28 "] while [0,1] \+ goes to [-sin " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 6 ", co s " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 47 " ]. Thus the \+ matrix has these as its columns:" }}{PARA 330 "> " 0 "" {MPLTEXT 1 0 66 "Rot:= matrix([[cos(theta),-sin(theta)],[sin(theta),cos(theta)]]); \n" }}}{EXCHG {PARA 333 "" 0 "" {TEXT -1 109 "To use the above, the an gle theta needs to be specified in advance. It is more flexible to \+ use a function:" }}{PARA 332 "> " 0 "" {MPLTEXT 1 0 74 "R:= (theta)-> \+ matrix([[cos(theta),-sin(theta)],[sin(theta),cos(theta)]]);\n" }{TEXT -1 41 "For instance, a rotation by 45 degrees is" }{MPLTEXT 1 0 10 "\n R(Pi/4);\n" }{TEXT -1 12 "And we see:\n" }{MPLTEXT 1 0 17 "linmap(R(Pi /4));\n" }}}{EXCHG {PARA 338 "" 0 "" {TEXT -1 27 "It should be clear t hat R(" }{XPPEDIT 18 0 "theta + phi" ",&%&thetaG\"\"\"%$phiGF$" } {MPLTEXT 1 0 0 "" }{TEXT -1 20 ") is the same as R(" }{XPPEDIT 18 0 " theta" "I&thetaG6\"" }{TEXT -1 3 ")R(" }{XPPEDIT 18 0 "phi" "I$phiG6\" " }{TEXT -1 29 "), that is, first rotate by " }{XPPEDIT 18 0 "phi" "I $phiG6\"" }{TEXT -1 15 " and then by " }{XPPEDIT 18 0 "theta" "I&the taG6\"" }{TEXT -1 16 ". Lets check: \n" }{MPLTEXT 1 0 41 "R(theta + p hi);\nevalm(R(theta)&*R(phi));\n" }{TEXT -1 40 "This reproves the stan dard formulas for " }{XPPEDIT 18 0 "cos(theta+phi " "-%$cosG6#,&%&thet aG\"\"\"%$phiGF'" }{TEXT -1 6 " and " }{XPPEDIT 18 0 " sin(theta+phi " "-%$sinG6#,&%&thetaG\"\"\"%$phiGF'" }{TEXT -1 3 " .\n" }}}}{SECT 1 {PARA 336 "" 0 "" {TEXT -1 55 "A Three Dimensional Example: Rotate th e hyperboloid " }{XPPEDIT 18 0 "z = -x^2 + y^2" "/%\"zG,&*$%\"xG\"\" #!\"\"*$%\"yG\"\"#\"\"\"" }}{EXCHG {PARA 0 "" 0 "" {TEXT 266 75 "First we define a matrix rotating by 45 degrees around the z axis\n-- see R (" }{XPPEDIT 267 0 "theta" "I&thetaG6\"" }{TEXT 268 16 ") ) just above : " }{MPLTEXT 1 0 111 "\nR:=(theta)-> matrix([[cos(theta),-sin(theta), 0], [sin(theta),cos(theta),0],[0,0,1]]);\nR(theta);\nRR:=R(Pi/4); " } {TEXT 271 36 "Rotation by Pi/4 around the z axis" }{MPLTEXT 1 0 15 " \n ID:= R(0); " }{TEXT 272 35 "Rotation by 0, the identity matrix." }{MPLTEXT 1 0 1 "\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 269 19 "Define ou r surface " }{XPPEDIT 270 0 "z=-x^2 + y^2" "/%\"zG,&*$%\"xG\"\"#!\"\"* $%\"yG\"\"#\"\"\"" }{TEXT -1 1 "\n" }{MPLTEXT 1 0 57 "x:= (s,t) -> s; \ny:= (s,t) -> t;\nz:= (s,t) -> -s^2 + t^2;\n" }{TEXT 264 18 "and make a vector " }{TEXT 281 1 "W" }{TEXT 282 26 " with these as components: " }{MPLTEXT 1 0 56 "\nW:=(s,t)->vector([x(s,t),y(s,t),z(s,t)]);\nwith( plots):\n" }}}{EXCHG {PARA 334 "> " 0 "" {MPLTEXT 1 0 123 "plot3d(W(s, t),s=-1.5..1.5, t=-1.5..1.5,view=-1.5..1.5, orientation=[30,80],grid=[ 25,25], style=patch,scaling=constrained);\n" }}}{EXCHG {PARA 339 "" 0 "" {TEXT -1 29 "This procedure just replaces " }{TEXT 279 1 "W" } {TEXT -1 14 " in the above " }{TEXT 295 6 "plot3d" }{TEXT -1 3 " by" } {TEXT 280 1 " " }{TEXT 284 1 "Q" }{TEXT 285 1 "W" }{TEXT -1 53 " to g et the hyperboloid after it has been moved by" }{TEXT 283 2 " Q" } {TEXT -1 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 149 "Hyperb:=proc(Q)\n plot3d(evalm(evalm(Q) &* W(s,t)),s=-1.5..1.5, t=-1.5..1.5, \norientati on=[30,80],grid=[25,25], style=patch,scaling=constrained);\nend;\n" }} }{EXCHG {PARA 335 "> " 0 "" {MPLTEXT 1 0 24 "Hyperb(ID);\nHyperb(RR); \n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 265 92 "Instead of just continuing like this, lets animate a sequence of 4 rotations by 45 degrees: " } {MPLTEXT 1 0 142 "\nH0:= Hyperb(ID): \nH1:= Hyperb(RR):\nH2:= Hyperb(R R^2):\nH3:= Hyperb(RR^3):\nH4:= Hyperb(RR^4):\n\ndisplay([H0, H1, H2, \+ H3, H4], insequence=true);\n" }}}}{SECT 1 {PARA 337 "" 0 "" {TEXT -1 51 "Another Three Dimensional Example: Rotate a torus" }}{EXCHG {PARA 0 "" 0 "" {TEXT 273 21 "We define the torus:\n" }{MPLTEXT 1 0 85 "x:= (s,t) -> sin(s);\ny:= (s,t) -> (3+cos(s))*cos(t);\nz:= (s,t) - > (3+cos(s))*sin(t);\n" }{TEXT 274 65 "and make a vector T defining ou r torus with these as components:\n" }{MPLTEXT 1 0 132 "T:=(s,t)->vect or([x(s,t),y(s,t),z(s,t)]);\nplot3d(T(s,t),s=0..2*Pi, t=0..2*Pi, scali ng=constrained,orientation=[34,86],style=patch);\n" }{TEXT 275 29 "Thi s procedure just replaces " }{TEXT 286 1 "T" }{TEXT 287 14 " in the ab ove " }{TEXT 293 6 "plot3d" }{TEXT 294 4 " by " }{TEXT 288 1 "Q" } {TEXT 290 1 "T" }{TEXT 289 48 " to get the torus after it has been \+ moved by " }{TEXT 291 1 "Q" }{TEXT 292 2 ".\n" }{MPLTEXT 1 0 131 "Toru s:=proc(Q)\nplot3d(evalm(evalm(Q) &* T(s,t)), s=0..2*Pi, t=0..2*Pi, sc aling=constrained,orientation=[34,86], style=patch);\nend;\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "Torus(RR);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Torus(RR^2);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 276 58 "Again we animate a sequence of 4 rotations by 45 degrees." } {TEXT -1 1 "\n" }{MPLTEXT 1 0 86 "T0:= Torus(ID): \nT1:= Torus(RR):\nT 2:= Torus(RR^2):\nT3:= Torus(RR^3):\nT4:= Torus(RR^4):" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 48 "display([T0, T1, T2, T3, T4], insequence=true) ;\n" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{MARK "0 1 0" 15 }{VIEWOPTS 1 1 0 1 1 1803 }