{VERSION 5 0 "SUN SPARC SOLARIS" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 286 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 291 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 292 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 296 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 297 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 300 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 301 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 302 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 303 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 304 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 305 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 306 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 307 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 308 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 309 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 310 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 311 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 312 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 313 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 314 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 315 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 316 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 317 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 318 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 114 " Math 2270\nMaple Projec t 2\nOctober 14, 20002. \n\nA Maple text version of this project may b e found at the web site" }}{PARA 0 "" 0 "" {TEXT -1 51 "http://www.mat h.utah.edu/~kapovich/teaching11.html/" }}{PARA 0 "" 0 "" {TEXT -1 59 " Go to that page and click on the \"2-nd maple assignment\". " }} {PARA 0 "" 0 "" {TEXT -1 78 "I recommend saving the file from the webs ite onto you home directory, and then" }}{PARA 0 "" 0 "" {TEXT -1 69 " opening it from Maple, as a ``Maple text'' document. You can then do " }}{PARA 0 "" 0 "" {TEXT -1 69 "the assignment by inserting the prope r commands as well as italicized" }}{PARA 0 "" 0 "" {TEXT -1 65 "textu al comments in answer to the various questions. In order to" }}{PARA 0 "" 0 "" {TEXT -1 69 "execute multiline commands from the version you get from the web, you" }}{PARA 0 "" 0 "" {TEXT -1 63 "should use the \+ ``Edit'' option in your maple window to ``join''" }}{PARA 0 "" 0 "" {TEXT -1 62 "execution groups which you have highlighted with your mou se. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 " I remind that in each problem you should use Maple " }}{PARA 0 "" 0 " " {TEXT -1 54 "to solve the problem, not pen-and-paper computations! \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 270 8 "PART \+ 1. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "I n the first part of the assignment we will study the geometric meaning of linear (hence matrix) maps" }}{PARA 0 "" 0 "" {TEXT -1 105 "f(x)=A x, from R^n to R^m. We will focus on maps from R^2 to R^2 to illustra te more general properties. " }}{PARA 0 "" 0 "" {TEXT -1 76 "Of cours e, computer graphics are most concerned with those maps and with R^3" }}{PARA 0 "" 0 "" {TEXT -1 52 "rotation maps and projection maps from \+ R^3 to R^2. " }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 67 " The very definition of a linear map, namely that f(u+v)= f(u)+f(v) " }}{PARA 0 "" 0 "" {TEXT -1 70 " and f(su)=sf(u), for all v ectors u,v and scalars s, has the following" }}{PARA 0 "" 0 "" {TEXT -1 25 " geometric consequences:\n" }}{PARA 0 "" 0 "" {TEXT -1 60 " Lin es are mapped to lines, and parallel lines are mapped to" }}{PARA 0 " " 0 "" {TEXT -1 66 "parallel lines: This is because a line can be des cribed as a set " }}{PARA 0 "" 0 "" {TEXT -1 74 "L=\{u + t v , where \+ t is in R\}, where u is a point on the line and v is " }}{PARA 0 "" 0 "" {TEXT -1 77 "a direction vector. Therefore the image set f(L): =\{f(u+tv)\}=\{f(u)+tf(v)\} " }}{PARA 0 "" 0 "" {TEXT -1 75 "is also \+ a line, going through f(u) with direction f(v). (If the directions" } }{PARA 0 "" 0 "" {TEXT -1 66 "f(v) turns out to be 0, then the line de generates into a point.) \n" }}{PARA 0 "" 0 "" {TEXT -1 67 "Therefore, line segments are mapped to line segments, polygons are" }}{PARA 0 " " 0 "" {TEXT -1 65 "mapped to polygons, and regions bounded by polygon s are mapped to" }}{PARA 0 "" 0 "" {TEXT -1 70 "regions bounded by pol ygons; if we know where the vertices go, we know" }}{PARA 0 "" 0 "" {TEXT -1 61 "everything. Let's use these facts to draw the images of \+ some" }}{PARA 0 "" 0 "" {TEXT -1 38 "polygonal regions under a matrix \+ map.\n" }}{PARA 0 "" 0 "" {TEXT -1 57 " First load the linear algeb ra and plotting libraries\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "with (plots):with(linalg):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 " verts0:=[[0,0],[1,0],[1,1],[0,1]]; \n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'verts0G7&7$\"\"!F'7$\"\"\"F'7$F)F)7$F'F)" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 30 " #corners of unit square\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "unitsquare:=polygonplot(verts0, col or=`yellow`):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 " #t his command make a polygon and colors\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " #the region inside it yellow. Make sure\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " #to end this command w ith a colon!\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "display(u nitsquare); #Now semicolon! this command\n" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6#-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\" \"\"F)F(7$F+F+7$F(F+-%'COLOURG6&%$RGBG$\"*++++\"!\")F3F(" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 " #shows the square\n" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "When you executed the sequence of \+ commands above you should have" }}{PARA 0 "" 0 "" {TEXT -1 67 "gotten \+ a picture of a yellow unit square. Now we will use a linear" }}{PARA 0 "" 0 "" {TEXT -1 70 "map with matrix A defined below to map this squ are to a parallelegram\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "A:=matr ix([[3,2],[-1,1]]); #the matrix of our\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7$7$\"\"$\"\"#7$!\"\"\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 " #random linear transf ormation\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "f:=x->evalm(A &*x); #our linear map\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6# %\"xG6\"6$%)operatorG%&arrowGF(-%&evalmG6#-%#&*G6$%\"AG9$F(F(F(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 273 10 "Problem 1." }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "1a) " } {TEXT 283 11 "Assignment:" }{TEXT -1 63 " For our map f(x)=Ax defined above, what are the images of the" }}{PARA 0 "" 0 "" {TEXT -1 68 "poi nts e1=[1,0] and e2=[0,1]? How do you find these images from the" }} {PARA 0 "" 0 "" {TEXT -1 22 "rows or columns of A?\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 293 11 "Solution. " }{TEXT -1 32 "To find the images \+ lets compute" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f(<1,0>);f(<0,1>);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7$\"\"$!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'v ectorG6#7$\"\"#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "These ar e nothing but the columns of the matrix A. " }}{PARA 0 "" 0 "" {TEXT -1 43 "-------------------------------------------" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "We use the ``map'' command below to see where the vertices of the unit" }}{PARA 0 "" 0 "" {TEXT -1 70 "square are sent \+ by f. The syntax is to put the mapping function in as" }}{PARA 0 "" 0 "" {TEXT -1 62 "the first argument, and the list of input points as \+ the second" }}{PARA 0 "" 0 "" {TEXT -1 63 "argument. The result of the command will be the list of output" }}{PARA 0 "" 0 "" {TEXT -1 66 "p oints. Check (not to hand in) that this is what happens below to" }} {PARA 0 "" 0 "" {TEXT -1 37 "the four points in the list verts0. \n" } }{PARA 0 "" 0 "" {TEXT -1 70 "Note that here we are conflating points \+ and vectors in R^2, namely we " }}{PARA 0 "" 0 "" {TEXT -1 66 "identif y each point P with its coordinates, i.e. with the vector" }}{PARA 0 "" 0 "" {TEXT -1 5 " -->" }}{PARA 0 "" 0 "" {TEXT -1 5 " OP\n" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "verts1:=map(f,verts0);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'ver ts1G7&-%'vectorG6#7$\"\"!F*-F'6#7$\"\"$!\"\"-F'6#7$\"\"&F*-F'6#7$\"\"# \"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "image1:=polygonpl ot(verts1, color=`red`):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display(\{unitsquare,image1\});\n" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\"\"$F)$!\" \"F)7$$\"\"&F)F(7$$\"\"#F)$\"\"\"F)-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F (-F$6$7&F'7$F5F(7$F5F57$F(F5-F86&F:F;F;F(" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "1b) " }{TEXT 282 11 "Assignment:" }{TEXT -1 66 " Explain where f(e1) and f(e2) are represented in the picture you \+ " }}{PARA 0 "" 0 "" {TEXT -1 49 "just made. (Mark them by hand on the picture.) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "By the way, if you click on the plot you can choose to" }} {PARA 0 "" 0 "" {TEXT -1 70 "have the projection ``constrained '', in \+ which case Maple will use the" }}{PARA 0 "" 0 "" {TEXT -1 70 "same sca les for the vertical and horizontal axis. Otherwise it scales" }} {PARA 0 "" 0 "" {TEXT -1 66 "the x and y-axes to make the picture fit \+ nicely onto your screen.\n" }}{PARA 0 "" 0 "" {TEXT 294 9 "Solution." }{TEXT -1 117 " f(e1) and f(e2) are represented by vectors which go a long the sides of the red parallelogram emanating from (0,0). " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "1c ) " }{TEXT 281 11 "Assignment:" }{TEXT -1 64 " If we compute f(f(x)): =f^2(x), then what will the matrix be for" }}{PARA 0 "" 0 "" {TEXT -1 67 "the resulting linear transformation? After answering that question ," }}{PARA 0 "" 0 "" {TEXT -1 64 "make the vertices for f(f(unitsquare )) as follows, and draw the " }}{PARA 0 "" 0 "" {TEXT -1 30 "correspon ding image polygons.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 295 8 "Solutio n" }{TEXT -1 122 ". The mapping f^2 is the composition of the mapping \+ f with itself, matrix of the composition is the product of matrices, \+ " }}{PARA 0 "" 0 "" {TEXT -1 30 "therefore the matrix of f^2 is" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalm(A&*A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7$7$\"\"(\"\")7$!\"%!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "verts2:=map(f,verts1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'verts2G7&-%'vectorG6#7$\"\"!F*-F'6#7$\"\"(! \"%-F'6#7$\"#:!\"&-F'6#7$\"\")!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "image2:=polygonplot(verts2,`color`=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "display(\{unitsquare,image1,image2 \});" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%)POLYG ONSG6$7&7$$\"\"!F)F(7$$\"\"(F)$!\"%F)7$$\"#:F)$!\"&F)7$$\"\")F)$!\"\"F )-%'COLOURG6&%$RGBGF(F($\"*++++\"!\")-F$6$7&F'7$$\"\"$F)F77$$\"\"&F)F( 7$$\"\"#F)$\"\"\"F)-F:6&F " 0 "" {MPLTEXT 1 0 17 "Ain:= inverse(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$AinG-%'m atrixG6#7$7$#\"\"\"\"\"&#!\"#F,7$F*#\"\"$F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g:=x->evalm(Ain&*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGf*6#%\"xG6\"6$%)operatorG%&arrowGF(-%&evalmG6#-%#&*G6$%$Ai nG9$F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "verts:=map(g, verts1);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&vertsG7&-%'vectorG6#7 $\"\"!F*-F'6#7$\"\"\"F*-F'6#7$F.F.-F'6#7$F*F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "image:=polygonplot(verts,`color`=green):display( image);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6#-%)PO LYGONSG6$7&7$$\"\"!F)F(7$$\"\"\"F)F(7$F+F+7$F(F+-%'COLOURG6&%$RGBGF($ \"*++++\"!\")F(" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Thus we indeed g ot the unit square back. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 272 10 "Problem 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 " Translations of objects are mappe d to translations of the image " }}{PARA 0 "" 0 "" {TEXT -1 67 "object s and scalings of objects are mapped to scalings of the image" }} {PARA 0 "" 0 "" {TEXT -1 11 "objects: \n" }}{PARA 0 "" 0 "" {TEXT -1 60 " First let's review what do we mean by an ``object'' and" }} {PARA 0 "" 0 "" {TEXT -1 71 "by translations and scalings of an object . An object is some set S " }}{PARA 0 "" 0 "" {TEXT -1 70 "of poin ts s. By the image of S we mean the collection of image " }} {PARA 0 "" 0 "" {TEXT -1 75 "points f(s) . We write f(S) for thi s image (like we did for the line " }}{PARA 0 "" 0 "" {TEXT -1 75 "L a nd its image f(L) in problem 1). Similarly if b is a translati on" }}{PARA 0 "" 0 "" {TEXT -1 73 "vector, then the object S+b means all points of the form s+b where s" }}{PARA 0 "" 0 "" {TEXT -1 74 "is in S. If c is a scalar, then the scaled (or dilated) set cS \+ means" }}{PARA 0 "" 0 "" {TEXT -1 49 "all points of the form cs , whe re s is in S.\n" }}{PARA 0 "" 0 "" {TEXT -1 69 " Now, if we appl y the linear map f to the translated object S+b we" }}{PARA 0 "" 0 "" {TEXT -1 80 "get all points of the form f(s+b)=f(s)+ f(b), where s i s in S (since f is " }}{PARA 0 "" 0 "" {TEXT -1 71 "linear), i.e. \+ exactly the set f(S)+f(b) , which is the translation of " }}{PARA 0 " " 0 "" {TEXT -1 73 "the image f(S) by the vector f(b). Similarly, \+ if we apply f to the " }}{PARA 0 "" 0 "" {TEXT -1 70 "scaled set cS \+ we get f(cS) to be the set of all points f(cs)=cf(s) " }}{PARA 0 " " 0 "" {TEXT -1 69 "(since f is linear), i.e. the scaling cf(S) of th e image set f(S).\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 " Here 's how to translate and scale the unit square from #1: " }}{PARA 0 " " 0 "" {TEXT -1 42 "Let's translate it by the vector b=[2,3]:\n" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "b:=[2,3];\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG7$\"\"#\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "trans:=x->evalm(x+b);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&transGf*6#%\"xG6\"6$%)operatorG%&arrowGF(-%&evalmG6#,&9$\"\" \"%\"bGF1F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "verts3:= map(trans,verts0);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'verts3G7&-% 'vectorG6#7$\"\"#\"\"$-F'6#7$F+F+-F'6#7$F+\"\"%-F'6#7$F*F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "image3:=polygonplot(verts3,color=`y ellow`):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "display(units quare,image3);\n" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\"\"\"F)F(7$F+F+7$F(F+-%'COLOURG6& %$RGBG$\"*++++\"!\")F3F(-F$6$7&7$$\"\"#F)$\"\"$F)7$F " 0 "" {MPLTEXT 1 0 8 "c:=0. 2;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "shrink:=x->evalm(c* x);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'shrinkGf*6#%\"xG6\"6$%)ope ratorG%&arrowGF(-%&evalmG6#*&%\"cG\"\"\"9$F1F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "verts4:=map(shrink,verts0);\n" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%'verts4G7&-%'vectorG6#7$$\"\"!F+F*-F'6#7$$\"\" #!\"\"F*-F'6#7$F/F/-F'6#7$F*F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "image4:=polygonplot(verts4,color=`red`):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display(\{image4,unitsquare\});\n" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLYGONSG6$7&7$$\" \"!F)F(7$$\"\"#!\"\"F(7$F+F+7$F(F+-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F( -F$6$7&F'7$$\"\"\"F)F(7$F;F;7$F(F;-F16&F3F4F4F(" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 274 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 5 "2a) " }{TEXT 277 11 "Assignment:" }{TEXT -1 61 " Describe where you expect the translated unit square, image3" }}{PARA 0 "" 0 "" {TEXT -1 69 "above, to be mapped by our linear map f from problem 1. Then use" }}{PARA 0 "" 0 "" {TEXT -1 69 "the ``m ap'' command and the ``trans'' command, as well as polygonplot" }} {PARA 0 "" 0 "" {TEXT -1 67 "and display, to make a picture of the ima ges of the unit square and" }}{PARA 0 "" 0 "" {TEXT -1 68 "its transla tion when they are mapped by f . Verify that the images" }}{PARA 0 " " 0 "" {TEXT -1 36 "differ by the expected translation.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 305 8 "Solution" }{TEXT -1 88 ". image3= unitsqu are + b, hence f(image3)= f(unitsquare + b)= image1+ f(b)= image1+ Ab ." }}{PARA 0 "" 0 "" {TEXT -1 105 "Thus we should expect f( image3) \+ to be obtained from image1 by parallel translation via the vector Ab. \+ " }}{PARA 0 "" 0 "" {TEXT -1 28 "Let's compute the vector Ab:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u:=evalm(A&*b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"uG-%'vectorG6#7$\"#7\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "Thus we should expect image1 and f(image 3) to differ by translation by the vector (12,1). Let's check it using MAPLE graphics:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "verts41 := map(f,verts3):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "image4 1:=polygonplot(verts41,color=`red`):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "display(\{image1, image41\});" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLYGONSG6$7&7$$\"#7\"\"!$\"\" \"F*7$$\"#:F*$F*F*7$$\"# " 0 "" {MPLTEXT 1 0 24 "transu:=x-> evalm(x+u); " } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'transuGf*6#%\"xG6\"6$%)operatorG%& arrowGF(-%&evalmG6#,&9$\"\"\"%\"uGF1F(F(F(" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "verts42:= map(transu, verts1):" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 45 "image42:=polygonplot(verts42,color=`green`): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "display([image1, image 42, image41]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 " 6%-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\"\"$F)$!\"\"F)7$$\"\"&F)F(7$$\"\"#F) $\"\"\"F)-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F(-F$6$7&7$$\"#7F)F57$$\"#: F)F(7$$\"# " 0 "" {MPLTEXT 1 0 23 "verts5:= map(f,verts4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'verts5G7&-%'vectorG6#7$$\"\"!F+F*-F'6#7$$\"\"'!\"\"$ !\"#F1-F'6#7$$\"#5F1F*-F'6#7$$\"\"%F1$\"\"#F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "image5:=polygonplot(verts5,color=`green`): " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "display(\{image1,image5\}); " }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLYGONSG 6$7&7$$\"\"!F)F(7$$\"\"'!\"\"$!\"#F-7$$\"#5F-F(7$$\"\"%F-$\"\"#F--%'CO LOURG6&%$RGBGF($\"*++++\"!\")F(-F$6$7&F'7$$\"\"$F)$F-F)7$$\"\"&F)F(7$$ F7F)$\"\"\"F)-F96&F;F " 0 "" {MPLTEXT 1 0 72 "Rot:=theta->matrix([[cos(theta),-si n(theta)],[sin(theta),cos(theta)]]);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$RotGf*6#%&thetaG6\"6$%)operatorG%&arrowGF(-%'matrixG6#7$7$-%$ cosG6#9$,$-%$sinGF3!\"\"7$F6F1F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "so to rotate the vector (2,3) by Pi/3, we would command\n" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "theta:=Pi/3;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG,$%#PiG#\"\"\"\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "evalm(Rot(theta)&*[2,3]);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7$,&\"\"\"F(*&#\"\"$\"\"#F(*$-%%sqrtG6#F+F( F(!\"\",&F-F(#F+F,F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 275 18 "Assignme nt for 3a)" }{TEXT 302 1 ":" }{TEXT -1 60 " Draw a picture of the unit square rotated by Pi/3 radians.\n" }}{PARA 0 "" 0 "" {TEXT 301 9 "Sol ution." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "h:= x-> evalm(Rot (theta)&*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hGf*6#%\"xG6\"6$%) operatorG%&arrowGF(-%&evalmG6#-%#&*G6$-%$RotG6#%&thetaG9$F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "verts6:=map(h,verts0):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "image6:=polygonplot(verts6,c olor=`green`): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display( [image6, unitsquare]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\"+++++]!#5$\"+SSDg')F-7 $$!+SSDgOF-$\"+/a-m8!\"*7$$!+SSDg')F-F+-%'COLOURG6&%$RGBGF($\"*++++\"! \")F(-F$6$7&F'7$$\"\"\"F)F(7$FDFD7$F(FD-F:6&F y and M: y--> z are linear transformations " }}{PARA 0 "" 0 "" {TEXT -1 70 "and N: x--> y=L(x) --> z= M(y) is their composition, \+ and the matrix " }}{PARA 0 "" 0 "" {TEXT -1 74 "of L is A, the matr ix for M is B, then the matrix for N is A&*B. \n" }}{PARA 0 "" 0 "" {TEXT 298 11 "Solution. " }{TEXT -1 50 "The matrix of reflection in the x-axis is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "R1:= matrix([[1,0],[0,-1]]);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R1G-%'matrixG6#7$7$\"\"\"\"\"!7$F+! \"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "because the image of (1, 0) is (1,0) and the image of (0,1) is (0, -1). The matrix of reflectio n across the x=y line is " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "R2:= matrix([[0,1],[1,0]]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%# R2G-%'matrixG6#7$7$\"\"!\"\"\"7$F+F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 248 "In general (see page 60 of the textbook) the reflection of the vecrtor v across the line through the unit vector u is given by the f ormula v-> 2(u.v) u-v. Thus to find reflection across the line y=-2x \+ we first have to find a vector w on this line:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 19 "w:= vector([1,-2]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG-%'vectorG6#7$\"\"\"!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "u:=evalm(w/(sqrt(dotprod(w,w)))); # unit vector \+ along y=2x line" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"uG-%'vectorG6#7 $,$*$-%%sqrtG6#\"\"&\"\"\"#F/F.,$F*#!\"#F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Now, let's compute the images of e1 and e2 under the refl ection R3 in the line y=2x:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "e1:= vector([1,0]); e2:= vector([0,1]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#e1G-%'vectorG6#7$\"\"\"\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#e2G-%'vectorG6#7$\"\"!\"\"\"" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 77 "j1:= evalm(2 * multiply(u,e1)*u - e1); j2:= ev alm(2 * multiply(u,e2)*u - e2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%# j1G-%'vectorG6#7$#!\"$\"\"&#!\"%F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#j2G-%'vectorG6#7$#!\"%\"\"&#\"\"$F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Thus the matrix of reflection across the line y=-2x is g iven by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "R3:= transpose( augment(j1, j2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R3G-%'matrixG6 #7$7$#!\"$\"\"&#!\"%F,7$F-#\"\"$F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "Now, let's define the transformations (reflections) correspondi ng to the matrices R1, R2, R3:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "r1:= x->evalm(R1&*x);r2:= x->evalm(R2&*x);r3:= x->evalm(R3&*x) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r1Gf*6#%\"xG6\"6$%)operatorG%& arrowGF(-%&evalmG6#-%#&*G6$%#R1G9$F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r2Gf*6#%\"xG6\"6$%)operatorG%&arrowGF(-%&evalmG6#-%#&*G6$%#R2 G9$F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r3Gf*6#%\"xG6\"6$%)ope ratorG%&arrowGF(-%&evalmG6#-%#&*G6$%#R3G9$F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "vertsr1:= map(r1, verts0); vertsr2:= map(r2, \+ verts0);vertsr3:= map(r3, verts0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%(vertsr1G7&-%'vectorG6#7$\"\"!F*-F'6#7$\"\"\"F*-F'6#7$F.!\"\"-F'6#7$ F*F2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(vertsr2G7&-%'vectorG6#7$\" \"!F*-F'6#7$F*\"\"\"-F'6#7$F.F.-F'6#7$F.F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(vertsr3G7&-%'vectorG6#7$\"\"!F*-F'6#7$#!\"$\"\"&#!\" %F0-F'6#7$#!\"(F0#!\"\"F0-F'6#7$F1#\"\"$F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "imager1:=po lygonplot(vertsr1,color=`green`): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "imager2:=polygonplot(vertsr2,color=`blue`): " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "imager3:=polygonplot(vertsr3 ,color=`red`): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "display( \{imager1,imager2,imager3\} );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\"\"\"F)F(7$F+$!\"\" F)7$F(F.-%'COLOURG6&%$RGBGF($\"*++++\"!\")F(-F$6$7&F'7$F(F+7$F+F+F*-F2 6&F4F(F(F5-F$6$7&F'7$$!+++++g!#5$!+++++!)FE7$$!+++++9!\"*$!+++++?FE7$F F$\"+++++gFE-F26&F4F5F(F(" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 59 "Finally, let's verify the statement about the compositi on. " }}{PARA 0 "" 0 "" {TEXT -1 52 "The composition is given by the p roduct of matrices:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "t:=Pi /4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"tG,$%#PiG#\"\"\"\"\"%" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "evalm((Rot(t)&* R1)&* Rot(-t ));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7$7$\"\"!\"\"\"7$F) F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "This is indeed the matrix o f reflection R2. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 67 "3c) You can scale by different factors in the x a nd y-directions. " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 284 11 "Ass ignment:" }{TEXT -1 55 " define a matrix expands the x-direction by 3 and the " }}{PARA 0 "" 0 "" {TEXT -1 99 "y-direction by 2? (This is a n analogue of the \"Procustes\" transformation discussed in the class) . " }}{PARA 0 "" 0 "" {TEXT -1 86 "Make a picture of what this transf ormation does to the unit square similarly to 1a). \n" }}{PARA 0 "" 0 "" {TEXT 308 9 "Solution." }{TEXT -1 14 " The matrix is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "P:=mat rix([[3,0],[0,2]]); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG-%'matri xG6#7$7$\"\"$\"\"!7$F+\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "p:= x-> evalm(P&*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pGf*6# %\"xG6\"6$%)operatorG%&arrowGF(-%&evalmG6#-%#&*G6$%\"PG9$F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "vertsP:= map(p, verts0);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%'vertsPG7&-%'vectorG6#7$\"\"!F*-F'6# 7$\"\"$F*-F'6#7$F.\"\"#-F'6#7$F*F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "imageP:=polygonplot(vertsP,color=`green`):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display(\{imageP,unitsquare \} );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLY GONSG6$7&7$$\"\"!F)F(7$$\"\"\"F)F(7$F+F+7$F(F+-%'COLOURG6&%$RGBG$\"*++ ++\"!\")F3F(-F$6$7&F'7$$\"\"$F)F(7$F:$\"\"#F)7$F(F=-F06&F2F(F3F(" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2 " }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "3d) Projections. " } {TEXT 285 11 "Assignment:" }{TEXT -1 50 " Write down the matrix which \+ projects points onto " }}{PARA 0 "" 0 "" {TEXT -1 68 "the line y=2x. \+ Illustrate what this projection map does to the unit" }}{PARA 0 "" 0 " " {TEXT -1 35 "square. Is there an inverse map? \n" }}{PARA 0 "" 0 " " {TEXT 309 10 "Solution. " }{TEXT -1 65 "The projection to the line L through a unit vector u is given by " }}{PARA 0 "" 0 "" {TEXT -1 101 "proj_L(v)= (v.u)u. The columns of the projection matrix are proj_L(e1 ), proj_L(e2). Therefore we get:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "Q:= transpose(augment( dotprod(e1,u)*u, dotprod(e2,u) *u));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"QG-%'matrixG6#7$7$#\"\"\" \"\"&#\"\"#F,7$F-#\"\"%F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "q:= x-> evalm(Q&*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qGf*6# %\"xG6\"6$%)operatorG%&arrowGF(-%&evalmG6#-%#&*G6$%\"QG9$F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "vertsQ:= map(q, verts0):" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "imageQ:=polygonplot(vertsQ, color=`green`):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display( \{imageQ,unitsquare\} );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\"+++++?!#5$\"+++++SF-7$ $\"+++++gF-$\"+++++7!\"*7$F.$\"+++++!)F--%'COLOURG6&%$RGBGF($\"*++++\" !\")F(-F$6$7&F'7$$\"\"\"F)F(7$FDFD7$F(FD-F:6&F " 0 "" {MPLTEXT 1 0 33 "Shear:=k->matrix([[1,k],[0,1]]);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&ShearGf*6#%\"kG6\"6$%)operatorG%&arrowGF( -%'matrixG6#7$7$\"\"\"9$7$\"\"!F1F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 286 11 "Assignment:" }{TEXT -1 66 " For k=1 expl ore what the shear map does to the unit square. What " }}{PARA 0 "" 0 "" {TEXT -1 84 "happens if you apply the shear map twice? Three times ? Make pictures using MAPLE. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 311 10 "Solution. " }{TEXT -1 30 "For k=1 the sh earing matrix is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "S1:= ev alm(Shear(1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#S1G-%'matrixG6#7$ 7$\"\"\"F*7$\"\"!F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "Appying sh ear n times is the same as multiplying the sher matrix n times by itse lf. Thus we get:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "S2:=eva lm(Shear(1)&*Shear(1)); S3:=evalm(Shear(1)&*Shear(1) &*Shear(1));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#S2G-%'matrixG6#7$7$\"\"\"\"\"#7$\" \"!F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#S3G-%'matrixG6#7$7$\"\"\" \"\"$7$\"\"!F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "These are noth ing but shears of the strength 2 and 3. Let's see what these shearing \+ maps do to the unit square" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "s1:= x-> evalm(S1&*x):s2:= x-> evalm(S2&*x):s3:= x-> evalm(S3&*x): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "vertsS1:= map(s1, verts0):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "vertsS2:= map(s2, verts0):ve rtsS3:= map(s3, verts0):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "imageS1:=polygonplot(vertsS1,color=`green`):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "imageS2:=polygonplot(vertsS2,color=`blue`):" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "imageS3:=polygonplot(vertsS3,color= `red`):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "display(\{imageS 1, imageS2, imageS3, unitsquare\} );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%)POLYGONSG6$7&7$$\"\"!F)F(7$$\"\"\"F)F(7$ $\"\"#F)F+7$F+F+-%'COLOURG6&%$RGBGF($\"*++++\"!\")F(-F$6$7&F'F*F07$F(F +-F26&F4F5F5F(-F$6$7&F'F*7$$\"\"%F)F+7$$\"\"$F)F+-F26&F4F5F(F(-F$6$7&F 'F*FDF--F26&F4F(F(F5" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 8 "PART 2. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 75 "This part of the project is about the fun damental subspaces associated with" }}{PARA 0 "" 0 "" {TEXT -1 64 "mat rices and about coordinates with respect to different bases. " }} {PARA 0 "" 0 "" {TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Here's the matrix we w ill work with. We will keep it moderately" }}{PARA 0 "" 0 "" {TEXT -1 65 "sized, but of course we could make it much bigger without causi ng" }}{PARA 0 "" 0 "" {TEXT -1 20 "Maple any problems. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "A:=matrix([[1, 1,1,2,6],[2,3,-2,1,-3],[3,5,-5,1,-8],[4,3,8,2,3]]);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"AG-%'matrixG6#7&7'\"\"\"F*F*\"\"#\"\"'7'F+\"\"$! \"#F*!\"$7'F.\"\"&!\"&F*!\")7'\"\"%F.\"\")F+F." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 21 "b:=vector([0,0,0,0]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG-%'vectorG6#7&\"\"!F)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "We will think of this as the matrix of a linear tran sformation " }{MPLTEXT 1 0 12 "L: R^5-> R^4" }{TEXT -1 1 "," } {MPLTEXT 1 0 10 " L(x)=Ax." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 11 "Problem 4. " }{TEXT -1 1 " " } {TEXT 287 11 "Assignment:" }{TEXT -1 48 " Compute rank and nullity of A using reduction " }}{PARA 0 "" 0 "" {TEXT -1 26 "of A to row echelo n form. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 312 10 "Solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "rref(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7&7'\"\"\"\"\"!\"\"&F)F( 7'F)F(!\"%F)!\"$7'F)F)F)F(\"\"%7'F)F)F)F)F)" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 50 "Therefore rank equals 3 and nullity equals 5-3=2. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 10 " Problem 5." }{TEXT -1 2 " " }{TEXT 288 11 "Assignment:" }{TEXT -1 35 " Find a basis for the kernel of A. " }}{PARA 0 "" 0 "" {TEXT -1 75 "T o solve this problem you can either use the \"rref\" command or \"lins olve\": " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 313 10 "Solut ion. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "linsolve(A,b,'r', \+ k); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7'&%\"kG6#\"\"#,&& F(6#\"\"\"!#6*&\"\"$F.F'F.!\"\"F,,&F'\"\"%*&\"#?F.F,F.F.,&F'F2*&\"\"&F .F,F.F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "This is the general s olution with the parameters k_1 and k_2. To find basis first assign k_ 1=1, k_2=0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "vec1:=[-5,4, 1,0,0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vec1G7'!\"&\"\"%\"\"\"\" \"!F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "and then take k_1=0, k_2 =1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "vec2:=[-1,3,0,-4,1]; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vec2G7'!\"\"\"\"$\"\"!!\"%\"\" \"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "The vectors vec1 and vec2 f orm a basis of the kernel. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT 261 10 "Problem 6." }{TEXT -1 2 " " } {TEXT 289 11 "Assignment:" }{TEXT -1 64 " Use Maple to find a basis v1 , v2, v3 for the column space of A " }}{PARA 0 "" 0 "" {TEXT -1 75 "(i .e. of the image of L). Recall that one way to find this basis is to do" }}{PARA 0 "" 0 "" {TEXT -1 71 "row operations, putting your matri x into reduced column echelon form. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 300 8 "Solution" }{TEXT -1 83 ". The reduced eachlon form of A h as 1-st, 2-nd and 4-th leading columns. Therefore " }}{PARA 0 "" 0 "" {TEXT -1 69 "the 1-st, 2-nd and 4-th columns of A form a basis of the \+ image of A: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "vec3:= col( A,1);vec4:= col(A,2);vec5:= col(A,4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vec3G-%'vectorG6#7&\"\"\"\"\"#\"\"$\"\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vec4G-%'vectorG6#7&\"\"\"\"\"$\"\"&F*" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%%vec5G-%'vectorG6#7&\"\"#\"\"\"F*F)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 11 "Problem 7. " }{TEXT -1 66 " Maple will compute bases for these spaces with single commands. " }}{PARA 0 "" 0 "" {TEXT -1 1 " " } {TEXT 291 11 "Assignment:" }{TEXT -1 69 " Compare the answers you've g otten above with Maple's answers: Since" }}{PARA 0 "" 0 "" {TEXT -1 70 "you know bases are not unique, you can pretty well guess that Mapl e is" }}{PARA 0 "" 0 "" {TEXT -1 68 "using methods close to the ones w e used, since its answers should be" }}{PARA 0 "" 0 "" {TEXT -1 31 "qu ite similar to some of yours:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 314 10 "Solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "nullspace(A);#nullspace=kernel basis" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$-%'vectorG6#7'\"\"\"!\"$\"\"!\"\"%!\"\"-F%6#7'F* !#6F(\"#?!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "This is exactly \+ the basis of the kernel of A which we found above. " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "colspace( A); #nice column space basis, i.e. a basis for the image of L. " }} {PARA 11 "" 1 "" {XPPMATH 20 "6#<%-%'vectorG6#7&\"\"\"\"\"!F)!\"$-F%6# 7&F)F(F)\"#<-F%6#7&F)F)F(!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "v:=convert(colspace(A),list); " }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%\"vG7%-%'vectorG6#7&\"\"\"\"\"!F+!\"$-F'6#7&F+F*F+\"#<-F'6#7&F+F+F *!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "The above command is mak ing a list of vectors in the basis. This keeps track of order in the s et " }{TEXT 265 1 "v" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 48 "o f vectors which form basis of the image of L. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 " v1:=v[1]; v2:= v[2]; v3:= v[3]; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v1G-%'vectorG6#7&\"\"\"\"\"!F*!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v2G-%'vectorG6#7&\"\"!\"\"\"F)\"#<" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v3G-%'vectorG6#7&\"\"!F)\"\"\"!\"*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "These vectors form a basis of the image of L. Compare these vectors with vectors from the problem 6. \+ " }}{PARA 0 "" 0 "" {TEXT -1 45 "MAPLE found a different basis for the image! " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "r1:=row(rref(A),1); #r_1,r_2,r_3 basis for rowspace(A )" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r1G-%'vectorG6#7'\"\"\"\"\"!\" \"&F*F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "r2:=row(rref(A), 2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r2G-%'vectorG6#7'\"\"!\"\"\" !\"%F)!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "r3:=row(rref( A),3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r3G-%'vectorG6#7'\"\"!F)F )\"\"\"\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "G:=convert( nullspace(A),list); #making a list" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%\"GG7$-%'vectorG6#7'!\"&\"\"%\"\"\"\"\"!F--F'6#7'!\"\"\"\"$F-!\"%F, " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 " #keeps track of ord er in a set" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "n1:=G[1];n2: =G[2]; #basis of kernel. " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 258 11 "Problem 8. " }{TEXT -1 1 " " }{TEXT 290 11 "Assignment:" }{TEXT -1 78 " Use Maple to verify tha t the vectors r1, r2, r3, n1, n2 form a basis of R^5. " }}{PARA 0 "" 0 "" {TEXT 266 5 "Hint:" }{TEXT -1 50 " use reduced row echelon form o r the determinant. " }}{PARA 0 "" 0 "" {TEXT 267 21 "Explain your solu tion" }{TEXT -1 83 ": namely, how the value of the determinant or shap e of the reduced echelon form is " }}{PARA 0 "" 0 "" {TEXT -1 192 "rel ated to the fact that we have a basis. Check that each of the vectors \+ r1, r2, r3 is orthogonal to each of the vectors n1,n2. I.e. the row-sp ace is orthogonal to the kernel of the matrix. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 315 10 "Solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "M:=augment(r1, r2, r3, n1, n2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG-%'matrixG6#7'7'\"\"\"\"\"!F+!\" \"!\"&7'F+F*F+\"\"$\"\"%7'\"\"&!\"%F+F+F*7'F+F+F*F3F+7'F*!\"$F0F*F+" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "rref(M);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#-%'matrixG6#7'7'\"\"\"\"\"!F)F)F)7'F)F(F)F)F)7'F)F)F( F)F)7'F)F)F)F(F)7'F)F)F)F)F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "det(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$X)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "Thus the matrix M is invertible, therefor e the vectors r1, r2, r3, n1, n2 form a basis in R^5. " }}{PARA 0 "" 0 "" {TEXT -1 31 "Now let's check orthogonality. " }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 58 "M1:= augment(r1, r2, r3); M2:= transpose(aug ment(n1, n2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#M1G-%'matrixG6#7' 7%\"\"\"\"\"!F+7%F+F*F+7%\"\"&!\"%F+7%F+F+F*7%F*!\"$\"\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#M2G-%'matrixG6#7$7'!\"\"\"\"$\"\"!!\"%\"\"\" 7'!\"&\"\"%F.F,F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "Multiplying \+ M2 and M1 will simultaneously compute the dot products (n1.r1), (n2.r 2),... " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalm(M2&*M1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7$7%\"\"!F(F(F'" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "We got zero matrix, therefore the \+ vectors are mutually orthogonal. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 259 10 "Problem 9." }{TEXT -1 3 " " }{TEXT 292 11 "Assignment:" }{TEXT -1 63 " Let us call E=\{e1,e2, e3,e4,e5\}, i.e. the set of standard basis" }}{PARA 0 "" 0 "" {TEXT -1 64 "vectors in R^5. Let us call our new basis S=\{r1,r2,r3,n1,n2 \}. " }}{PARA 0 "" 0 "" {TEXT -1 29 "Find the transition matrices " } {MPLTEXT 1 0 9 "P_\{E <-S\}" }{TEXT -1 4 " and" }{TEXT 264 1 " " } {MPLTEXT 1 0 8 "P_\{S<-E\}" }{TEXT -1 8 ", which " }}{PARA 0 "" 0 "" {TEXT -1 78 "convert S coordinates to E-coordinates and vice-versa. Re call that the matrix " }{MPLTEXT 1 0 9 "P_\{E <-S\}" }{TEXT -1 1 " " } }{PARA 0 "" 0 "" {TEXT -1 36 "is very easy to find and the matrix " } {MPLTEXT 1 0 8 "P_\{S<-E\}" }{TEXT -1 36 " is easily obtained from the matrix " }{MPLTEXT 1 0 12 "P_\{E <-S\}. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 316 8 "Solution" }{TEXT -1 85 ". The t ransition matrix P_\{E <-S\} is the matrix which has the vectors r1,r2 ,r3,n1,n2 " }}{PARA 0 "" 0 "" {TEXT -1 34 "as columns, i.e. it is our \+ matrix " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "M:=augment(r1, r 2, r3, n1, n2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG-%'matrixG6#7 '7'\"\"\"\"\"!F+!\"\"!\"&7'F+F*F+\"\"$\"\"%7'\"\"&!\"%F+F+F*7'F+F+F*F3 F+7'F*!\"$F0F*F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "The transitio n matrix P:=P_\{S<-E\} is the inverse of the matrix M:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "P:= inverse(M);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"PG-%'matrixG6#7'7'#\"$)H\"$X)#\"$V$F,#\"$=\"F,#\" $s\"F,#!#VF,7'F-#\"$V%F,#!#dF,#\"$K#F,#!#eF,7'F1F:#!#oF,#\"$t\"F,#\"$o \"F,7'#\"#VF,#\"#eF,#!# " 0 "" {MPLTEXT 1 0 18 " a:= <0,1,-4,0,-3>:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "Scoor inatesofa:= evalm(P&*a); #coordinates of a in S-basis" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/ScoorinatesofaG-%'vectorG6#7'\"\"!\"\"\"F)F)F) " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 317 16 "10b) Solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "b:= <1,0,0,0,0>:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "Scoorinatesofb:= evalm(P&*b);#coordinates of b in S-basis" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/ScoorinatesofbG -%'vectorG6#7'#\"$)H\"$X)#\"$V$F+#\"$s\"F+#\"#VF+#!$=\"F+" }}}}{MARK " 93 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }