{VERSION 2 3 "APPLE_PPC_MAC" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 0 1 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 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 259 "" 0 "" {TEXT -1 10 "Appendix: " }}{PARA 0 "" 0 "" {TEXT -1 406 " The ideas of the course can be understood in the co ntext of linear algebra. This Appendix to the notes is to remind the r eader that the ideas of this course are ideas about solving linear equ ations -- about linear algebra. In class, I would typically spend one \+ day pointing the student to this reference and invite them to look at \+ old notes they already transcribed that might be about linear algebra. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 " T he first part of the Appendix is about the geometry of " }{XPPEDIT 18 0 "R^n" ")%\"RG%\"nG" }{TEXT -1 148 ". In Appendix 2 the idea is about how the adjoint of a linear operator is defined. Note, for the purpos es of this worksheet we take the space to be " }{XPPEDIT 18 0 "R^n" ") %\"RG%\"nG" }{TEXT -1 13 ", instead of " }{XPPEDIT 18 0 "C^n" ")%\"CG% \"nG" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 12 "Exercise 2.1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "A:=matrix([[1,2],[-1,0]]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "B:=transpose(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "dotprod(A&*[x,y],[u,v]);\ndotprod([ x,y],B&*[u,v]);\nis(\"\"=\");" }}}{PARA 257 "" 0 "" {TEXT -1 14 "Exerc ise 2.2.c" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "A:=matrix([[2,3 ],[2,3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "linsolve(A,[3 ,2]);" }}}{PARA 0 "" 0 "" {TEXT -1 25 "You wish Maple would say " } {TEXT 256 22 "There is not solution!" }{TEXT -1 68 ", instead of stand ing mute. Maybe the code is not working; try this." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 18 "linsolve(A,[3,3]);" }}}{PARA 258 "" 0 "" {TEXT -1 14 "Exercise 3.1.b" }}{PARA 0 "" 0 "" {TEXT -1 101 "We find t he null space of A and of A*. The answer is in the form of a basis for the nullspace of each" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "A: =matrix([[1,2,1],[2,4,2],[0,0,0]]);\nNA:=nullspace(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "Astar:=transpose(A);\nNAstar:=nulls pace(Astar);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "eq1:=dotpro d([f,g,h],NAstar[1]);\neq2:=dotprod([f,g,h],NAstar[2]);\nsolve(\{eq1=0 ,eq2=0\},\{f,g,h\});" }}}{PARA 0 "" 0 "" {TEXT -1 198 "You should beli eve that [1, 2, 0] is orthogonal to the nullspace of A* and [1, 2, 1] \+ is not. If you believe the Fredholm alternative theroems, one of these is in the range of A and the other is not." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 20 "linsolve(A,[1,2,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "linsolve(A,[1,2,1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 260 "" 0 "" {TEXT -1 10 "Exercise 4" }} {PARA 0 "" 0 "" {TEXT -1 231 "The point for this section is to make up a way to get the inverse of A that will generalize even when A is not a matrix. Here is the method suggested in case A has an inverse. We p ut off getting the generalized inverse until later." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "A:=matrix([[1,2],[-1,0]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "x:=linsolve(A,[1,0]);\ny:=linsolve( A,[0,1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "G:=transpose(m atrix([x,y]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "u:=evalm( G&*[3,4]);\nevalm(A&*u);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "inverse(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1 0" 14 }{VIEWOPTS 1 1 0 1 1 1803 }