{VERSION 3 0 "APPLE_PPC_MAC" "3.0" } {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 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 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 "Bulle t Item" 0 15 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 3 3 0 0 0 0 0 0 15 2 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 1 14 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 256 "" 0 "" {TEXT -1 54 "Linear Algebra, Infinite Dimens ional Spaces, and Maple" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "Jim Herod, Georgia Tech" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 12 "S ection 9: " }{TEXT -1 31 "The Finite Dimensional Paradigm" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 271 " In this sec tion, we return to the representation of a matrix. This time, we use t he ideas that have come before, and make a representation that will be repeated for more general linear transformations on infinite dimensio nal spaces. We have for self adjoint matrices" }}{PARA 0 "" 0 "" {TEXT -1 26 " A = " }{XPPEDIT 18 0 "sum(lambda[i] ,i=1..n)" "6#-%$sumG6$&%'lambdaG6#%\"iG/F);\"\"\"%\"nG" }{TEXT -1 3 " \+ < " }{XPPEDIT 18 0 "x[i]" "6#&%\"xG6#%\"iG" }{TEXT -1 2 ", " } {XPPEDIT 18 0 "phi[i]" "6#&%$phiG6#%\"iG" }{TEXT -1 3 " > " }{XPPEDIT 18 0 "phi[i]" "6#&%$phiG6#%\"iG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 11 "where each " }{XPPEDIT 18 0 "lambda[i]" " 6#&%'lambdaG6#%\"iG" }{TEXT -1 22 " is an eigenvalue and " }{XPPEDIT 18 0 "phi[i]" "6#&%$phiG6#%\"iG" }{TEXT -1 83 " is an eigenvector. We \+ illustrate how to make this decomposition with two examples." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{PARA 0 " " 0 "" {TEXT -1 14 "First Example:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "A:=matrix([[3,0,-1],[0,3,0],[-1,0,3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "eigenvects(A);" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 66 "u1:=vector([1,0,1]); \nu2:=vector([-1, 0, 1] );\nu3:=vector([0,1,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "u:=vector([x,y,z]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "re p:=(2)*dotprod(u,u1)*u1/norm(u1,2)^2 + (4)*dotprod(u,u2)*u2/norm(u2,2) ^2\n + (3)*dotprod(u,u3)*u3/norm(u3,2)^2 ;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "evalm(rep);\nevalm(A&*u);" }}}{PARA 0 "" 0 "" {TEXT -1 14 "Second example" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "B:=matrix([[-3/2,1/2,0],[1/2,-3/2,0],[0,0,0]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "eigenvects(B);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "u1:=vector([-1,1,0]); \nu2:=vector([0,0, 1]);\nu3: =vector([1,1,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "u:=vec tor([x,y,z]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 119 "rep:=(-2) *dotprod(u,u1)*u1/norm(u1,2)^2 + (0)*dotprod(u,u2)*u2/norm(u2,2)^2\n \+ + (-1)*dotprod(u,u3)*u3/norm(u3,2)^2 ;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "evalm(rep);\nevalm(B&*u);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "In this section 9, we also gave the generalized inv erse for matrices. We illustrate this with the above examples." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "The gener alized inverse for A." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "u1: =vector([1,0,1]); \nu2:=vector([-1, 0, 1]);\nu3:=vector([0,1,0]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "u:=vector([x,y,z]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 136 "GenInvA:=(1/2)*dotprod(u,u1 )*u1/norm(u1,2)^2 \n +(1/4)*dotprod(u,u2)*u2/norm(u2,2)^2\n \+ + (1/3)*dotprod(u,u3)*u3/norm(u3,2)^2;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 126 "col1:=evalm(subs(\{x=1,y=0,z=0\},GenInvA));\ncol2: =evalm(subs(\{x=0,y=1,z=0\},GenInvA));\ncol3:=evalm(subs(\{x=0,y=0,z=1 \},GenInvA)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Adagger:= transpose(matrix([col1,col2,col3]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalm(A &* Adagger &* A);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 " " {TEXT -1 30 "The generalized inverse for B." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "u1:=vector([-1,1,0]); \nu2:=vector([0,0, 1]);\nu 3:=vector([1,1,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "u:=v ector([x,y,z]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 134 "GenInvB :=(-1/2)*dotprod(u,u1)*u1/norm(u1,2)^2 \n +(0)*dotprod(u,u2)*u2/no rm(u2,2)^2\n + (-1)*dotprod(u,u3)*u3/norm(u3,2)^2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "col1:=evalm(subs(\{x=1,y=0, z=0\},GenInvB));\ncol2:=evalm(subs(\{x=0,y=1,z=0\},GenInvB));\ncol3:=e valm(subs(\{x=0,y=0,z=1\},GenInvB)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "col3:=vector([0,0,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Bdagger:=transpose(matrix([col1,col2,col3]));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalm(B &* Bdagger &* B);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 265 "We provide here Maple syntax t o compute generalized inverses that does not depend on being able to c ompute eigenvalues and eigenvectors. The following defines a function \+ defined on matrices. Naturally, one should investigate the limits of s uch numerical procedures." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "MPinv:=M-> map( limit,\n evalm(transpose(M) &* (M &*transpose(M) +t *&* ())^(-1)), t=0, right);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "MPinv(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "MPinv(B) ;" }}}{PARA 0 "" 0 "" {TEXT -1 73 "This last method using limits certa inly does not require square matrices." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "C:=matrix([[1,0],[0,a],[0,0]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "MPinv(C);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 13 "Section 10: " }{TEXT -1 56 "Bounded Linear Maps from the Spac e to the Complex Plane." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 " We will find the Riesz representation for a lin ear transformation from " }{XPPEDIT 18 0 "L^2" "6#*$%\"LG\"\"#" } {TEXT -1 135 "[0 ,1] to the complex plane. In this case, the represent ation will produce the Green's function for a two-point boundary-value problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " The two-point boundary-value problem is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 " y '' \+ + 3 y ' + 2 y = f, with y(0) = y(1) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "For each f, this equation will \+ produce a function y in " }{XPPEDIT 18 0 "L^2" "6#*$%\"LG\"\"#" } {TEXT -1 65 "[0 ,1]. In fact, this equation defines a linear map from \+ f 's in " }{XPPEDIT 18 0 "L^2" "6#*$%\"LG\"\"#" }{TEXT -1 134 "[0 ,1] \+ into the same space. Now, we choose x in [0 ,1]. Define L(f) to be y(x ), where y is the solution of the boundary-value problem." }}{PARA 0 " " 0 "" {TEXT -1 56 " By the Reisz Representation Theorem, there is g in " }{XPPEDIT 18 0 "L^2" "6#*$%\"LG\"\"#" }{TEXT -1 14 "[0,1] so t hat " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " \+ y(x) = L(y)(x) = " }{XPPEDIT 18 0 "int(f( t)*g(t),t=0..1)" "6#-%$intG6$*&-%\"fG6#%\"tG\"\"\"-%\"gG6#F*F+/F*;\"\" !\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "To record that the g we find goes with the x that wa s chosen, we write" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 37 " y(x) = " }{XPPEDIT 18 0 "i nt(f(t)*G(x,t),t=0..1)" "6#-%$intG6$*&-%\"fG6#%\"tG\"\"\"-%\"GG6$%\"xG F*F+/F*;\"\"!\"\"\"" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "dsolve(\{diff(y(x),x,x)+3*diff(y(x),x)+2*y(x)=f(x),y( 0)=0\},y(x));" }}}{PARA 0 "" 0 "" {TEXT -1 23 "We find the solution is " }}{PARA 0 "" 0 "" {TEXT -1 16 " " }}{PARA 0 "" 0 "" {TEXT -1 28 " " }{XPPEDIT 18 0 "y(x) = (-in t(f(u)*exp(2*u),u = 0 .. x)*exp(-2*x)*exp(x)+int(f(u)*exp(u),u = 0 .. \+ x)-_C2*exp(-2*x)*exp(x)+_C2)/exp(x)" "6#/-%\"yG6#%\"xG*&,**(-%$intG6$* &-%\"fG6#%\"uG\"\"\"-%$expG6#*&\"\"#F3F2F3F3/F2;\"\"!F'F3-F56#,$*&\"\" #F3F'F3!\"\"F3-F56#F'F3FA-F,6$*&-F06#F2F3-F56#F2F3/F2;F;F'F3*(%$_C2GF3 -F56#,$*&\"\"#F3F'F3FAF3-F56#F'F3FAFNF3F3-F56#F'FA" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "(-int(f(u)*exp(2*u),u=0..x)* exp(-x)+int(f(u)*exp(u),u=0..x)-_C2*exp(-x)+_C2)/exp(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "subs(x=1,%);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 19 "C2:=solve(%=0,_C2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(_C2=C2,%%%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{PARA 0 "" 0 "" {TEXT -1 67 " Ex amining the output of the last calculation, we can create G." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 267 "G:=proc(x,u)\n if u < x then \n (-exp(2*u-x-1)+exp(2*u-x)+exp(u-1)-exp(u)+\n \+ exp(2*u-x-1)-exp(u-x)-exp(2*u-1)+exp(u))*exp(-x)/(exp(-1)-1)\n \+ else\n (exp(2*u-x-1)-exp(u-x)-exp(2*u-1)+exp(u))*exp(-x )/(exp(-1)-1)\n fi\n end;" }}}{PARA 0 "" 0 "" {TEXT -1 154 "In the process of checking that this is a Green's function for th e two-point boundary-value problem, we recall several properties of th e Green's Function." }}{PARA 15 "" 0 "" {TEXT -1 141 "As a function of x, G satisfies the differential equation. To verify this calculation, we take z1 to be G for x < t and z2 to be G for t < x." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 299 "z1:=x->(exp(2*u-x-1)-exp(u-x)-exp( 2*u-1)+exp(u))*exp(-x)/(exp(-1)-1);\nz2:=x->(-exp(2*u-x-1)+exp(2*u-x)+ exp(u-1)-exp(u)+\n exp(2*u-x-1)-exp(u-x)-exp(2*u-1)+exp(u))* exp(-x)/(exp(-1)-1);\ndiff(z1(x),x,x)+3*diff(z1(x),x)+2*z1(x): simplif y(%);\ndiff(z2(x),x,x)+3*diff(z2(x),x)+2*z2(x):simplify(%);" }}}{PARA 15 "" 0 "" {TEXT -1 57 "As a function of x, G satisfies the boundary c onditions. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "z1(0); z2(1); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot3d(G,0..1,0..1,axes =NORMAL,orientation=[-115,45]);" }}}{PARA 0 "" 0 "" {TEXT -1 38 "We ve rify a solution in case f(t) = 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 211 "x:='x': t:='t':\nint((-exp(2*u-x-1)+exp(2*u-x)+exp(u -1)-exp(u)+\n exp(2*u-x-1)-exp(u-x)-exp(2*u-1)+exp(u))*exp(-x)/(exp( -1)-1),u=0..x)\n+\nint((exp(2*u-x-1)-exp(u-x)-exp(2*u-1)+exp(u))*exp(- x)/(exp(-1)-1),u=x..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 " solG:=simplify(expand(%));" }}}{PARA 0 "" 0 "" {TEXT -1 89 "We verify \+ this is a solution by comparing with the solution that Maple computes \+ un-aided." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "dsolve(\{diff(y (x),x,x)+3*diff(y(x),x)+2*y(x)=1,y(0)=0,y(1)=0\},y(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solM:=rhs(%);" }}}{PARA 0 "" 0 "" {TEXT -1 36 "We subtract the two to compare them." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "simplify(solG-solM);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(solG,x=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 54 }{VIEWOPTS 1 1 0 1 1 1803 }