{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 "" 1 14 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 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Warning" 2 7 1 {CSTYLE "" -1 -1 "" 0 1 0 0 255 1 0 0 0 0 0 0 1 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple O utput" 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 "" 0 256 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 } {PSTYLE "" 0 257 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 }{PSTYLE "" 0 258 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 {PARA 256 "" 0 "" {TEXT -1 11 "Math 2270-2" }}{PARA 257 "" 0 " " {TEXT -1 20 "Tuesday September 18" }}{PARA 258 "" 0 "" {TEXT 256 37 "Finding bases for image(f) and ker(f)" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 68 "We work with the vectors from problem 24 in section 3.3; page 131. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(lin alg):" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected names nor m and trace have been redefined and unprotected\n" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 165 "At:=matrix(5,5,[1,2,3,2,1,3,6,9,6,3,3,2,4,1 ,2,8,4,9,1,5,\n0,4,5,5,1]); \n#this is the \"transpose\" of the matrix \n#I want, i.e. the columns and rows have been reversed." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#AtG-%'matrixG6#7'7'\"\"\"\"\"#\"\"$F+F*7'F, \"\"'\"\"*F.F,7'F,F+\"\"%F*F+7'\"\")F1F/F*\"\"&7'\"\"!F1F4F4F*" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "A:=transpose(At); #this is t he matrix I want.\n#the image of f(x)=Ax is the subspace of #24 page 1 31" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7'7'\"\"\"\"\" $F+\"\")\"\"!7'\"\"#\"\"'F/\"\"%F17'F+\"\"*F1F3\"\"&7'F/F0F*F*F47'F*F+ F/F4F*" }}}{PARA 0 "" 0 "" {TEXT -1 376 "If I was doing 24, I am looki ng for a basis of subspace image(f), where f(x)=Ax. In 24 I want my b asis to be a subset of the original 5 column vectors. These 5 vector s in the \"column space\" may be dependent, in which case I would like to throw away redundant vectors until I am left with a basis. I can \+ discover dependencies from the kernel( f), which I get from rref(A): " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "rref(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7'7'\"\"\"\"\"$\"\"!!\"\"F)7'F*F*F(F)F +7'F*F*F*F*F*F-F-" }}}{PARA 0 "" 0 "" {TEXT -1 234 "Backsolve to find \+ a basis for kef(f): (Think of A and rref(A) augmented by a zero colum n for the right hand side.) Check that you have a basis for ker(f), b ecause this is something you need to know how to do in lots of applica tions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "How d oes your work compare to Maple:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "nullspace(A); #basis for the kernel of f,\n#also called the nu llspace of A" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<%-%'vectorG6#7'!\"$\" \"!\"\"\"F)F*-F%6#7'F(F*F)F)F)-F%6#7'!\")F)F)F*\"\"$" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 538 "Actually, since column dependencies are really \"the sam e\" as solutions to Ax=0, they are unchanged as you do elementary row operations. Therefore since in the reduced matrix I have the the sec ond column equal to 3 times the first, the same fact is true in the or iginal matrix! Since in the reduced matrix the 4th column is 3 times \+ the third minus the first, the same is true in the original matrix! S imilarly, the 5th column is dependent on the first and third columns. \+ Thus an acceptable basis for the span of my 5 column vectors is" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "imbasis:=\{col(A,1),col(A,3)\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%(imbasisG<$-%'vectorG6#7'\"\"\"\"\"#\"\"$F+F*-F'6#7'F,F+\"\"%F*F+" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 202 "A nic er basis would have vectors with more zero entries. We can obtain thi s by doing elementary column operations to A, since these do not chang e the span of the columns! This is the same as computing" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "transpose(rref(At));" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#-%'matrixG6#7'7'\"\"\"\"\"!F)F)F)7'F)F(F)F)F)7'# F(\"\"##\"\"&\"\"%F)F)F)7'#!\"\"F-F.F)F)F)7'F,#F(F0F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 26 "This is how maple does it:" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 67 "colspace(A); #basis for column space of A,\n#i .e. for the image of f" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$-%'vectorG 6#7'\"\"\"\"\"!#F(\"\"##!\"\"F+F*-F%6#7'F)F(#\"\"&\"\"%F1#F(F3" }}} {PARA 0 "" 0 "" {TEXT -1 232 "It's always easy to tell that the basis \+ you get this way is actually a basis; the independence follows easily, and for example it is also easy to express your basis obtained by del etion in terms of this col op derived basis. Do it." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "56 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }