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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 "" 
{TEXT 261 20 "Beispiel zu 6.25 (d)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}{PARA 7 "" 1 "" {TEXT -1 
80 "Warning, the protected names norm and trace have been redefined an
d unprotected\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Gegeben seien \+
folgende Vektoren mit Eintr\344gen aus dem K\366rper " }{XPPEDIT 18 0 
"Q;" "6#%\"QG" }{TEXT -1 25 "  der rationalen  Zahlen:" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 109 "v1 := vector([0, 1, -1, 1]);v2 := vector([1, \+
0, 1, 1]);v3 := vector([1, 2, 1, 1]);v4 := vector([1, 1, 1, 1]);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v1GK%'vectorG6#7&\"\"!\"\"\"!\"\"F*
Q)pprint546\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v2GK%'vectorG6#7&
\"\"\"\"\"!F)F)Q)pprint556\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v3G
K%'vectorG6#7&\"\"\"\"\"#F)F)Q)pprint566\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#v4GK%'vectorG6#7&\"\"\"F)F)F)Q)pprint576\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Sei U=<v1,v2,v3,v4> , der von den \+
Vektoren v1,v2,v3,v4 erzeugte Untervektorraum von  " }{XPPEDIT 18 0 "Q
^`4x1`;" "6#)%\"QG%$4x1G" }{TEXT -1 3 "  ." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 71 "Die Vektoren v1,v2,v3,v4 als Spalten zusammengefasst zu e
iner Matrix M:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "M:=concat(v1,v2,v
3,v4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MGK%'matrixG6#7&7&\"\"!
\"\"\"F+F+7&F+F*\"\"#F+7&!\"\"F+F+F+7&F+F+F+F+Q)pprint586\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 9 "Au
fgabe: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
42 "Erg\344nze die linear unabh\344ngigen Vektoren  " }{XPPEDIT 18 0 "
u1 = vector([3, 4, 2, 4]);" "6#/%#u1G-%'vectorG6#7&\"\"$\"\"%\"\"#F*" 
}{TEXT -1 4 " ,  " }{XPPEDIT 18 0 "u2 = vector([5, 5, 4, 6]);" "6#/%#u
2G-%'vectorG6#7&\"\"&F)\"\"%\"\"'" }{TEXT -1 29 "  aus U zu einer Basi
s von U." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "u1:=vector([ 3, 4, 2
, 4]);u2:=vector([5, 5, 4, 6]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#
u1GK%'vectorG6#7&\"\"$\"\"%\"\"#F*Q)pprint596\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#u2GK%'vectorG6#7&\"\"&F)\"\"%\"\"'Q)pprint606\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "u1 und u2  liegen tats\344chlich i
n U. Probe !" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "#linsolve(M,u1),lin
solve(M,u2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Sind u1,u2 tats
\344chlich linear unabh\344ngig ? Probe!" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "#concat(u1,u2);rank(%);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 182 "Als erstes  berechne ich selbst eine Basis von U durch S
paltenumformumgen an M (hier nicht ausgef\374hrt) oder mit Hilfe von M
aple eine Basis des Spaltenraums von M der ja gerade U ist:" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Z
RM:=colspace(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ZRMG<%K%'vector
G6#7&\"\"\"\"\"!F+\"\"#Q)pprint886\"KF'6#7&F+F*F+F+Q)pprint89F.KF'6#7&
F+F+F*!\"\"Q)pprint90F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Die Ba
sisvektoren bezeichnen wir mit w1,w2,w3:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 76 "w1:=vector([1, 0, 0, 2]);w2:=vector([0, 1, 0, 0]);w3:
=vector([0, 0, 1, -1]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#w1GK%'ve
ctorG6#7&\"\"\"\"\"!F*\"\"#Q)pprint916\"" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#w2GK%'vectorG6#7&\"\"!\"\"\"F)F)Q)pprint926\"" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#w3GK%'vectorG6#7&\"\"!F)\"\"\"!\"\"Q)pprint936
\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Als Spalten zu einer Matrix
 zusammengesetzt:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "B:=concat(op(Z
RM));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"BGK%'matrixG6#7&7%\"\"\"
\"\"!F+7%F+F*F+7%F+F+F*7%\"\"#F+!\"\"Q)pprint946\"" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 72 "Erg\344nzung von
 (u1,u2) zu einer Basis von U mit Hife des Austauschlemmas:" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Dazu st
elle ich u dar als Linearkombination der  Basis B von U." }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 15 "linsolve(B,u1);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#K%'vectorG6#7%\"\"$\"\"%\"\"#Q)pprint956\"" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 15 "Demnach ist u=3" }{XPPEDIT 18 0 "w[1];" "6#&%\"wG6
#\"\"\"" }{TEXT -1 2 "+4" }{XPPEDIT 18 0 "w[2];" "6#&%\"wG6#\"\"#" }
{TEXT -1 2 "+2" }{XPPEDIT 18 0 "w[3];" "6#&%\"wG6#\"\"$" }{TEXT -1 3 "
  ." }}{PARA 0 "" 0 "" {TEXT -1 251 "Alle Vektoren in dieser Linearkom
bination haben von 0 verschiede Koeffizienten und k\366nnen daher durc
h u1  ersetzt werden,  ohne die Basiseigenschaft zu zerst\366ren. Nach
 der Ersetzung etwa von v1 durch u1 liegt  die Basis (u1,w2,w3) vor, d
ie u1 enth\344lt." }}{PARA 0 "" 0 "" {TEXT -1 44 "Nun ist u2 mit der n
euen Basis darzustellen:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 
"C:=concat(u1,w2,w3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CGK%'matr
ixG6#7&7%\"\"$\"\"!F+7%\"\"%\"\"\"F+7%\"\"#F+F.7%F-F+!\"\"Q)pprint966
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "linsolve(C,u2);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#K%'vectorG6#7%#\"\"&\"\"$#!\"&F)#\"\"#
F)Q)pprint976\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 223 "Auch hier kan
n jetzt u2 gegen jeden der Basisvektoren ausgetauscht werden. Nat\374r
lich werfen wir nicht u1 wieder weg. Ich tausche w2 aus und erhalte di
e Basis (u1,u2,w3) von U , die beide vorgegebenen Vektoren u1,u2 enth
\344lt." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 52 "Erg\344n
zung von (u1,u2) zu einer Basis von U mit Hife " }{TEXT -1 0 "" }
{TEXT 259 31 "elementarer Zeilenumformungen: " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Danach bilde ich \+
 die Matrix [" }{XPPEDIT 18 0 "u;" "6#%\"uG" }{TEXT -1 214 "1,u2, w1,w
2,w3] , berechne (Teilschritt I) eine SSF nur (!) f\374r die ersten be
iden Spalten (die u-s), r\344ume mit deren Eckeintr\344gen nach rechts
 aus (Teilschritt II) und berechne dann eine SSF f\374r die neuen w-Sp
alten:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 27 "uw:=concat(u1,u2,w1,w2,w3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#uwGK%'matrixG6#7&7'\"\"$\"\"&\"\"\"\"\"!F-7'\"\"%F+F-F,F-7'\"\"#F/
F-F-F,7'F/\"\"'F1F-!\"\"Q)pprint986\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "uw_I:=addcol(uw,1,2,-5/3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%uw_IGK%'matrixG6#7&7'\"\"$\"\"!\"\"\"F+F+7'\"\"%#!\"
&F*F+F,F+7'\"\"##F2F*F+F+F,7'F.#!\"#F*F2F+!\"\"Q)pprint996\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "uw_II_1:=addcol(uw_I,1,3,-1/
3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%(uw_II_1GK%'matrixG6#7&7'\"\"$\"\"!F+F+F+7'\"\"%#!\"&
F*#!\"%F*\"\"\"F+7'\"\"##F4F*#!\"#F*F+F27'F-F6F5F+!\"\"Q*pprint1006\"
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "uw_II_2:=addcol(addcol(
uw_II_1,2,3,-4/5),2,4,3/5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(uw_I
I_2GK%'matrixG6#7&7'\"\"$\"\"!F+F+F+7'\"\"%#!\"&F*F+F+F+7'\"\"##F1F*#!
\"'\"\"&#F1F5\"\"\"7'F-#!\"#F*#\"\"'F5#F:F5!\"\"Q*pprint1016\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "uw_III:=addcol(addcol(uw_II_
2,3,4,1/3),3,5,5/6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'uw_IIIGK%'m
atrixG6#7&7'\"\"$\"\"!F+F+F+7'\"\"%#!\"&F*F+F+F+7'\"\"##F1F*#!\"'\"\"&
F+F+7'F-#!\"#F*#\"\"'F5F+F+Q*pprint1026\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 30 "Normiert sieht es sch\366ner aus:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 43 "uw_III:=mulcol(mulcol(uw_III,2,-3),3,-5/6);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'uw_IIIGK%'matrixG6#7&7'\"\"$\"\"!F
+F+F+7'\"\"%\"\"&F+F+F+7'\"\"#!\"#\"\"\"F+F+7'F-F0!\"\"F+F+Q*pprint103
6\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 97 "Jetzt stellt die dritte Spalte  von uw_III  eine E
rg\344nzung von (u1,u2)  zu einer Basis von U dar." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Wie kann man hier
 eine Probe machen ?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 115 "Man kann z.B. die reduzierten Spaltenstufenfor
men  von B und uw_III vergleichen. Die reduzierte SSF von uw_III ist:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "transpose(rref(transpos
e(submatrix(uw_III,1..4,1..3))));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#K
%'matrixG6#7&7%\"\"\"\"\"!F)7%F)F(F)7%F)F)F(7%\"\"#F)!\"\"Q*pprint1076
\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "Letzteres ist genau die Ma
trix B. Die Spalten von uw_III und B erzeugen also (warum ?!) denselbe
n Untervektorraum, und zwar U." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 213 "Eine andere M\366glichkeit zu \374berpr
\374fen, ob tats\344chlich die neue Basis U erzeugbesteht z.B. darin z
u kl\344ren, ob alle Vektoren in U liegen. Da sie linear unabh\344ngig
 sind, m\374ssen Sie aus Dimensionsgr\374nden U erzeugen !" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "2 1 0" 80 }{VIEWOPTS 1 
1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }