{VERSION 5 0 "HP RISC UNIX" "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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {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 "" 0 256 1 {CSTYLE "" -1 -1 "helvetica" 1 24 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 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 261 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 263 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 264 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 57 "Beispiel zur Berechnung \+ von Eigenwerten\nund Eigenvektoren" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "restart: with(linalg): unprotect(D) :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "A := matrix(4,4,[1,2,3 ,4,5,6,7,8,8,7,6,5,4,3,2,1]);" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 50 "Als erstes berechnen wir die Determinante von A - " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 33 "E und bestimmen ihre Nullstell en:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "A - lambda*E = evalm(A - lambda);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "charakteristisches_Polynom := sort(det(A-lambda));" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "solve(charakteristisches_P olynom);" }}}{EXCHG {PARA 258 "" 0 "" {TEXT -1 163 "Es gibt drei Nulls tellen, also die drei Eigenwerte 0, 18 und -4. Wir beginnen mit der Nu ll.\nDie Eigenvektoren zum Eigenwert 0 erf\374llen das lineare Gleichu ngssystem" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "G := geneqns(A , [x[1], x[2], x[3], x[4]]): G[1]; G[2];G[3];G[4];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "linsolve(A, [0$4]); " }}}{EXCHG {PARA 259 " " 0 "" {TEXT -1 164 "Die L\366sung h\344ngt also ab von zwei Parameter n; wir erhalten eine Basis des L\366sungsraums,\nindem wir jeweils ein en der beiden auf eins und den anderen auf null setzen:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "v1 := subs(_t[1]=1, _t[2]=0, %);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "v2 := subs(_t[1]=0, _t[2]= 1, %%);" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 93 "F\374r den Eigenwert 18 m\374ssen wir das homogene lineare Gleichungssystem mit Matrix A-1 8E l\366sen." }{MPLTEXT 1 0 1 "\000" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "A - 18*E = evalm(A-18);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "G := geneqns(evalm(A-18), [x[1], x[2], x[3], x[4]]): \+ G[1]; G[2];G[3];G[4];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "li nsolve(A-18, [0$4]); " }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 75 "Der L \366sungsraum ist also eindimensional und wird beispielsweise erzeugt \+ von" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "v 3 := subs(_t[1]=5, %);" }}}{EXCHG {PARA 262 "" 0 "" {TEXT -1 29 "Bleib t noch der Eigenwert -4:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "A + 4*E = evalm(A + 4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "G := geneqns(evalm(A+4), [x[1], x[2], x[3], x[4]]) : G[1]; G[2];G[3];G[4];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 " linsolve(A+4, [0$4]);" }}}{EXCHG {PARA 263 "" 0 "" {TEXT -1 50 "Der Ei genraum wird also beispielsweise erzeugt von" }{MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "v4 := subs(_t[1]=1, %);" }}} {EXCHG {PARA 264 "" 0 "" {TEXT -1 130 "Damit haben wir vier linear una bh\344ngige Eigenvektoren v1, v2, v3 und v4 gefunden;\nbez\374glich di eser Basis hat die Matrix A die Form" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "D = diag(0,0,18,-4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1 0 0" 36 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }