西南大学《线性代数》（英文版）课件-第9部分.pdf

4.1?A?A?HO?5“3cn,L1?,?9?S?5|?n.8U,ay35“AK?AX=X.(0.1)A n?,X n?,P.e(0.1)k),K A?AAA?(eigenvalue);uA?AAA?(eigenvector).5“A?A?A Pnn,e3 Pn?A=0,0 P,(0.2)K 0 A?A?,uA?0?A?.X?A=1111!,=11!.duA=1111!11!=22!=2 11!=2,d 2 A?A?,A?u 2?A?.5“A?5KA?u 0?A?E,0?A?.yyy e A?u 0?A?,=A=0.?0 6=k P,K k 6=0 A(k)=k(A)=k(0)=0(k).d,k A?u 0?A?.555A?A?w,k.5“K KKK?kA?A?A:?A?uA?0?A?,KA=0.lA?w,?u A C?0.?L7?:_?=3,K?A L:A=12323212!du?3=eCgC?.d3 R23 v A=.l?A vkA?,vkA?.5“Lc?f,?k?kA?,k?vkA?.X?P?kA?A?XJk?,N?A?A?A?kwN?f.?A=1214!,A kA?A?0 A?A?,A?u 0?A?A=0,6=0,0 P(0E A)=0,6=0,0 P g5|(0E A)X=0?),0 P|0E A|=0,(0E A)X=0?),0 P5“du|0E A|=?0 1210 4?=20 50+6,(0.3)d,|0E A|=0 20 50+6=0 0 2 5+6=0?.r 2 5+6?A?AAA?,No?d(0.3)?u|E A|=2 5+6.?A?AAA?|E A|.5“o(0 A?A?,A?u 0?A?m0A?|E A|=0 3 P?,(0E A)X=0?)?Lu?n?A.dkn?A P?n?,K10 A?A?0A?|E A|=0 3 P?;2 A?u 0?A?(0E A)X=0?).5“A?5?A=(aij)nn,K?A?5f()=|E A|=?a11a12a1na21 a22a2n.an1an2 ann?.A?X 1?n g.|1?n 1 g7,g?n?.l?n 1 g?X(a11+a22+ann).f()?u f(0),dA?(1)n|A|.uA?kXe/:f()=|E A|=n(a11+a22+ann)n1+(1)n|A|.5“A?5Ke n?A k n A?1,2,n,K11+2+n=a11+a22+ann;212n=|A|.yyy d,A?kXe)f()=(1)(2)(n).|X?X=?(.5“?A?A?A?A?A?Xe O A?A?|E A|;?O|E A|3 P kvk.XJ3 P vk,K A vkA?,l?vkA?.XJ|E A|3 Pk,K?A?A?.d?X1n.u A?zA?j,g5|(jE A)X=0?:)X:1,t.u A?u j?A?|?8k11+ktt|k1,kt P,?0.5:A?.5“f 1?A=1214!P?,A?A?A?.)k,A?A?|E A|=?121 4?=2 5+6=(2)(3).d A?A?2,3.uA?2.)g5|(2E A)X=0:1212!1200!,?)x1=2x2,x2gd.l?k:)X1=21!.5“fd,A?u 2?A?k11|k1 P,k16=0.aq/,uA?3,(3E A)X=0?:)X2=11!.d,A?u 3?A?k22|k2 P,k26=0.5“f 2?A=222214241,P?,A?A?A?.)k,A?A?|E A|=?2222+1424+1?=?2222+140 3 3?=?2422+5400 3?=(3)?242+5?=(3)(2+3 18)=(3)2(+6).d A?A?3(?-),6.5“fuA?3,)g5|(3E A)X=0:122244244 122000000,?)x1=2x2+2x3,x2,x3gd.l?:)X1=210,2=201.d A?uA?3?A?k11+k22|k1,k2 P,k1,k2?0.5“fuA?6,g5|(6E A)X=0?:)X3=122d A?uA?6?A?k33|k3 P,k36=0.5“f 3?A=1111!.r A w R?,A kvkA?rA w C?,A?A?A?.)k,A?A?|E A|=?111 1?=2 2+2.du?O =(2)2 4 1 2=4 0,d 2 2+2 vk,l?A vkA?.2 2+2 kE 1+i,1 i.E?A?A?.5“fuA?1+i,(1+i)E A)X=0?:)X1=i1!.d,A?u 1+i?A?k11|k1 P,k16=0.uA?1 i,(1 i)E A)X=0?:)X2=i1!.d,A?u 1 i?A?k22|k2 P,k26=0.555?kA?A?w=?k.E?okA?A?.5“f 4?0 A?A?,y:(1)e k?,K k0 Ak?A?;(2)e A _,K10 A1?A?.)0 A?A?,?3?A=0.uA2=A(A)=A(0)=0(A)=20,20 A2?A?.da,k0 Ak?A?.e A _,K A vkA?.=06=0.uA=0=A1(A)=A1(0)=A1=10,10 A1?A?.5“f 5?()=amm+am1m1+a1+a0,0 A?A?.o(0)?(A)=amAm+am1Am1+a1A+a0E,?A?.)0 A?A?,?3?A=0.duk0 Ak?A?.l?(A)=(amAm+am1Am1+a1A+a0E)=(amm0+am1m10+a10+a0)=(0).5“yn?1,2,s A?A?,1,2,s A Ou 1,2,s?A?,K|1,2,s5.yyyuA?s 8B.?s=1,duA?5,(.b?(u s 1 1,=s 1?A?A?5|?|5.s A?/.?k11+k22+ks1s1+kss=0u5“yA(k11+k22+ks1s1+kss)=0(0.4)k1A1+k2A2+ks1As1+ksAs=0(0.5)k111+k222+ks1s1s1+ksss=0(0.6)qw,kk1s1+k2s2+ks1ss1+ksss=0(0.7)(0.3)-(0.4)?k1(1 s)1+k2(2 s)2+ks1(s1 s)s1=05“d8Bb?,1,2,s15,uk1(1 s)=0,k2(2 s)=0,ks1(s1 s)=0.du 1,2,s,i s6=0,i=1,2,s 1.uk1=k2=ks1=0.d kss=0,=?ks=0.l?1,2,s5.5“f?nw,?A?A?|?|5.,aq/y2?(:n?1,2,s A?A?,11,12,1t1u 1?5?A?;21,22,2t2u 2?5?A?;s1,s2,stsu s?5?A?,K|11,12,1t1,21,22,2t2,s1,s2,sts5.5“4.2?q?zHO?5“3,B?Xe48?:F1=1,F2=1,Fn=Fn1+Fn2(n 3,n N)3y“n,ON(?,z?+,B?k?A.ggg,?KKK:TTT?.k,35“S?):A=1110!,FnFn1!=A Fn1Fn2!.du F2F1!=11!,u FnFn1!=An2 11!,8(An2.5“KKK e,?O?(J.?=diag1,2,n=12.n?NO.k=diagk1,k2,kn=k1k2.kn.5“O?g u?A,XU?_?P 9?A=PP1,K|?NO?A55z A?O.dkAk=(PP1)(PP1)(PP1)=P(P1P)(P1P)P1=PkP1.o?/Q?K,?qVg:?A,B?.e3?_?P,?P1AP=B,K A B q.P A B.e B?,K A q?z.5“q?5?qX?dX:g5:A A5:A B=B AD45:A B,B C=A C qC:1?;A?.=,A B=|A|=|B|A B=|E A|=|E B|yyy e A B,K3_?P?B=P1AP.u|B|=|P1AP|=|P1|A|P|=|A|.|E B|=|P1EP P1AP|=|P1|E A|P|=|E A|.5“?U?z?KKK A oU?qQ?_?P?P=(1,2,n)u|1,2,n5.l A=PP1u,?AP=P A(1,2,n)=(1,2,n)12.n(A1,A2,An)=(11,22,nn)Ai=ii,i=1,2,n.i A?u i?A?i=1,2,n.5“?U?z?(n?A?z?k n 555?AAA?.?A?z,q?A?A?.XXX?555?AAA?QQQ?rA?A?A?8?5,ouA?A?35Q?Y?.n?1,2,s A?A?,1,2,s A Ou 1,2,s?A?,K|1,2,s5.5“f?nw,?A?A?|?|5.,aq/y2?(:n?1,2,s A?A?,11,12,1t1u 1?5?A?;21,22,2t2u 2?5?A?;s1,s2,stsu s?5?A?,K|11,12,1t1,21,22,2t2,s1,s2,sts5.5“?u?A,O A?A?|E A|.3 P A?1,2,s.uz i,i=1,2,s,g5|(iE A)X=0?:)X i1,i2,iti.e t1+ts n,K A U?z.e t1+ts=n,K A?z,PP=(11,12,1t1,s1,s2,sts)=12.n,k P1AP=,=A=PP1.e?A k n?A?,K A?z.5“f 1?.d?,IOAn2 11!,A=1110!.|E A|=?111?=(1)1=2 1.)2 1=0?A kA?1=1+52,2=1 52.uA?1=1+52,1+52 1111+52!=512111+52!512100!5“fu)x1=1+52x2?g5|(1E A)X=0?:)X1=1+521!=11!uA?2=152,aq?(2E A)X=0?:)X2=21!uP=(1,2)=1211!,=1002!5“dP1=15 1211!l?An2=Pn2P1=15 n11 n121n12 2n11n21 n221n22 2n21!q FnFn1!=An2 11!=15 n1 n2n11 n12!u?Fn=15(1+52)n(1 52)n!,n 2.?n=1,?y F1v.5“f 2?A=00111a100,?a,A?z.)k,A?A?.|EA|=?011 1a10?=(1)?11?=(+1)(1)2.d A?A?1=1,2=1.uA?1.)g5|(E A)X=0:E A=10112a101 10102a 1000,?r(1E A)=2.l?5|(1E A)X=0?:)X.5“fuA?1,k2E A=10110a101 10100a+1000e?A U?z,KuA?2,?A 7Lk5?A?.?5|(2E A)X=0?:)X.d r(2E A)=1.a+1=0,=a=1.5“f 3?3?A?A?1,1,2.|A+2A+E|.)du A kn?A?,?A?z.l?3_?P?P1AP=112.(0.1)duq?1?,|A|=1 1 2=2.uA=|A|A1=2A1.d(0.5)A=PP1,l?A1=(PP1)1=P1P1.u A=2P1P1,d P1AP=21=221.5“f?P1(A+2A+E)P=P1AP+2P1AP+E=221+224+111=114u|A+2A+E|=|P1(A+2A+E)P|=4.5“SIO?4.3?5HO?5“NSN1S2IO?SIO?3c,S?n?m?5.n?m 2?3?m?2.2?3?mk,?,?k?IX,=?.3?n?mUk?9?.5?2?3?m?Vg5?u 2?3?S:e =a1a2!,=b1b2!,K(,)=a1b1+a2b2.5“SIO?n?S?=a1a2.an,=b1b2.bn R?n?,(,)=a1b1+a2b2+anbn?SSS.555 er,?,KS?(,)=T.5“SIO?S?5|S?,N?ySk:?555(,)0 (,)=0 =0;555555(+,)=(,)+(,),(c,)=c(,),c R;55