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2013数学建模国赛B题思考过程和模型还有代码(适合初学者)

于 2021-05-06 发布
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代码说明:

这个是2013数学建模国赛B题的一个完整的解答过程,包括了对题目的思考,模型的建立,代码的编写,论文的写法。当然还包括了最终的结果-----完整的报告。

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    北京航空航天大学矩阵理论学习笔记,总结版,学霸总结,可以放心下载使用北京航空航天大学张京蕊工程系统工程系月录§0补充公式§1 Jordan(约当)标准形(简介)§2线性变换与矩阵.24§3欧式空间与QR分解.48§4常用矩阵分解●鲁D●●·,,,,,74§5范数与级数.81§6广义逆A..97§7直积拉直及应用105矩阵理论A笔记北京航空航天大学张京蕊工程系统工程系§0补充公式令A=(a)mxn∈C",风x)=4o+a1x0定义f(4)=a0+a1A+…+amAm,其中I=l若g(x)=bo+b1x+…+bkx,(x)g(x)=g(x)(x),则f(4)“g(A)=g(A)f(A)分块公式A10令A,A1,A2为方阵00 A(2)f(A),fx)为多项式令A=,A1,4为方阵AO(2)f(4)相似关系:A∽B,(PAP=B)则:(1)(P1AP)=P!AP,(k=0,1,2,(2)f(PAP)=PfA)P,f(x)为多项式许尔公式( schur):每个复方阼,A-(a)nxm都相似丁上三角形。共113页矩阵理论A笔记第1页北京航空航天大学张京蕊工程系统工程系即:P-1AP=其中41,,的次序可以任意指定Pf:用归纳法n=1时成立可以设为(n=1阶方阵成立对于n阶方阵A=(an)2×n设特征值为A,…,n取为对应的特征向量,记为a1≠0,A1=1ax1把a1扩展为可逆方阵Q=(a1,02,xn)22e又:g(a,a,…,.)=(Qa,Qba2,,Qan)其中Qe1,aQ0Q4=QA(a1a2,…an)2-I(Aa,,AAQ=(Qa,、+)…,(*)其中A1为(n-1所阶0人:0 A为由假设,对于A1必有(n-1)阶P,可推出PAPEg知n阶方阵A,适合A=0,则A+|=1共113页矩阵理论A笔记第2页北京航空航天大学张京蕊工程系统工程系Pf:A=0→任意特征值A=0→>=0即全体特征值为00,,00由需要P1AP=→PAP+7=1pAP+PP|=P(4+1)P=14+1→A+1=-1注(1)若AB(相似),则AB有相同特征值A,可引入记号:谱集(4)={2,2,…,λ}(全体特征值,含重复)A∽B→o()=o(B)(2)A∽B→1-A=1-B-(2-4元一2)…(-n),特征多项式PAP=B=A-A=p(1-A)P=A-B引理:若A0A2,则M-A|-|M1-4|-1-A1|2-A2→ar(4)=o(A)∪a(42k+1,Ak-2,…n1f(x2)设B,f(x)为多项式,则f(B)=o f(,)引理:若n阶方阵A的谱集(4)=1,42,…},则)的全体特社值为)2,…,),x)为多项式Pf:由许尔定理,A∽B→f(4)∽f(B)f(x)的全体特征值为(A1)(42),,()},fx)为多项式例如:4为A的特征值→x为4的特征值。(x)=x)共113页矩阵理论A笔记第3页北京航空航天大学张京蕊工程系统工程系引理:令B,f(x)=x-B|=(x-41)(x-12)….(x-n)则fB)=(B-1D(B-21)…(B-A1D=0Pf:当n=2时,B=0x2f(x)=(x-1)(x-2)000→f(B)-(B-41)(B-21)(2-元)0(00∴得证★ Cayley公式:设n阶方阵A的特征多项式为f(x)=|x-A|=a+a1x+…,+x则f4)=anl+a14+…,+4=0Pf:由许尔PAP=B=→P(4)P=fp3P)=f(B)=0(引理)定义若多项式x)使(4)=0,则称(x)为A的个零化式结论方阵A的特征多项式)=1x1-4为A的一个零化式g特征多项式fx)=x2可知:f(A)=A2+1=+I=00-1Hx)=|xI-A|=(x-)(x+i,(i=√-1,t2=-1)f(A)=(A-i)(4+i1=0也可取P=则PPAP=,对角形共113页矩阵理论A笔记第4页北京航空航天大学张京蕊工程系统工程系g:知A则A"=0Onxn由 Cayley特征多项式:f(x)=x"→f(4)=4"=0Ex 1. A=求P使得PP为对角阵,并验证 Cayley定理2.A=cd/,求fx)=x1-4验证f4)-0补充知识( schur公式、 Cayley公式)应用由A"=-(a0I+a1A+1A·AanA+a142+…+a.,A把①代入②→Am1=(-)+(+)4+…+(+)41可知:任何和(m≥n)都可写成,4,,A的线性组合任何多项式g(A),可写成lA,…,4的组合。Fg:若A|≠0,fx)=xI-A|=a0+a1x+…+x",ao=|-A|≠则A可用A的多项式表示∵a1A+a242+…+an21A-+A"--a072A(a1+a24+…+an-142+A)Aa1+…+an1A"2+A-1零化式定义:若g(x)=b+b1x+…+bnx,使得g(4)=bn+b14+…+bn4m=0,称g(x)为方阵A的零化式注:方阵A的零化式有无穷多个∴取特征多项式x)则4)=0任取式M(x),f(A(4)=0→f(x)(x)也是零化式极小式定义:在方阵A的零化式集合中,去次数最小的且首项系数为1的零化式m(x),称它为A的极小式共113页矩阵理论A笔记第5页北京航空航天大学张京蕊工程系统工程系注:极小式唯一性质:①极小式m(x)必为特征多项式fx)=|xI-A的因式。②特征多项式fx)=|x1-A的每个单因子(x-4)也是极小式的因子)f(x)=|x1-4=(x-x)(x-2)则极小式m(x)=(x-x)(x-2)y…(x-,),且1≤l1≤m1,1≤l2≤m2,…,1≤l≤n,41,A2…,n互不相同210EgA=020,B=020,求极小式mA(),m()解:(1)|xI-A|=(x-2)(x-1)极小式为:(x-2)(x-1)或(x-2)(x-1)计算:(4-2/)4-1)=000010k≠000000∴极小式为m4(x)=(x-2)2(x-1)(2)|-B|-=(x-2)2(x-1)00000计算:(B-2)B-1)=000010=000-1八000∴极小式为m(x)=(x-2)(x-1)Eg求下列极小式m(x)4604-60(1)A=-3-50,(2)B=2-303-6100210(3)C,(4)D=000010002000解:(1)特征多项式|x7-A|-(x-1)(x+2)极小式为:(x-1)(x+2)或(x-1)(x+2)共113页矩阵理论A笔记第6页北京航空航天大学张京蕊工程系统工程系验证:(4-D(A+2D=0∴极小式为m(x)-(x-1)(x+2)(3)解法如下引理:A1,A2的极小式为m1(x),m2(x)A10的极小式m(x)等丁m1(x),m2(x)的最小公倍式0A2(此引力可推广到A1,42,43)0100极小式为(x-1)2,0010极小式为(x-1)0取最小公倍式(x-1)2为C的极小式。460(5)F-/40,A1=020|,A00 A0123-6101O引理;设D=,则D的极小式m(x)O验证:先证D的性质(右推公式)设A-(an)xn=(a1,2,…,n)则有AD=(0,01,a2,,.m1)AD2=(0,0,∞1,,x12)AD=(0,….0.,a1,,axn)单位向量技巧:∵AI=A(en,e2…,en)=(el,leAen)=A=(a1, a2,. a,)∴Ae1=01,Ae2=(2,.,A→AD=A(0,e1,e2,…,en-1)=(0,a1,a2…,an-)同理AD2=(AD)D=(0,.01,.12)可知:D-1-(D)Dy2-(0.,0,,e1)≠0D"=(D)D1=0,而特征多项式(x)=|x1-D|=x,极小式为某个x共113页矩阵理论A笔记第7页
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