第05讲 矩阵的转置和逆.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,矩 阵,2.3,矩阵的转置 对称矩阵,2.4,可逆矩阵的逆矩阵,第,5,讲,第,2,章,2.3,矩阵的转置 对称矩阵,定义,2.11,把矩阵,A,=(,a,ij,),m,n,的行列依次互换得到,n,m,矩阵,称为,A,的,转置矩阵,记作,A,T,矩阵的转置运算满足以下运算律,:,(1)(,A,T,),T,=,A,;,(2)(,A,+,B,),T,=,A,T,+,B,T,;,(3)(,k,A,),T,=,k,A,T,(,k,是数量,);,(4)(,A,B,),T,=,B,T,A,T,;,(5),A,T,=,A,(,A,1,A,2,A,n,),T,=,A,n,T,A,2,T,A,1,T,证明,(4)(,AB,),T,=,B,T,A,T,。,设,A,=(,a,ij,),m,n,A,T,=(,a,T,ji,),n,m,B,=(,b,ij,),n,s,B,T,=(,b,T,ji,),s,n,，,则,(,A,B,),T,与,B,T,A,T,都是,s,m,矩阵，且,故,(,A,B,),T,=,B,T,A,T,。,j=,1,s;i=,1,m,定义,2.12,则 称,A,为,对称矩阵,；,则 称,A,为,反对称矩阵。,n,阶反对称矩阵,A,的主对角元都为零，,因为 由,a,ii,=,a,ii,即得,a,ii,=,0(,i=,1,2,n,),。,A,为对称矩阵的充要条件是,A,T,=,A,;,A,为反对称矩阵的充要条件是,A,T,=,A,。,例,1,设,A,是,m,n,矩阵，则,A,T,A,和,A,A,T,都是对称矩阵。,因为,A,T,A,是,n,阶矩阵,且,(,A,T,A,),T,=,A,T,(,A,T,),T,=,A,T,A,;,同理,A A,T,是,m,阶对称矩阵。,必须注意,，两个对称矩阵,A,和,B,的乘积不一定是对称矩阵。因为，,(,A,B,),T,=,B,T,A,T,=,B,A,而,B,A,不一定等于,AB,。,例,2,设,A,B,分别是,n,阶对称和反对称矩阵，则,AB,+,BA,是反对称矩阵。,因为,(,A B,+,B A,),T,=,B,T,A,T,+,A,T,B,T,=(,B,),A,+,A,(,B,)=,(,AB,+,BA,),。,2.4,可逆矩阵的逆,定义,2.13,设,A,为,n,阶方阵，若存在,n,阶方阵,B,使得,BA,=,AB,=,I,则称矩阵,A,是,可逆,的，称,B,为,A,的,逆矩阵,，记作,B,=,A,1,.,(,或,B,是可逆的且,A,=,B,1,),如单位矩阵,I,是可逆的，且,I,1,=,I,因为,I,I,=,I,显然,A,B,是平等的，,B,也是,A,的逆，,定理,2.2,若,A,是可逆矩阵，则,A,的逆矩阵是唯一的。,证,设,B,C,都是,A,的逆矩阵，,即,BA,=,AB,=,CA,=,AC,=,I,，则,B,=,BI,=,B,(,AC,),=(,BA,),C,=,IC,=,C,所以，,A,的逆矩阵是唯一的,可见，,A,可逆的必要条件是,|A|,0.,即,A,是非奇异的。,下面来证明，,|A|,0,也是,A,可逆的充分条件。,为此，引入伴随矩阵的概念。,定义,2.14,设,A,=(,a,ij,),n,n,A,ij,是,det,A,中,a,ij,的代数余,子式,称,cof,A,=(,A,ij,),n,n,为,A,的,代数余子式矩阵,，,其转置矩阵,A,*,=(,cof,A,),T,称为,A,的,伴随矩阵,，,记作,A,*,。,由教材,pp.5960,的例题,6,，可知,AA,*,=,A,*,A,=,|A|I,当,|A|,0,时，,定理,2.3,矩阵,A,可逆的充要条件是,A,0,。且,且,A,可逆时，其逆为,证,必要性：,若,A,可逆，则存在,B,使得,AB,=,I,于是,AB,=,A,B,=,I,=1,故,A,0,。,充分性,:,用构造性证法。若,A,0,由,AA*,=,A*A,=|,A,|,I,，,推论,1,设,A,，,B,都是,n,阶矩阵,且,AB,=,I,则,BA,=,I,即,A,，,B,都可逆，并互为逆矩阵。,证,由,AB,=,I,得,AB,=,A,B,=,I,=1,故,A,0,，,B,0,即,A,，,B,都可逆。,AB,=,I,A,1,(,AB,),A,=,A,1,IA,=,I,即,BA,=,I,推论,2,对角阵、上（下）三角阵可逆的充要条件是,主对角元全部不为零。,注意,:,A,B,都可逆，而,A,+,B,不一定可逆，,即使,A,+,B,可逆,一般地,(,A,+,B,),1,A,1,+,B,1,。,例,1,是否可逆？若可逆,求其逆矩阵。,解：,A,=4,0,A,可逆,(,非奇异,),。,A,11,=,3,A,12,=,4,A,13,=5,A,21,=3,A,22,=0,A,23,=,1,A,31,=1,A,32,=4,A,33,=,3,C,=0,故,C,不可逆。,例,2,解：,B,=,ad-,bc,当,ad-,bc,0,时,B,可逆,其逆矩阵为,可逆矩阵的运算性质（,A,B,为,n,阶可逆矩阵，数,k,0),A,*,=,A,n,-1,(,A,1,),1,=,A,(,k,A,),1,=,k,1,A,1,(,A,B,),1,=,B,1,A,1,(,A,T,),1,=(,A,1,),T,A,1,=,A,1,(,A,1,A,2,A,k,),1,=,A,k,1,A,2,1,A,1,1,(,A,k,),1,=(,A,1,),k,A,k,(,A,1,A,2,A,k,均可逆,);,证,(,k,A,)(,k,1,A,1,)=(,kk,1,)(,AA,1,)=1,I,=,I,。,(,AB,)(,B,1,A,1,)=,A,(,BB,1,),A,1,=,AIA,1,=,AA,1,=,I,。,因为,AA,*,=,A,I,A,A,*,=,A,n,A,*,=,A,n-1,。,由,AA,1,=,I,得,(,AA,1,),T,=(,A,1,),T,A,T,=I,；,(,A,T,),1,=(,A,1,),T,。,由,AA,1,=I,得,|,A,|,A,1,|,=1,|,A,|,0,A,1,=,A,1,。,例,3,设方阵,B,为幂等矩阵,(,即,B,2,=,B,),A,=,I,+,B,证明：,A,是可逆阵，且,A,1,=(3,I,A,)/2,。,证,由,B,=,A,I,B,2,=(,A,I,),2,=,A,2,2,A,+,I,及,B,2,=,B,=,A,I,得,A,2,2,A,+,I,=,A,I,A,2,3,A,=,A,(,A,3,I,)=,2,I,即,A,(3,I,A,),/,2=,I,A,可逆，且,A,1,=(3,I,A,)/2,。,例,4,主对角元都是非零数的对角阵是可逆的，且,注意：,例,5,设,A,为,n,阶可逆对称,(,反对称,),矩阵，则,A,1,也是,对称,(,反对称,),的。,证,设,A,T,=,A,，则,设,A,T,=-,A,，则,例,6,已知,A,为非零,n,阶实矩阵，当,A,*,=,A,T,时，证明,:,A,为可逆矩阵。,证,要证,A,可逆，即证,A,0,。,由,A,T,A,=,A,*,A,=,A,I,，知,当,A,*,=,A,T,时，,A,0,A,T,A,0,例,7,若,A,B,C,D,均为,n,阶矩阵,且,ABCD,=,I,(,n,阶单位阵,),，,以下哪个成立？,BCDA,=,I,；,(B),CABD,=,I,；,(C),BACD,=,I,；,(D),CBAD,=,I,；,(E),BCAD,=,I,；,(F),CDAB,=,I,。,解：,根据矩阵乘法 满足结合律和定理,2.3,的推论，由于,ABCD,=,I,A,(,BCD,)=,I,(,BCD,),A,=,I,(A),成立。,(,AB,)(,CD,)=,I,(,CD,)(,AB,),=,I,CDAB,=,I,(F),成立。,例,8,已知,A,=diag(1,2,1),且,A*BA,=2,BA,8,I,求,B,。,解：,由,A*BA,2,BA,=,8,I,得,=,4,(,I+A,),1,=4 diag(2,1,2),1,(,A*,2,I,),BA,=,8,I,B,=,8(,A*,2,I,),1,A,1,=,8(,A,(,A*,2,I,),1,=,8(,A A,*,2,A,),1,=,8(,2,I,2,A,),1,=4diag(2,1,1,2,1,),，,所以，,B,=diag(2,4,2),例,9,设,A,可逆，且,A,*,B,=,A,1,+,B,，证明,B,可逆，当,时，求,B,。,解：,由,A,*,B,=,A,1,+,B,=,A,1,+,I B,得,(,A*,I,),B,=,A,1,，,因为,|,A*,I,|,B|,=,|A,1,|,0,，,所以，,|,B,|,0,，,B,可逆。,B,=,(,A*,I,),1,A,1,=(,A,(,A*,I,),1,=(,|,A,|,I,A,),1,例,10,已知：,n,阶矩阵,A,，,B,均可逆，证明：,(1)(,AB,)*,=,B,*,A,*;,(2)(,A*,)*,=|A|,n-2,A,证明：,作业,Ex.33,Ex.36,Ex.38,Ex.40(3),Ex.45,习题讲解,Ex.11,已知,为正交矩阵，试求,a,b,c,d,e,的值,.,解：,因为,Q,为正交矩阵,所以,