南航双语矩阵论matrix theory第4章部分习题参考答案

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南京航空航天大学Matrix-Theory双语矩阵论期末考试2015

南京航空航天大学Matrix-Theory双语矩阵论期末考试2015

NUAALet 3P (the vector space of real polynomials of degree less than 3) defined by(())'()''()p x xp x p x σ=+.(1) Find the matrix A representing σ with respect to the ordered basis [21,,x x ] for 3P .(2) Find a basis for 3P such that with respect to this basis, the matrix B representing σ is diagonal.(3) Find the kernel (核) and range (值域)of this transformation. Solution: (1)221022x x x x σσσ===+()()() 002010002A ⎛⎫⎪= ⎪ ⎪⎝⎭----------------------------------------------------------------------------------------------------------------- (2)101010001T ⎛⎫ ⎪= ⎪ ⎪⎝⎭(The column vectors of T are the eigenvectors of A)The corresponding eigenvectors in 3P are 1000010002T AT -⎛⎫⎪= ⎪ ⎪⎝⎭(T diagonalizes A ) 22[1,,1][1,,]x x x x T += . With respect to this new basis 2[1,,1]x x +, the representingmatrix of σis diagonal.------------------------------------------------------------------------------------------------------------------- (3) The kernel is the subspace consisting of all constant polynomials.The range is the subspace spanned by the vectors 2,1x x +-----------------------------------------------------------------------------------------------------------------------Let 020012A ⎛⎫⎪= ⎪ ⎪-⎝⎭.(1) Find all determinant divisors and elementary divisors of A .(2) Find a Jordan canonical form of A .(3) Compute At e . (Give the details of your computations.) Solution: (1)110020012I A λλλλ-⎛⎫ ⎪-=- ⎪ ⎪-⎝⎭,(特征多项式 2()(1)(2)p λλλ=--. Eigenvalues are 1, 2, 2.)Determinant divisor of order 1()1D λ=, 2()1D λ=, 23()()(1)(2)D p λλλλ==-- Elementary divisors are 2(1) and (2)λλ-- .---------------------------------------------------------------------------------------------------------------------- (2) The Jordan canonical form is100021002J ⎛⎫ ⎪= ⎪ ⎪⎝⎭--------------------------------------------------------------------------------------------------------------------------(3) For eigenvalue 1, 010010011I A ⎛⎫⎪-=- ⎪ ⎪-⎝⎭ , An eigenvector is 1(1,0,0)T p = For eigenvalue 2, 1102000010I A ⎛⎫⎪-= ⎪ ⎪⎝⎭, An eigenvector is 2(0,0,1)T p =Solve 32(2)A I p p -=, 331100(2)00000101A I p p --⎛⎫⎛⎫⎪ ⎪-== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭we obtain that3(1,1,0)T p =-101001010P ⎛⎫ ⎪=- ⎪ ⎪⎝⎭, 1110001010P -⎛⎫⎪= ⎪ ⎪-⎝⎭ 1At J e Pe P -=22210100110001000101000010tt t t e e te e ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭22220000t t t t t t e e e e tee ⎛⎫-⎪= ⎪ ⎪-⎝⎭ --------------------------------------------------------------------------------------------------------------------Suppose that ∈R A and O I A A =--65.(1) What are the possible minimal polynomials of A ? Explain.(2) In each case of part (1), what are the possible characteristic polynomials of A ? Explain.Solution:(1) An annihilating polynomial of A is 256x x --.The minimal polynomial of A divides any annihilating polynomial of A. The possible minimal polynomials are6x -, 1x +, and 256x x --.---------------------------------------------------------------------------------------------------------------(2) The minimal polynomial of A divides the characteristic polynomial of A. Since A is a matrix of order 3, the characteristic polynomial of A is of degree 3. The minimal polynomial of A and the characteristic polynomial of A have the same linear factors. Case 6x -, the characteristic polynomial is 3(6)x - Case 1x +, the characteristic polynomial is 3(1)x + Case 256x x --, the characteristic polynomial is 2(1)(6)x x +- or 2(6)(1)x x -+-------------------------------------------------------------------------------------------------------------------Let 120000A ⎛⎫=⎪⎝⎭. Find the Moore-Penrose inverse A +of A .Solution: ()12011200000A PG ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭1()(1,0)T T P P P P +-==, 111()250T T G G GG +-⎛⎫⎪== ⎪ ⎪⎝⎭110112(1,0)2055000A G P +++⎛⎫⎛⎫ ⎪⎪=== ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭也可以用SVD 求.------------------------------------------------------------------------------------------------------------------Part II (选做题, 每题10分)请在以下题目中(第6至第9题)选择三题解答. 如果你做了四题,请在题号上画圈标明需要批改的三题. 否则,阅卷者会随意挑选三题批改,这可能影响你的成绩.Let 4P be the vector space consisting of all real polynomials of degree lessthan 4 with usual addition and scalar multiplication. Let 123,,x x x be three distinct real numbers. For each pair of polynomials f and g in 4P , define 31,()()i i i f g f x g x =<>=∑.Determine whether ,f g <> defines an inner product on 4P or not. Explain.Let n n A ⨯∈R . Show that if x x A =)(σis the orthogonal projection fromn R to )(A R , then A is symmetric and the eigenvalues ofA are all 1’s and 0’s.n n A ⨯∈C . Show that x x A H is real-valued for all n C x ∈if and only if Ais Hermitian.Let n n B A ⨯∈C , be Hermitian matrices, and A bepositive definite. Show thatAB is similar to BA , and is similar to a real diagonal matrix.若正面不够书写,请写在反面.123()()()x x x x x x ---. Then ,0f f <>=. But 0f ≠. This does not define an inner product. For any x , ()()x x T A R A N A ⊥-∈=, ()x x 0T A A -=. Hence, T T A A A =. Thus. T A A =.From above, we have 2A A =. This will imply that λλ-2is an annihilating polynomial of A. The eigenvalue of A must be the roots of 02=-λλ. Thus, the eigenvalues of A are1’s and 0’s.See Thm 7.1.1, page 182. 也可以用其它方法.Since A is nonsingular, 1()AB A BA A -=. Hence, A is similar to BASince A is positive definite, there is a nonsingular hermitian matrix P such that H A PP =. 1()H H AB PP B P P BP P -==Since H P BP is Hermitian, it is similar to a real diagonal matrix.is similar to H AB P BP , H P BP is similar to a real diagonal matrix. Thus AB is similar to a real diagonal matrix.。

南航双语矩阵论期中考试卷mid-term-exam(2014)优选全文

南航双语矩阵论期中考试卷mid-term-exam(2014)优选全文

Mid-term Exam of Matrix Theory (2014)Preferentially Selected Five Questions (5×20 )Q1.Given A ∈P n ×n ,consider the following questions.1)If A is invertible,prove that A −1can be represented by the polynomial of A with degree less than n .2)For any positive integer k ∈N ,prove that A k can be represented by the polynomial of A with degree less than n .3)Especially A = 11221,find the representative polynomials of A −1and A 2014as men-tioned in 1)and 2).Q2.Denote A a linear transformation in R 3,α1,α2,α3the basis of R 3.Suppose that the representation matrix of A with respect to α1,α2,α3is A = 12020−2−2−1.1)Show that β1=α1,β2=α1+α2,β1=α1+α2+α3also form a basis of R 3.2)Determine the representative matrix of A with respect to β1,β2,β2.3)Find the eigenvalues and eigenvectors of A .Q3.Denote R [x ]3to be the vector space of zero and polynomials with degree less than 3.1)Determine the dimension of R [x ]3and give a basis of R [x ]3.2)Define the linear transformation D on R [x ]3,D (f (x ))=f (x ),∀f (x )∈R [x ]3.Show R (D )and ker(D ).3)Prove that D is not diagonalizable.4)Define the inner product on R [x ]3,(f,g )= 1−1f (x )g (x )dx,∀f (x ),g (x )∈R [x ]3,please Gram-Schmidt orthogonalize the basis given in 1).Q4.1)To the best of your knowledge about λ−matrix,determine if the following two matrices are similar or not,and give reason,1A =210021002 ,B = 2a 002a 002 .2)Denote V ={ a 11a 12a 21a 22∈R 2×2|a 11=a 22}.i)Find a basis of V and show the dimension.ii)Arbitrarily given A = a 11a 12a 21a 22 and B = b 11b 12b 21b 22in V ,define (A,B )=a 11b 11+2a 12b 12+a 21b 21.Please show that (A,B )is an inner product on V .Q5.Given A ∈C m ×n and b ∈C m ,please prove1)there exists a real number α>0such that A H A +αI is nonsingular;2)the solution to the least square problem min x ∈C n{ Ax −b 2+α x 2}is x ∗=(A H A +αI )−1A H b ,where · stands for the 2−norm in C m .Q6.Given A ∈R n ×n ,summarize the necessary and sufficient conditions of A to be di-agonalizable,and prove at least one of them.Determine if the matrix A given in Q2is diagonalizable or not.If yes,please explain why,if not,please give the Jordan canonical form of A .2。

研究生矩阵理论课后答案4,5章习题

研究生矩阵理论课后答案4,5章习题

2 1 − 2 3 1 0 4 1 1 0 −1 2 1 −1 0
0 5 0 1 1 0 4 1 1 0 −1 −2 0 −2 0

1 1 1 −2 −1 −1
0 5 0 1 1 0 4 1 1 0 1 2 0 2 0
同一向量的三种范数之间的大小关系 习题#5-4:对n维线性空间的任意向量x成 习题#5维线性空间的任意向量x #5
‖x‖∞ ≤‖x‖2 ≤‖x‖1 ≤ n‖x‖∞ ≤ n‖x‖2 ≤ n‖x‖1 ≤ n2‖x‖∞ ≤ …

证: |,…,|x ‖x‖∞= max{|x1|, ,|xn|} ≤(Σi=1n|xi|2)1/2 = ‖x‖2 |+…+|x ≤((|x1|+ +|xn|)2)1/2 = ‖x‖1 |,…,|x ≤ n max{|x1|, ,|xn|} = n‖x‖∞
习题#5是正定矩阵,x ,x∈ 习题#5-6A∈Cn×n是正定矩阵,x∈Cn #5
是向量范数. •证明:‖x‖=(x*Ax)1/2 是向量范数. 证明:‖x‖=(x
解1:因A是正定Hermite矩阵A,故存在可逆矩阵B 是正定Hermite矩阵A,故存在可逆矩阵B Hermite矩阵A,故存在可逆矩阵 使得A=B B.则 的上述表示式可写为: 使得A=B*B.则x的上述表示式可写为: (Bx)) ‖x‖=(x*Ax)1/2 =((Bx)*(Bx))1/2 =‖Bx‖2 其中‖‖ 是向量2 范数.再注意可逆矩阵B 其中‖‖2 是向量2-范数.再注意可逆矩阵B的性 Bx=0,即可直接推出非负性 即可直接推出非负性. 质:x=0 ⇔ Bx=0,即可直接推出非负性. ‖kx‖=‖B(kx)‖2=|k|‖Bx‖2=|k|‖x‖ 推出齐次性;三角不等式则由下式推出: 推出齐次性;三角不等式则由下式推出: ‖x+y‖=‖B(x+y)‖2≤‖Bx‖2+‖By‖2

南航矩阵论课后习题答案

南航矩阵论课后习题答案

南航矩阵论课后习题答案南航矩阵论课后习题答案矩阵论是数学中的一个重要分支,广泛应用于各个领域,包括物理学、工程学、计算机科学等等。

南航的矩阵论课程是培养学生数学思维和解决实际问题的重要环节。

在课后习题中,学生需要运用所学的矩阵理论知识,解答各种问题。

下面是南航矩阵论课后习题的一些答案和解析。

1. 已知矩阵A = [1 2 3; 4 5 6; 7 8 9],求A的逆矩阵。

解析:要求一个矩阵的逆矩阵,需要先判断该矩阵是否可逆。

一个矩阵可逆的充要条件是其行列式不为零。

计算矩阵A的行列式,得到det(A) = -3。

因此,矩阵A可逆。

接下来,我们可以使用伴随矩阵法求解逆矩阵。

首先,计算矩阵A的伴随矩阵Adj(A),然后将其除以行列式的值,即可得到逆矩阵。

计算得到A的伴随矩阵为Adj(A) = [-3 6 -3; 6 -12 6; -3 6 -3]。

最后,将伴随矩阵除以行列式的值,即可得到矩阵A的逆矩阵A^-1 = [-1 2 -1; 2 -4 2; -1 2 -1]。

2. 已知矩阵A = [2 1; 3 4],求A的特征值和特征向量。

解析:要求一个矩阵的特征值和特征向量,需要先求解其特征方程。

特征方程的形式为|A - λI| = 0,其中A为给定矩阵,λ为特征值,I为单位矩阵。

计算得到特征方程为|(2-λ) 1; 3 (4-λ)| = (2-λ)(4-λ) - 3 = λ^2 - 6λ + 5 = 0。

解这个二次方程,得到特征值λ1 = 1,λ2 = 5。

接下来,我们可以求解对应于每个特征值的特征向量。

将特征值代入(A - λI)x = 0,即可求解出特征向量。

对于特征值λ1 = 1,解得特征向量x1 = [1; -1];对于特征值λ2 = 5,解得特征向量x2 = [1; 3]。

3. 已知矩阵A = [1 2; 3 4],求A的奇异值分解。

解析:奇异值分解是将一个矩阵分解为三个矩阵的乘积:A = UΣV^T,其中U和V是正交矩阵,Σ是对角矩阵。

矩阵理论第4章习题解答 (2)

矩阵理论第4章习题解答 (2)

矩阵理论第四章习题解答1. 习题1问题描述已知矩阵A和B定义如下:A = [1, 2, 3][4, 5, 6][7, 8, 9]B = [9, 8, 7][6, 5, 4][3, 2, 1]求矩阵C = A + B。

解答我们可以直接对A和B对应位置的元素进行相加,得到矩阵C。

A +B = [1+9, 2+8, 3+7][4+6, 5+5, 6+4][7+3, 8+2, 9+1]计算结果为:[10, 10, 10][10, 10, 10]2. 习题2问题描述已知矩阵A和B定义如下:A = [1, 2, 3][4, 5, 6][7, 8, 9]B = [9, 8, 7][6, 5, 4][3, 2, 1]求矩阵D = A - B。

解答我们可以直接对A和B对应位置的元素进行相减,得到矩阵D。

A -B = [1-9, 2-8, 3-7][4-6, 5-5, 6-4][7-3, 8-2, 9-1]计算结果为:[-2, 0, 2][4, 6, 8]3. 习题3问题描述已知矩阵A和B定义如下:A = [1, 2, 3][4, 5, 6][7, 8, 9]B = [2, 0, 1][1, 2, 1][0, 1, 2]求矩阵E = A * B。

解答我们可以通过矩阵乘法的定义来计算E。

矩阵乘法的定义为:矩阵C的第i行第j列的元素等于矩阵A的第i行与矩阵B的第j列对应元素的乘积之和。

对于矩阵A和B,可以计算得到矩阵E。

E = [1*2+2*1+3*0, 1*0+2*2+3*1, 1*1+2*1+3*2][4*2+5*1+6*0, 4*0+5*2+6*1, 4*1+5*1+6*2][7*2+8*1+9*0, 7*0+8*2+9*1, 7*1+8*1+9*2]计算结果为:E = [4, 7, 8][10, 13, 16][16, 19, 22]4. 习题4问题描述已知矩阵A定义如下:A = [1, 2, 3][4, 5, 6][7, 8, 9]求矩阵F = A^T,其中A^T表示A的转置矩阵。

04南航戴华《矩阵论》第四章l矩阵的因子分解

04南航戴华《矩阵论》第四章l矩阵的因子分解

对初等下三角矩阵,当i <j 时,有
1 0 1 T T li 1,i Li (li ) L j (l j ) I li ei l j e j 1 l j 1, j 0 0 lni 0 1 l nj
4.1.3 Householder矩阵
取u = v = w, σ=2,并且w是单位向量,即 ||w|| =1,初等矩阵
H (w) E(w, w,2) I 2ww
H
(4.1.7)
称为Householder矩阵或初等Hermite矩阵。
定理4.1.2 Householder矩阵H(w)具有如下性质:
(U H AU U 1 AU B)
则称A正交(酉)相似于B。 定理4.5.1(Schur定理) 任何一个n 阶复矩阵A都酉相 似于一个上三角矩阵,即存在一个n 阶酉矩阵U 和 一个n阶上三角矩阵 R 使得
U AU R
H
(4.5.1)
其中R 的对角元是A 的特征值,它们可以按要求的 次序排列。
上(下)三角矩阵的性质
• 什么是矩阵的LU分解? • 矩阵的LU分解是否存在?如果存在, LU分解 是否唯一? • 如何计算矩阵的LU分解? • LU分解有什么应用?
定理4.3.1(LU分解定理)设 A 是 n 阶非奇异矩 阵,则 存在唯一的单位下三角矩阵L和上三角矩 阵U使得 A LU 的充分必要条件是A的所有顺序主子式均非零, 即
H H
mn
,则
AH A与AAH的特征值均为非负实数 ;
(2)
A A与AA 的非零特征值相同,并 且非零特征 值的个数(重特征值按重数计算 )等于rank( A).

南京航空航天大学MatrixTheory双语矩阵论期末考试

南京航空航天大学MatrixTheory双语矩阵论期末考试
(2) The Jordan canonical form is
--------------------------------------------------------------------------------------------------------------------------
(2) Find a basis for such that with respect to this basis,thematrixBrepresenting is diagonal.
(3) Find thekernel(核)andrange(值域)of this transformation.
Solution:
南京航空航天大学Matrix-Theory双语矩阵论期末考试
———————————————————————————————— 作者:
———————————————————————————————— 日期:
Part I (必做题,共5题,70分)
第1题(15分)
得分
Let denote the set of all real polynomials of degree less than 3 withdomain(定义域) .The addition and scalar multiplication are defined intheusual way.Definean inner product on by
第2题(15分)
得分
Let be the linear transformation on (the vector space of real polynomials of degree less than 3) defined by

矩阵理论第4章习题解答

矩阵理论第4章习题解答

第四章 习题解答1. 证明:实对称矩阵A 的所有特征值在区间[],a b 上的充要条件是对任何0a λ<,0λ-A E 是正定矩阵;而对任何0a λ<,0λ-A E 是负定矩阵.证:因为A 为实对称矩阵,所以存在正交矩阵Q ,使得{}12,,n diag λλλ T A =Q Q ,其中特征值[],i a b λ∈.{}010200,,n diag λλλλλλλ--- T A -E =Q Q ,所以对于00,0i a λλλ∀<->知A 为正定矩阵;00,0i b λλλ∀>-<知A 为负定矩阵. 2. 设A ,B 都是实对称矩阵, A 的一切特征值在区间[],a b 上, B 的一切特征值在区间[],c d 上. 证明: A+B 的特征值必在区间[],a c b d ++.证:设A ,B 的特征值分别为()()()12n b A A A a λλλ≥≥≥≥≥ , ()()()12n d B B B c λλλ≥≥≥≥≥ ,又因为A ,B 为实对称矩阵,所以A ,B 为Hermite 矩阵,由定理18知,A+B 的特征值()k λ+A B ,1,2,,k n ∀= . 有()()()()()1k n k k λλλλλ+≤+≤+A B A B A B .即()()()()()()()1k k n k k k a c c d b dλλλλλλλ+≤+≤+≤+≤+≤+≤+A A B A B A B A 3 设P 是酉矩阵,()1,,n A diag a a = ,证明PA 的特征值μ满足不等式m M μ≤≤,其中,{}min i im a =,{}max i iM a =.证:因为P 是酉矩阵,所以HP P E =,又因为()()HH H H PA PA A P PA A A ==,所以由Browne 定理知,PA 的特征值μ满足不等式minminmaxmaxiiiiμ=≤≤=而minmin i iia m ==,maxmax i iia M ==,所以 m M μ≤≤.4.用圆盘定理证明9121081110401001-⎡⎤⎢⎥⎢⎥⎢⎥-⎢⎥⎣⎦A =至少有两个实特征值. 证: A 的4个盖尔圆为{}1|94G z z =-≤,{}2|82G z z =-≤, {}3|41G z z =-≤,{}4|11G z z =-≤,它们构成的两个连通区域部分为1123S G G G = , 24S G =, 易知1S 与2S 都关于实轴对称, 因为实矩阵的复特征值必成对共轭出现, 所以2S 中含有A 的一个特征值, 而1S 中至少含有A 的一个实特征值, 因此A 中至少有两个实特征值. 5 参见课本135页中的例1. 6 用圆盘定理估计7-168-1678885⎡⎤⎢⎥-⎢⎥⎢⎥--⎣⎦A =的特征值和A 的谱半径, 然后选取一组正数123,,p p p 对A 的特征值作更细的估计. 解: A 的3个特征值在它的2个盖尔圆724z -≤,516z +≤得并集中, 且()31r A ≤. 因为矩阵A 有相同的主对角元素,所以,无法通过选取正数123,,p p p 给出更精细的估计.7证明141414141525115161636161171737⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A =的谱半径()1r <A . 证: 113:||44S z -≤,223:||55S z -≤,333:||66S z -≤,433:||77S z -≤,故矩阵A 的盖尔,圆盘位于单位圆内且只与单位圆交于1,又因为||0E A -≠,所以知()1r <A .8. 证明14141141251515161636161717147⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A =的谱半径()1r =A . 证: 113:||44S z -≤,223:||55S z -≤,333:||66S z -≤,443:||77S z -≤,故矩阵A 的盖尔,圆盘位于单位圆内且只与单位圆交于1,又因为()det 10=I -A , 所以()1r =A . 9.举例说明:(1)在有两个盖儿圆构成构成的连通部分中,可以在每一个盖儿圆中恰有一个特征值. (2)不一定每个盖尔圆中必有一个特征值.解:(1)如122-1⎛⎫ ⎪⎝⎭A =,故250λλ-=-=E A,1,2λ=(2)如1-0.80.50⎛⎫ ⎪⎝⎭A =,故20.40λλλ-=-+=E A,(1,211.2λ=±11.设()n,nij a =∈C A ,满足()1,2,,ij ij j ia a i n ≠>=∑ 则(1)A 可逆; (2)1det .nii ij j i i a a ≠=⎛⎫≥- ⎪⎝⎭∑∏A 证:(1)因为A 为严格对角占优矩阵,由定理4知,A 可逆。

矩阵论第四章作业答案

矩阵论第四章作业答案

第四章作业答案
第六题
证明:只需证明新定义的范数满足矩阵范数的四个性质即可!
非负性与齐次性是显然的。

(3)三角不等式性:
||||||()||||||||||||||||||||||b a a a a b b X Y B X Y C BXC BYC BXC BYC X Y +=+=+≤+=+
(4)相容性:
111111||||||||||||||||||||||||||||||||,||||1
||||||||||||||||||||b a a a a a a a a b a a b b
XY BXYC BXCC B BYC BXC C B BYC C B XY BXC BYC X Y ------==≤+++∴≤≤+ 都小于
所以是范数
第七题 如果看过第六题不会做第七题我也没办法了
第八题
证明:(2)如果A 可逆,则||||0A ≠ 11111||||||||||||||||
||||||||
E AA A A A A ----==≤∴≤
第九题
证明:(1)
2
2||||()()1||||1H U UU E U ρρ===∴=
(2)P83有证明
第十,十一题后面有详细证明,这里就不类
述了
第十四题
解:矩阵的1范数就是矩阵的列向量各数绝对值之和中最大的那个数矩阵的无穷范数就是矩阵的行向量各数绝对值之和中最大的那个数。

矩阵论 Matrix4-1

矩阵论 Matrix4-1

A–1 = A+;A–1L = (AHA)–1AH = A+; A–1R = AH(AAH)–1 =A+;若 A+,则A+是A{1}。
2、M-P 广义逆的惟一性 定理4.9 (P.98) 如果A有M-P广义逆,则A的M-P
广义逆是惟一的。
3、M-P广义逆的存在性及其求法
定理4.10(P.99)任何矩阵都有M-P广义逆。 求法: • 设A有满秩分解 A = BC,则有 A+ = CH (CCH )–1 (BH B)–1 BH 。 • (定理4.11) 设A有奇异值分解 : 1 r 0 H 0 H r ,则 A U V A V U 0 0 0 0


例题4
1 1 1 1 0 0 0 (1, 1) 0 2 1

Ir Ir Ar r n 0 ( I r , 0); ( I r , 0) 0 ; Ar ( F )行满秩, 0 ? Ar I r A A I , 0 A 0 0 r r r r ,0
§ 4. 1 矩阵的左逆与右逆
一、满秩矩阵和单侧逆 必要条件 1、左逆和右逆的定义 右逆存在 m r ( AC) r ( A) n 左逆存在 n r ( BA) r ( A) m 定义4.1 (P.93)
• A Cmn, B Cnm,BA=In,则称矩阵B 1 为矩阵A的左逆,记为 B = AL • A Cmn , C Cnm,AC=Im,则称矩阵C 1 为矩阵A的右逆,记为 C = AR
例题1 求矩阵A的左逆:A = A的右逆?不存在!

南京航空航天大学Matrix-Theory双语矩阵论期末考试2015

南京航空航天大学Matrix-Theory双语矩阵论期末考试2015

NUAALet 3P (the vector space of real polynomials of degree less than 3) defined by(())'()''()p x xp x p x σ=+.(1) Find the matrix A representing σ with respect to the ordered basis [21,,x x ] for 3P .(2) Find a basis for 3P such that with respect to this basis, the matrix B representing σ is diagonal.(3) Find the kernel (核) and range (值域)of this transformation. Solution: (1)221022x x x x σσσ===+()()() 002010002A ⎛⎫⎪= ⎪ ⎪⎝⎭----------------------------------------------------------------------------------------------------------------- (2)101010001T ⎛⎫ ⎪= ⎪ ⎪⎝⎭(The column vectors of T are the eigenvectors of A)The corresponding eigenvectors in 3P are 1000010002T AT -⎛⎫⎪= ⎪ ⎪⎝⎭(T diagonalizes A ) 22[1,,1][1,,]x x x x T += . With respect to this new basis 2[1,,1]x x +, the representingmatrix of σis diagonal.------------------------------------------------------------------------------------------------------------------- (3) The kernel is the subspace consisting of all constant polynomials.The range is the subspace spanned by the vectors 2,1x x +-----------------------------------------------------------------------------------------------------------------------Let 020012A ⎛⎫⎪= ⎪ ⎪-⎝⎭.(1) Find all determinant divisors and elementary divisors of A .(2) Find a Jordan canonical form of A .(3) Compute At e . (Give the details of your computations.) Solution: (1)110020012I A λλλλ-⎛⎫ ⎪-=- ⎪ ⎪-⎝⎭,(特征多项式 2()(1)(2)p λλλ=--. Eigenvalues are 1, 2, 2.)Determinant divisor of order 1()1D λ=, 2()1D λ=, 23()()(1)(2)D p λλλλ==-- Elementary divisors are 2(1) and (2)λλ-- .---------------------------------------------------------------------------------------------------------------------- (2) The Jordan canonical form is100021002J ⎛⎫ ⎪= ⎪ ⎪⎝⎭--------------------------------------------------------------------------------------------------------------------------(3) For eigenvalue 1, 010010011I A ⎛⎫⎪-=- ⎪ ⎪-⎝⎭ , An eigenvector is 1(1,0,0)T p = For eigenvalue 2, 1102000010I A ⎛⎫⎪-= ⎪ ⎪⎝⎭, An eigenvector is 2(0,0,1)T p =Solve 32(2)A I p p -=, 331100(2)00000101A I p p --⎛⎫⎛⎫⎪ ⎪-== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭we obtain that3(1,1,0)T p =-101001010P ⎛⎫ ⎪=- ⎪ ⎪⎝⎭, 1110001010P -⎛⎫⎪= ⎪ ⎪-⎝⎭ 1At J e Pe P -=22210100110001000101000010tt t t e e te e ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭22220000t t t t t t e e e e tee ⎛⎫-⎪= ⎪ ⎪-⎝⎭ --------------------------------------------------------------------------------------------------------------------Suppose that ∈R A and O I A A =--65.(1) What are the possible minimal polynomials of A ? Explain.(2) In each case of part (1), what are the possible characteristic polynomials of A ? Explain.Solution:(1) An annihilating polynomial of A is 256x x --.The minimal polynomial of A divides any annihilating polynomial of A. The possible minimal polynomials are6x -, 1x +, and 256x x --.---------------------------------------------------------------------------------------------------------------(2) The minimal polynomial of A divides the characteristic polynomial of A. Since A is a matrix of order 3, the characteristic polynomial of A is of degree 3. The minimal polynomial of A and the characteristic polynomial of A have the same linear factors. Case 6x -, the characteristic polynomial is 3(6)x - Case 1x +, the characteristic polynomial is 3(1)x + Case 256x x --, the characteristic polynomial is 2(1)(6)x x +- or 2(6)(1)x x -+-------------------------------------------------------------------------------------------------------------------Let 120000A ⎛⎫=⎪⎝⎭. Find the Moore-Penrose inverse A +of A .Solution: ()12011200000A PG ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭1()(1,0)T T P P P P +-==, 111()250T T G G GG +-⎛⎫⎪== ⎪ ⎪⎝⎭110112(1,0)2055000A G P +++⎛⎫⎛⎫ ⎪⎪=== ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭也可以用SVD 求.------------------------------------------------------------------------------------------------------------------Part II (选做题, 每题10分)请在以下题目中(第6至第9题)选择三题解答. 如果你做了四题,请在题号上画圈标明需要批改的三题. 否则,阅卷者会随意挑选三题批改,这可能影响你的成绩.Let 4P be the vector space consisting of all real polynomials of degree lessthan 4 with usual addition and scalar multiplication. Let 123,,x x x be three distinct real numbers. For each pair of polynomials f and g in 4P , define 31,()()i i i f g f x g x =<>=∑.Determine whether ,f g <> defines an inner product on 4P or not. Explain.Let n n A ⨯∈R . Show that if x x A =)(σis the orthogonal projection fromn R to )(A R , then A is symmetric and the eigenvalues ofA are all 1’s and 0’s.n n A ⨯∈C . Show that x x A H is real-valued for all n C x ∈if and only if Ais Hermitian.Let n n B A ⨯∈C , be Hermitian matrices, and A bepositive definite. Show thatAB is similar to BA , and is similar to a real diagonal matrix.若正面不够书写,请写在反面.123()()()x x x x x x ---. Then ,0f f <>=. But 0f ≠. This does not define an inner product. For any x , ()()x x T A R A N A ⊥-∈=, ()x x 0T A A -=. Hence, T T A A A =. Thus. T A A =.From above, we have 2A A =. This will imply that λλ-2is an annihilating polynomial of A. The eigenvalue of A must be the roots of 02=-λλ. Thus, the eigenvalues of A are1’s and 0’s.See Thm 7.1.1, page 182. 也可以用其它方法.Since A is nonsingular, 1()AB A BA A -=. Hence, A is similar to BASince A is positive definite, there is a nonsingular hermitian matrix P such that H A PP =. 1()H H AB PP B P P BP P -==Since H P BP is Hermitian, it is similar to a real diagonal matrix.is similar to H AB P BP , H P BP is similar to a real diagonal matrix. Thus AB is similar to a real diagonal matrix.。

南京航空航天大学07-08矩阵论答案(B)

南京航空航天大学07-08矩阵论答案(B)
T

= 5; A
= 23 ;
T 1 2
∵ λ ( A A) = {3, 5,15} , ∴ A 2 = [λmax ( A A)] = 15 。
的特征向量, (2)设 x ∈ C 是 A 相应于特征值 λ 的特征向量,∴ Ax = λ x , x ≠ 0 , )
n
两 边 取 矩 阵 范 数 导 出 的 C 上 向 量 范 数 可 得 : λ x = λ x = Ax ≤ A x ,

Ik
0 , In−k
使得 PAP
H
A11 = 0
=B, A22 − A A A12 0
H 12 −1 11
H − ∵ A11 > 0, A22 − A12 A111 A12 > 0,∴ B > 0, 从而有 A > 0 。
5 ∆ 1 = 5 > 0, ∆ 2 = 1 > 0 , ∆ 3 = A − B = 1 − t 2 > 0 4
即−
2 2 成立。 <t< 时 A > B 成立。 5 5
H
矩阵, (2)∵ A 是 Hermite 矩阵,∴ 存在酉矩阵 U ,使得 U AU = diag{λ1 , λ2 ,⋯ , λn } , ) 由此可知: 由此可知: λmin ( A) I ≤ A ≤ λmax ( A) I ,
共 3 页 ∴ ∀x ∈ C n , x ≠ 0 ,有 λmin ( A) ≤ R( x ) =
−1
第 3 页
x H Ax ≤ λmax ( A) 。 xH x
− 存在,构造可逆矩阵 (3)∵ A11 > 0,∴ A11 存在,构造可逆矩阵 P = ) − A H A− 1 12 11

矩阵论课后参考答案(第一二三四

矩阵论课后参考答案(第一二三四

矩阵为 A

1 1
18 22
15 20



T
在 基 1 (1,2,1) , 2 (3,1,2),

1
21,2)下的矩阵。
解:由题可知1,2,3 与1,2,3 时空间 L(F 3) 的两组基,则存在一个
过渡矩阵 C 使得
3 -1 2
2 1 2
1 0 0
0 1 0
0 r 2(2)r1 1
0 r3(1)r10
1
0
3 5 -1
2 5 0
1 2 -1
0 1 0
0 0 1
1r2



5 (1)r 3
1 3 2 1 0
0
1 3 2 1 0 0
r2r30 1 0 1 0 1 r3(1)r20 1 0 1 0 1
2
1 0 0 0 0 1 0 0
0 0 0 0
0
0
0
0

1
0
0
0


0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1
(3)解:同上理,对 AT A 分析可知其为一个上下成负对称的矩阵,
且对角元全为 0,则其维数为
dim(V ) (n 1) (n 2) 1 (n 1)((n 1) 1) n(n 1)
2
2
其基为 n(n 1) 个 n n 阶的矩阵,故基可写为
2
0 1 0 0 0 0 1 0
1 0
0 0
0 0
所以V1 V2 {0} 。
2)明显V1 V2 Fn

南航双语矩阵论matrix theory第3章部分习题参考答案

南航双语矩阵论matrix theory第3章部分习题参考答案

1
Exercise 8
Let S be the subspace of C[a, b] spanned by e x , xe x , and x 2 e x . Let D be the differentiation operation of S, i.e., D( f ) f ' . Find the matrix representing D with respect to [e x , xe x , x2 e x ] Solution
3
This will imply that k1 must be zero since n 1 ( v) 0 . By repeating the process above, we obtain that k0 , k1 , , kn 1 must be all zero. This proves that
Solution
(1) 1
1
1
( x)
1 / 2 0
( x) 1 0 1
Hence, A
1 1/ 2 1 0
1/ 2 1 1/ 2 0
4
If B is nonsingular, then AB B1 ( BA) B . AB and BA are similar.
A P diБайду номын сангаасg(1 , 2 , , n ) P1 then Ak P diag(1k , 2k ,
, n . Hence, diag(1 , 2 ,
, n k ) P1 .
Ak O implies that i 0 for i 1, 2,
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)
If i is a root of p( ) 0 , then p(i ) 0 . We obtain that eigenvalue of C T with eigenvector x (1, i ,, in 2 , in 1 )T .
Exercise 16
Let be an orthogonal transformation on a Euclidean space V (an inner product space over the real number field). If W is a -invariant subspace of V, show that the orthogonal complement of W is also -invariant. Proof Let V W W , where W is -invariant. Let {u1 , u2 ,, uk } be an orthonormal basis for
0 1 T C x 0 0 0 0 1 0 0 0 0 0 0 an 0 an 1 0 an 2 1 a1
T
i i 1 2 2 i i i n2 n 1 n 1 i i i n 1 n n 1 a a a p ( i n i n 1 i 1 i i
C T x i x . Then i is an
(b) If p( ) has n distinct roots, then all roots of p( ) are eigenvalues of C T . We obtain that the characteristic polynomial of C T and p( ) have the same n distinct roots. And also they have the same degree and the same leading coefficient. Hence, the characteristic polynomial of C T is the same as p( ) . Since C and C T have the same characteristic polynomial, we know that p( ) is the characteristic polynomial of C.
of
AT
are the nonzero vectors in
N ( AT ) {k (0,1)T | k R} . The intersection
N ( AT ) N ( A) {(0,0)T } . Hence, Aand AT do not have eigenvectors in common.
characteristic polynomials. Hence, Aand AT have the same eigenvalues. Aand AT do not necessarily have the same eigenvectors. For example,
2 1 2 0 T Let A , then A . 0 2 1 2 The eigenvectors of A are the nonzero vectors in N ( A) {k (1,0)T | k R} . The eigenvectors
Exercise 11
Let [u1 , u 2 ,, u n ] be an orthonormal basis for R n and let A be a linear combination of rank
T T 1 matrices u1u1 , u2uT 2 ,…, u n u n . If T T A c1u1u1 c2u2uT 2 cn u n u n
Show that A is a symmetric matrix with eigenvalues c1, c2 ,, cn and that u i is an eigenvector belonging to ci for each i. Proof
T T AT (c1u1u1 c2u2uT 2 cn u n u n )
Exercise 6
Let Q be a unitary or orthogonal matrix. (a) Show that if is an eigenvalue of Q, then 1 (b) Show that |det(Q)|=1. Proof (a) Let x be an eigenvector
k 1 for k 1, 2,3 .If the eigenvalue of Q are all real, then
k 1 or 1 k 1, 2,3 .
Since det(Q) 1 , 12 3 1 . Hence, the possible triples of eigenvalues (1 , 2 , 3 ) with
ci ui (uT i u i ) ci u i
Hence, u i is an eigenvector belonging to ci for each i.
Exercise 13
Let A and B be n n matrices. Show that (a) If is a nonzero eigenvalue of AB, then it is also an eigenvalue of BA. (b) If 0 is an eigenvalue of AB, then 0 is also an eigenvalue of BA. Proof (a) If is a nonzero eigenvalue of AB, then there is a nonzero vector x such that ABx x . From ABx x , we see that Bx 0 . Since BA( Bx) Bx , we obtain that Bx 0 is an eigenvector of BA corresponding to the eigenvalue .Hence, is also an eigenvalue of BA. (b) If 0 is an eigenvalue of AB,then det( AB) 0 . Hence, det( BA) det( AB) 0 . Thus, 0 is an eigenvalue of BA.ofQ源自corresponding
to
. Then
Qx x .
x x Qx x
hence, 1 (b) QH Q I , det(Q H ) det(Q) 1 , det(Q) det(Q) 1 . Hence, |det(Q)|=1.
Exercise 9
T T T T T c1 (u1u1 ) c2 (u2uT 2 ) cn (u n u n )
T T c1u1u1 c2u2uT 2 cn u n u n A
Hence, A is a symmetric matrix.
T T Aui (c1u1u1 c2 u 2 uT 2 cn u n u n )u i T T c1u1 (u1 ui ) c2u 2 (uT 2 u i ) cn u n (u n u i )
1
1 2 3 are
(1,-1,-1),(1,1,1). (b) If the eigenvalues 2 and 3 are complex, then 2 and 3 must be the conjugate of each
other. 2 3 =1 . Hence, 1 =1 since 12 3 1 .
(The matrix C is called the companion matrix of p( x) .) (a) Show that if i is a root of p( ) 0 then i is an eigenvalue of C T with eigenvector x (1, i ,, in 2 , in 1 )T . (b) Use part (a) to show that if p( ) has n distinct roots then p( ) is the characteristic polynomial of C. (The result is true even if all the eigenvalues of p( ) are not distinct.) Hint: C and C T have the same characteristic polynomial. Proof (a)
i 1 j 1 i 1 j 1 n n n n
Hence, tr ( AB) tr ( BA) .
2
tr ( AB BA) tr ( AB) tr ( BA) 0 , tr ( I ) n
It is impossible to have matrices A and B such that AB BA I .
Exercise 15
Let p( x) x n a1 x n 1 an be a polynomial of degree n 1 , and let
0 1 C 0 0 an 0 0 0 an 1 1 0 0 an 2 0 0 1 a1 0 0 0
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