南航戴华《矩阵论》第五章Hermite矩阵与正定矩阵

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戴华《矩阵论》第一章线性空间与内积空间

这说明，维数是有限维线性空间的唯一的本质特征。在同构的意义下，n维向量空间Pn并不只是线性空间V 的一个特殊例子，而是所有的n维线性空间的代表。即每一个
1 0 C1 0 1 1 0 0 1 0 1 1 0 0 1 1 0
而基 ( III ) 到基 ( II ) 的过渡矩阵为
1 1 C2 1 1 1 1 1 0 1 1 0 0 1 0 0 0
所以
( A , A2 , A3 , A4 ) ( E11 , E12 , E21 , E22 )C1 1 ( B1 , B2 , B3 , B4 ) ( E11 , E12 , E21 , E22 )C2
dim(V1 V2 ) dim(V1 ) dim(V2 ) dim(V1 V2 ).
在维数公式中，和空间的维数不大于子空间维数之和。那么何时等号成立呢？
V1 , V2 是数域 P 上线
性空间 V 的两个有限维子空间，则它们的交与和
例1.4.6 设 S , K 分别是 n 阶实对称矩阵和反对称矩阵的全体。显然容易证明 S , K 均为线性空间 R nn 的子
( III )
显然
1 A1 0 0 E11 E22 1 1 0 ( E11 , E12 , E21 , E22 ) 0 1
类似地，
1 A2 0 0 E11 E22 1 1 0 ( E11 , E12 , E21 , E22 ) 0 1 0 1 ( E11 , E12 , E21 , E22 ) 1 0
证明：
1 0 取1= 0 0
0 1 3= 0 0 2= 0 1 1 0

南航双语矩阵论matrixtheory第五章部分习题参考答案

第五章部分习题参考答案#2. Find determinant divisors and elementary divisors of each of the following matrices.(a) 1000100015432λλλλ-⎛⎫ ⎪-⎪ ⎪- ⎪+⎝⎭ (b)001010100000λλλλ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭Solution（a ） 100010()0015432A λλλλλ-⎛⎫ ⎪- ⎪= ⎪- ⎪+⎝⎭det (())A λ4322345λλλλ=++++100det 10101λλ-⎛⎫⎪-=- ⎪ ⎪-⎝⎭. Hence, the determinant divisors are 123()()()1D D D λλλ===,4324()2345D λλλλλ=++++. Invariant divisor are 123()()()1d d d λλλ===,4324()2345d λλλλλ=++++Unfortunately, it is not easy to factorize 4324()2345d λλλλλ=++++ by hand. With the help of Maple or Matlab, we can see that ()A λ has four distinct linear elementary divisors. (b) 44()D λλ=, 123()()()1D D D λλλ===. There is a unique elementary divisor 4λ #3. Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , a a B a εε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭ be n n ⨯ matrices, where 0ε≠. Show that A and B are similar.Proof The Smith normal forms of both I A λ- and I B λ-are11()n a λ⎛⎫ ⎪⎪ ⎪ ⎪-⎝⎭. A and B have the same set of elementary divisors. Hence they are similar to each other. #4. Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , 11a a B a ε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭be n n ⨯ matrices, where 0ε≠. Show that A and B are NOT similar. ProofThe determinant of I A λ- is ()n a λ- . The determinant of I B λ- is ()n a λε--. A and B have distinct characteristic polynomials. Hence, they are not similar.#11. How many possible Jordan forms are there for a 66⨯ complex matrix with characteristic polynomial 42(2)(1)x x +-?Solution The possibilities for the sets of elementary divisors are { 42(2),(1)x x +-}, {4(2),(1),(1)x x x +--}{32(2),(2),(1)x x x ++-}, {3(2),(2),(1),(1)x x x x ++--} {222(2),(2),(1)x x x ++-}, {22(2),(2),(1),(1)x x x x ++--},{22(2),(2),(2),(1)x x x x +++-}, {2(2),(2),(2),(1),(1)x x x x x +++--}{2(2),(2),(2),(2),(1)x x x x x ++++-}, {(2),(2),(2),(2),(1),(1)x x x x x x ++++--}. For each set of elementary divisors, there is a Jordan canonical form up to similarity. There are 10 Jordan canonical forms up to similarity.#12. Classify up to similarity all 33⨯ complex matrices A such that 3A I =. Solution An annihilating polynomial of A is 321(1)()()x x x x ωω-=---, where ω A is diagonalizable.The possibilities for the minimal polynomial of A are1x -, x ω-, 2x ω-;(1x -)(x ω-), (x ω-)(2x ω-), (1x -)(2x ω-);2(1)()()x x x ωω---Up to similarity, all 33⨯ complex matrices A are100010001⎛⎫ ⎪ ⎪ ⎪⎝⎭, 000000ωωω⎛⎫⎪ ⎪ ⎪⎝⎭, 222000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 10001000ω⎛⎫⎪ ⎪ ⎪⎝⎭, 1000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 22000000ωωω⎛⎫ ⎪⎪ ⎪⎝⎭, 2000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭;221000000ωω⎛⎫⎪ ⎪ ⎪⎝⎭,210001000ω⎛⎫⎪ ⎪ ⎪⎝⎭21000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭#14. If N is a nilpotent (幂零的) 33⨯ matrix over C , prove that 21128A I N N =+- satisfies2A I N =+, i.e., A is a square root of I N +. Use the binomial series for 1/2(1)t + to obtain asimilar formula for a square root of I N +, where N is any nilpotent n n ⨯ matrix over C .Use the result above to prove that if c is a non-zero complex number and N is a nilpotent complex matrix, then cI N +has a square root. Now use the Jordan form to prove that every non-singular complex n n ⨯ matrix has a square root.Solution If N is an n n ⨯ matrix and k N O =, then k x is an annihilating polynomial for N . The minimal polynomial of N must be of the form p x , where p n ≤ and p k ≤ since the minimal polynomial of a matrix divides its characteristic polynomial. Thus, n N O =.(1) If N is a nilpotent 33⨯ matrix, then 3N O =. By straightforward computation, we can verify that 2A I N =+.(2) If N is an n n ⨯ nilpotent matrix, n N O =.1/22111111(1)(1)((1)1)122222(1)122!(1)!n n t t t t n -----++=+++++- 1/22111111(1)(1)((1)1)122222()22!(1)!n n I N I N N N n -----++=++++-(3) Since1N c is a nilpotent matrix, 1I N c + has a square root 1/21()I N c+. cI N + has a square root 1/21/21()c I N c+.(4) Suppose that 12121()0()000()r d d d r J J P AP J J λλλ-⎛⎫ ⎪⎪==⎪ ⎪ ⎪⎝⎭. Then each ()k d k J λ has asquare root 1/2()k d k J λ since ()k d k J λ is of the form k I N λ+, where 0k λ≠ because A is nonsingular and N is nilpotent.Let 121/211/2211/2()000()000()r d d d r J J B P P J λλλ-⎛⎫⎪⎪=⎪ ⎪⎪⎝⎭, then 2B A =. Hence, A has a squareroot.#20. Prove that the minimal polynomial of a matrix is equal to the characteristic polynomial if andonly if the elementary divisors are relatively prime in pairs.Proof Suppose that a Jordan canonical form of A is1212()000()000()r d d d r J J J J λλλ⎛⎫⎪ ⎪=⎪ ⎪ ⎪⎝⎭(where 12,,,r λλλ are not necessarily distinct. Each ()i d i J λ is a Jordan block.)The minimal polynomial of A is the same as that of J . The characteristic polynomial of A is the same as that of J . The elementary divisors of A are 11()d λλ-, , ()rd r λλ-The minimal polynomial of ()i d i J λ is ()i d i λλ-. The minimal polynomial of J is the least common multiple (最小公倍式) of 11()d λλ-, , ()rd r λλ-. The characteristicpolynomial of J is 1212()()()()rd d d r p λλλλλλλ=--- .The least common divisor of 11()d λλ-, , ()rd r λλ- is equal to the product of11()d λλ-, , ()r d r λλ- if and only if ()j dj λλ-and ()k d k λλ-are relatively prime forj k ≠. Thus the minimal polynomial of a matrix is equal to the characteristic polynomial ifand only if the elementary divisors are relatively prime in pairs.。

戴华《矩阵论》线性空间与内积空间PPT精品文档

组 Ax 的一个基础解系，为 Ax b 的一
个特解。
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向量的线性相关性：
线性代数中关于向量的线性组合、线性表示、线性相关、线性无关、秩等定义和结论都可以推广到一般线性空间。
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证明：取k1 ，k2 ，k3∈R，令 k11+k22+k33
k 1 1 00 0 k2 1 11 0 k3 1 01 0 0 00 0 则有k1-k2=0, k2 +k3=0
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从而 ( B 1 , B 2 , B 3 , B 4 ) ( E 1 1 ,E 1 2 ,E 2 1 ,E 2 2 ) C 2
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由题，在基 1,2,3下的坐标为 x(3,2,4)T
而且，基 1,2,3 到基 1,2,3的过渡矩阵为
1 2 4
所以
P
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0 0 1
1 2 4 3 2 3
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例1.3.5 已知矩阵空间 R 2 2 的两组基：
1 0
（a,bR）都有a+bi=(1,i)( a ),所以(a,b) T即为k的坐
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b
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例 1.3.2 实数域 R上的线性空间R [x]n中的向量组 1，x， x2 ,… xn-1

矩阵论教学大纲南航

矩阵论教学大纲南航矩阵论教学大纲南航矩阵论作为数学中的一个重要分支，具有广泛的应用领域，对于提高学生的数学思维能力和解决实际问题具有重要意义。

南航作为一所具有优秀数学学科传统的高校，其矩阵论教学大纲设计也备受关注。

首先，矩阵论教学大纲应该明确教学目标。

在南航的矩阵论教学中，学生应该通过学习矩阵的基本概念和性质，掌握矩阵运算的方法和技巧，了解矩阵的特征值和特征向量的相关理论，掌握矩阵的相似性和对角化等重要概念和方法。

同时，教学大纲还应该注重培养学生的数学建模能力和解决实际问题的能力，通过实例和应用案例的讲解，引导学生将矩阵论的知识应用到实际问题中。

其次，矩阵论教学大纲应该合理安排教学内容。

南航的矩阵论教学大纲可以从矩阵的基本概念和性质开始，逐步引入矩阵的运算法则和矩阵的特殊类型，如对角矩阵、上三角矩阵等。

在此基础上，可以进一步介绍矩阵的特征值和特征向量的相关理论和计算方法，并引入矩阵的相似性和对角化的概念。

此外，还可以通过实例和应用案例，讲解矩阵论在线性方程组、最小二乘法、网络分析等领域的应用，以培养学生的应用能力和解决实际问题的能力。

再次，矩阵论教学大纲应该注重教学方法的创新和教学手段的多样性。

南航可以通过采用多媒体教学、案例分析、互动讨论等教学方法，激发学生的学习兴趣和主动性。

同时，可以鼓励学生参与小组讨论和实践项目，培养学生的团队合作和创新能力。

此外，南航还可以充分利用现代教育技术手段，如网络教学平台、虚拟实验室等，提供丰富的教学资源和学习工具，使学生能够在不同的学习环境中进行自主学习和实践探索。

最后，矩阵论教学大纲应该注重评估和反馈机制的建立。

南航可以通过定期的作业、小测验和期末考试等方式，对学生的学习情况进行评估和反馈。

同时，可以建立学生和教师之间的互动平台，及时解答学生的疑问和问题，引导学生进行深入的学习和思考。

此外，南航还可以组织学术交流和学术竞赛等活动，激发学生的学习热情和竞争意识，提高学生的学习效果和能力。

研究生矩阵理论课后答案第5章

按范数收敛
定义:赋范空间V的序列{x(n)|n=1,2,…}按范数 ‖‖α收敛于aV,如果 limn‖x(n)-a‖α=0 命题:对赋范空间V的任意两个等价向量范数 ‖‖α, ‖‖β, 都有 limn‖x(n)-a‖α=0 limn‖x(n)-a‖β=0 (即按任意两个向量范数的收敛实质上等价) 因 0 limn‖x(n)-a‖α d limn‖x(n)-a‖β 0 limn‖x(n)-a‖β(1/c)limn‖x(n)-a‖α
1=|yk|(i=1n|yi|p)1/p =‖y‖p n1/p (*) (i|yi|=|xi|/|xk|1) 1=limp1limp‖y‖p limpn1/p=n0=1 1=limp‖y‖p=limp‖x‖p/‖x‖ ‖x‖=limp‖x‖p
同一向量的三种范数之间的大小关系
Frobenius 矩阵范数
例5.2.2:矩阵的Frobenius范数定义为 ‖A‖F=(i=1mj=1n|aij|2)1/2. (ACmn的向量2-范数蕴含前3条公理)不难证明4 条范数公理全部满足.因非负性和齐次性是显然的;③的证明见课本.我们只讲④的证明. ‖AB‖F2=i=1mj=1n|k=1paikbkj|2 i=1mj=1n((k=1p|aik|2)(k=1p|bkj|2))(C-S不等
则

n
1 ak 1 bk a k bk a b p q q b p a
1 a k bk a b k 1 pa
p

n k 1
ak
p
1 qb
q

b k 1 k
n q
1 1 ab ab q xn|}=|k‖x‖; ‖x+y‖= max{|x1+y1|,…,|xn+yn|} max{|x1|+|y1|,…,|xn|+|yn|} max{|x1|,…,|xn|}+max{|y1|,…,|yn|} =‖x‖+‖y‖

南航双语矩阵论matrix theory第5章部分习题参考答案

第五章部分习题参考答案Exercise 2Find determinant divisors and elementary divisors of each of the following matrices.(a) 1000100016733λλλλ-⎛⎫ ⎪- ⎪ ⎪- ⎪--+⎝⎭ (b)001010100000λλλλ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭ Solution（a ） 100010()0016733A λλλλλ-⎛⎫ ⎪-⎪= ⎪- ⎪--+⎝⎭det (())A λ432323376(1)(46)λλλλλλλλ=+--+=-++-222(1)(56)(1)(2)(3)λλλλλλ=-++=-++There is a 3x3 submatrix whose determinant is100det 10101λλ-⎛⎫ ⎪-=- ⎪ ⎪-⎝⎭. Hence, the determinant divisors are 123()()()1D D D λλλ===,4324()3376D λλλλλ=+--+. Invariant divisor are 123()()()1d d d λλλ===,4324()3376d λλλλλ=+--+ The elementary divisors of ()A λ are 2(1)λ-, 2λ+, 3λ+(b) 44()D λλ=, 123()()()1D D D λλλ===. There is a unique elementary divisor 4λExercise 3Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , a a B a εε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭be n n ⨯ matrices, where 0ε≠. Show that A and B are similar.Proof The Smith normal forms of both I A λ- and I B λ-are11()n a λ⎛⎫ ⎪⎪ ⎪ ⎪-⎝⎭.A andB have the same set of elementary divisors. Hence they are similar to each other.Exercise 4Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , 11a a B a ε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭be n n ⨯ matrices, where 0ε≠. Show that A and B are NOT similar.ProofThe determinant of I A λ- is ()n a λ- . The determinant of I B λ- is ()n a λε--. A and B have distinct characteristic polynomials. Hence, they are not similar.Exercise 6For each of the following matrices, find the Jordan Canonical Form J and matrix P such that 1P AP J -=(a) 040140122-⎛⎫ ⎪- ⎪ ⎪--⎝⎭(b)134478677-⎛⎫⎪- ⎪ ⎪-⎝⎭Solution(a)40140122I A λλλλ⎛⎫ ⎪-=-+ ⎪ ⎪-+⎝⎭Determinant divisors are 33()det()(2)D I A λλλ=-=+, 2()det()(2)D I A λλλ=-=+,1()1D λ= Invariant divisors are 23()(2)d λλ=+, 2()(2)d λλ=+, 1()1d λ= Elementary divisors are 2(2)λ+, (2)λ+A Jordan canonical form is 210020002-⎛⎫⎪- ⎪ ⎪-⎝⎭. Let123(,,)p p p P =, then1121233222p p p p p p p A A A =-=-=-Solving (2)x 0A I +=, that is, 123240012001200x x x -⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪-= ⎪⎪ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭, we obtain that {1p ,3p } form abasis for (2)((0,0,1),(2,1,0))T T N I A Span -= . Let 1(2,1,0)(0,0,1),p T T a b =+ To obtain 2p , we solve the system212(2)p p a A I a b ⎛⎫⎪-== ⎪ ⎪⎝⎭, that is,1232402120120y a y a y b -⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪-= ⎪⎪ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭. This system is consistent only if a b =. Let 1a b ==. Then 1(2,1,1)p T =, We solve the system above for vector2(1,0,0)p T =. Take 3(0,0,1)p T =.210100101P ⎛⎫ ⎪= ⎪ ⎪⎝⎭(b) 134478677I A λλλλ--⎛⎫ ⎪-=-+- ⎪ ⎪--⎝⎭23()det()(1)(3)D I A λλλλ=-=+-, 2()1D λ= , 1()1D λ=Invariant divisors are 23()(1)(3)d λλλ=+-, 2()1d λ=, 1()1d λ= Elementary divisors are 2(1)λ+, (3)λ-A Jordan canonical form is 110010003-⎛⎫⎪- ⎪ ⎪⎝⎭. Let123(,,)p p p P =, then11212333p p p p p p p A A A =-=-= 3(1,2,2)p T =, 1(1,2,1)p T =, 2(1,1,0)p T =--111212102P -⎛⎫⎪=- ⎪ ⎪⎝⎭Exercise 8Show that if p A I = for some positive integer p , then A is similar to a diagonal matrix over the complex number field.Proof Since p A I =, 1p x - is an annihilating polynomial. The minimal polynomial ()m x of A must divide 1p x -. Since the polynomial 1p x - has only single roots(单根)，()m x has only single roots. Therefore, by Theorem 5.2.7 (see lecture notes p124), matrix A is diagonalizable.Exercise 9Prove that if an n n ⨯ matrix satisfies 256A A I -=then A is diagonalizable.Proof Since 256A A I -=, 256x x -- is an annihilating polynomial of A . . The minimal polynomial ()m x of A must divide 256x x --. Since 256(6)(1)x x x x --=-+, the minimal polynomial must be a product of distinct linear factors. By Theorem 5.2.7 (see lecture notes p124), matrix A is diagonalizable.Exercise 10Show that if A is nonsingular, then 1A - can be written as a polynomial of A .Proof Let 1110()n n n p x c c c λλλ--=++++ be the characteristic polynomial of A . The constant term of ()p x must not be zero since A is nonsingular. By Cayley-Hamilton Theorem,()p A O =. That is, 1110n n n A c A c A c I O --++++= . Thus,1121101()n n n A A c A c I c ----=-+++ , which is a polynomial of A .Exercise 11How many possible Jordan forms are there for a 66⨯ complex matrix with characteristic polynomial 42(2)(1)x x +-?Solution The possibilities for the sets of elementary divisors are { 42(2),(1)x x +-}, {4(2),(1),(1)x x x +--}{32(2),(2),(1)x x x ++-}, {3(2),(2),(1),(1)x x x x ++--} {222(2),(2),(1)x x x ++-}, {22(2),(2),(1),(1)x x x x ++--},{22(2),(2),(2),(1)x x x x +++-}, {2(2),(2),(2),(1),(1)x x x x x +++--}{2(2),(2),(2),(2),(1)x x x x x ++++-}, {(2),(2),(2),(2),(1),(1)x x x x x x ++++--}. For each set of elementary divisors, there is a Jordan canonical form up to similarity. There are 10 Jordan canonical forms up to similarity.Exercise 12Classify up to similarity all 33⨯ complex matrices A such that 3A I =.Solution An annihilating polynomial of A is 321(1)()()x x x x ωω-=---, where ω= A is diagonalizable.The possibilities for the minimal polynomial of A are1x -, x ω-, 2x ω-;(1x -)(x ω-), (x ω-)(2x ω-), (1x -)(2x ω-);2(1)()()x x x ωω---Up to similarity, all 33⨯ complex matrices A are100010001⎛⎫ ⎪ ⎪ ⎪⎝⎭, 000000ωωω⎛⎫⎪ ⎪ ⎪⎝⎭, 222000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 10001000ω⎛⎫⎪ ⎪ ⎪⎝⎭, 1000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 22000000ωωω⎛⎫ ⎪⎪ ⎪⎝⎭, 2000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭;221000000ωω⎛⎫⎪ ⎪ ⎪⎝⎭,210001000ω⎛⎫⎪ ⎪ ⎪⎝⎭21000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭Exercise 14If N is a nilpotent (幂零的) 33⨯ matrix over C , prove that 21128A I N N =+- satisfies2A I N =+, i.e., A is a square root of I N +. Use the binomial series for 1/2(1)t + to obtain asimilar formula for a square root of I N +, where N is any nilpotent n n ⨯ matrix over C .Use the result above to prove that if c is a non-zero complex number and N is a nilpotent complex matrix, then cI N +has a square root. Now use the Jordan form to prove that every non-singular complex n n ⨯ matrix has a square root.Solution If N is an n n ⨯ matrix and k N O =, then k x is an annihilating polynomial for N . The minimal polynomial of N must be of the form p x , where p n ≤ and p k ≤ since the minimal polynomial of a matrix divides its characteristic polynomial. Thus, n N O =.(1) If N is a nilpotent 33⨯ matrix, then 3N O =. By straightforward computation, we canverify that 2A I N =+.(2) If N is an n n ⨯ nilpotent matrix, n N O =.1/22111111(1)(1)((1)1)122222(1)122!(1)!n n t t t t n -----++=+++++- 1/22111111(1)(1)((1)1)122222()22!(1)!n n I N I N N N n -----++=++++-(3) Since1N c is a nilpotent matrix, 1I N c + has a square root 1/21()I N c+. cI N + has a square root 1/21/21()c I N c+.(4) Suppose that 12121()00()000()r d d d r J J P AP J J λλλ-⎛⎫ ⎪⎪==⎪ ⎪ ⎪⎝⎭. Then each ()k d k J λ has asquare root 1/2()k d k J λ since ()kd k J λ is of the form k I N λ+, where 0k λ≠ because A is nonsingular and N is nilpotent.Let 121/211/2211/2()000()000()r d d d r J J B P P J λλλ-⎛⎫⎪⎪=⎪ ⎪⎪⎝⎭, then 2B A =. Hence, A has a squareroot.Exercise 20Prove that the minimal polynomial of a matrix is equal to the characteristic polynomial if andonly if the elementary divisors are relatively prime in pairs.Proof Suppose that a Jordan canonical form of A is1212()000()000()r d d d r J J J J λλλ⎛⎫ ⎪⎪= ⎪ ⎪⎪⎝⎭(where 12,,,r λλλ are not necessarily distinct. Each ()id i J λ is a Jordan block.)The minimal polynomial of A is the same as that of J . The characteristic polynomial of A is the same as that of J . The elementary divisors of A are 11()d λλ-, , ()rd r λλ-The minimal polynomial of ()id i J λ is ()id i λλ-. The minimal polynomial of J is theleast common multiple (最小公倍式) of 11()d λλ-, , ()rd r λλ-. The characteristicpolynomial of J is 1212()()()()rd d d r p λλλλλλλ=--- .The least common divisor of 11()d λλ-, , ()rd r λλ- is equal to the product of11()d λλ-, , ()r d r λλ- if and only if ()j dj λλ-and ()k d k λλ-are relatively prime for j k ≠. Thus the minimal polynomial of a matrix is equal to the characteristic polynomial ifand only if the elementary divisors are relatively prime in pairs.。

Hermite区间矩阵的全正定性

Total Positive Deﬁniteness of Hermite Interval MatricesJunwei Shao,Xiaorong HouFaculty of Science,Ningbo University,Zhejiang,PRCE-mail:junweishao@,houxiaorong@AbstractThe deﬁnition of total positive deﬁniteness of Hermite interval matrices is given and asuﬃcient and necessary criterion is derived to verify this property.Key words:Interval matrices;Hermite interval matrices;Total positive deﬁnitenessAMS subject classiﬁcation(2000):15A18,65Y991IntroductionDetermining the total positive deﬁniteness(and related properties)of symmetric interval matrices plays important roles in global optimization problems[1][2].For cases in real number ﬁeld,papers[3][4]gave a suﬃcient and necessary criterion.In this paper,we mainly discussed the problem in complex numberﬁeld,and gave the deﬁnition of total positive deﬁniteness of Hermite interval matrices and its suﬃcient and necessary criterion which only needed to check positive deﬁniteness of4n−1(n−1)!Hermite vertex matrices for a n×n Hermite interval matrix.2Some notationsWeﬁrst introduce following deﬁnitions and notations according to paper[4].For a square real matrix A,we denote its transpose by A T,its spectral radius byρ(A), and its absolute matrix by|A|=(|A ij|),where A ij represents the element in i-th row and j-th column of Matrix A.For a n×n real symmetric matrix,eigenvalues are all real,we denote them from large to small byλ1(A)≥λ2(A)≥···≥λn(A).And matrix inequalities as A≤B or A<B are to be understood componentwise.Let A c and∆be n×n real matrices,and∆≥0,we deﬁne the following setA I=[A,A]=[A c−∆,A c+∆]={A:A c−∆≤A≤A c+∆}(1) as n×n real interval matrix.When some matrix A in A I satisfying A ij=(A)ij or A ij=(A)ij, i,j=1,...,n,then we call A a vertex matrix of A I.From this deﬁnition,we know that A I has at most2n2vertex matrices.When A c,∆are symmetric matrices,we call A I a real symmetric interval matrix,it has at most2n(n+1)/2symmetric vertex matrices.With each real symmetric matrix A I=[A c−∆,A c+∆],we shall associate the symmetric interval matrixA I s=[A c−∆ ,A c+∆ ](2)whereA c=12A c+A T c,∆ =12∆+∆here we know,if A∈A I,then12A+A T∈A I s.Now we introduce following setY={z:z∈R n,|z j|=1,j=1,...n},its cardinal number is2n.For each z∈Y,we deﬁne T z as the n×n diagonal matrix whose diagonal vector is z,and letA z=A c−T z∆T z(3)Then,for each pair i,j,we have(A z)ij=(A c−∆)ij if z i z j=1;and(A z)ij=(A c+∆)ij if z i z j=−1.So A z∈A I,in fact,A z is a vertex matrix of A I.Since A−z=A z,{A z|z∈Y} have at most2n−1diﬀerent matrices(exactly2n−1when∆>0).When A I is symmetric,A z is also symmetric.Now we deﬁne function f:R n×n→R1f(A)=minx=0x T Axx T x(4)When f(A)≥0,we say that A is positive semideﬁnite(that is,for each x,x T Ax≥0, this deﬁnition conforms to the usual one).Similarly,when f(A)>0,we say that A is positive deﬁnite(that is,for each x=0,x T Ax>0).Real interval matrix A I is total positive (simi)deﬁnite if every A∈A I is positive(simi)deﬁnite.For the total positive(semi)deﬁniteness of real interval matrices,we have the following elegant theorem.Theorem1.[4]Let A I be a interval matrix,then the following assertions are equivalent: (i)A I is total positive(semi)deﬁnite;(ii)A I s is total positive(semi)deﬁnite;(iii)A z is total positive(semi)deﬁnite for each z∈Y.From theorem1,it only needs to check positive(semi)deﬁniteness of2n−1vertex matrices to determine the positive(semi)deﬁniteness of a interval matrix.This result in paper[4]is an improvement on the one in paper[3],in the latter,it needs to check positive(semi)deﬁniteness of2n(n−1)/2vertex matrices.Further,in paper[5],it was proved that the problem of deter-mining the total positive deﬁniteness of a symmetric interval matrix is HP-hard.Now we consider interval matrices in complex numberﬁeld and corresponding results.3Main resultsFirst we introduce corresponding deﬁnitions in the complex numberﬁeld.For an n×n complex matrix C=(c rs),we denote its Hermite transpose by C H,when C=C H,we call C a Hermite matrix,it is well known that its eigenvalues are all real numbers, we denote them from large to small byλ1(C)≥λ2(C)≥...≥λn(C).Let A I,B I be n×n real interval matrices,whereA I=[A,A],B I=[B,B].We deﬁne an n×n complex interval matrix asC I=A I+iB I={A+Bi:A∈A I,B∈B I}(5)When A,B are vertex matrices of A I,B I respectively,we call A+Bi a vertex matrix of C I.It can be seen that C I has2n2×2n2=22n2vertex matrices.When A I,B I are real symmetric interval matrices,and B=−B,we can deﬁne the following n×n Hermite interval matrix:C I H={A+Bi:A∈A I,B∈B I,A is symmetric,B is antisymmetric}(6) It has at most2n(n+1)/2×2n(n−1)/2=2n2Hermite vertex matrices.Now we deﬁne the function g:C IH→R1asg(C)=minx=0x H Cxx H x(7)In fact,g(C)=λn(C).When g(C)≥0,we say that C is positive semideﬁnite(that is,foreach x,x H Cx≥0,this conﬁrms the usual deﬁnition),similarly,when g(C)>0,we say that C is positive deﬁnite(that is,for each x=0,we have x H Cx>0).A Hermite interval matrix is total positive(semi)deﬁnite if each C∈C IHis positive(semi)deﬁnite.Theorem2.The minimum eigenvalues of Hermite matrices in an n×n Hermite intervalmatrix C IHobtain the minimal value at some Hermite vertex matrix.Proof.Suppose A+Bi is some Hermite matrix in C IH ,and X+Y i=(x1+y1i,x2+y2i,...,x n+y n i)T is the eigenvector which has modular1correspond to the smallest eigenvalueλn(A+Bi). Let U be the n-dimension all-1column vector,thenλn(A+Bi)=(X+Y i)H(A+Bi)(X+Y i)=U T((A+Bi)◦((X+Y i)(X+Y i)H))U(8)Where A◦B represents the Hardmard production of matrices A and B,that is,the matrix of productions of correspond elements of A and B.Next,we construct a matrix Q=(u rs+iv rs)n×n as follows:u rs=A rs,x r x s+y r y s≥0A rs,x r x s+y r y s<0v rs=B rs,x r y s−x s y r≥0B rs,x r y s−x s y r<0From the deﬁnition,we know that Q is a Hermite matrix and satisﬁes:u rr=A rr,v rr=B rr,u rs(x r x s+y r y s)≤A rs(x r x s+y r y s),v rs(x r y s−x x y r)≤B rs(x r y s−x x y r).therefore((u rs+iv rs)(x r+iy r)(x s−iy s))=u rs(x r x s+y r y s)+v rs(x r y s−x s y r)≤A rs(x r x s+y r y s)+B rs(x r y s−x s y r)= ((A rs+iB rs)(x r+iy r)(x s−iy s)),where (x)represents the real part of number x.Thus such Q we constructed satisﬁes (Q◦((X+Y i)(X+Y i)H))≤ ((A+Bi)◦((X+Y i)(X+Y i)H)),where (W n×n)=( (W ij))n×n.Note thatλn(A+Bi)=min{Z H(A+Bi)Z: Z =1,Z∈C n},thus we haveλn(Q)=min{Z H QZ: Z =1,Z∈C n}≤(X+Y i)H Q(X+Y i)=U T(Q◦((X+Y i)(X+Y i)H))U=U T (Q◦((X+Y i)(X+Y i)H))U≤U T ((A+Bi)◦((X+Y i)(X+Y i)H))U=U T((A+Bi)◦((X+Y i)(X+Y i)H))U=(X+Y i)H(A+Bi)(X+Y i)=λn(A+Bi)(9)Since A+Bi is an arbitrary Hermite matrix in C IH ,it follows that the minimum of minimaleigenvalues of Hermite matrices in C IH is obtained at some vertex Hermitematrix.Since g(C)=λn(C),we haveCorollary1.n×n Hermite interval matrix C IHis total positive(semi)deﬁnite if and only if its2n2Hermite vertex matrices are positive(semi)deﬁnite.In fact,the number of vertex matrices can be reduced to4n−1(n−1)!in Theorem2and its corollary.To prove it,we need the following auxiliary statement.First we deﬁne the sign of matrix A=(a ij)as sign(A)=(sign(a ij))and we set sign0=1.Lemma.The cardinal number of the setS n={(sign(XX T+Y Y T),sign(XY T−Y X T)):X,Y∈R n}is at most4n−1(n−1)!.Proof.We use induction on n.For n=1,S1={(sign(x21+y21),sign0)}={(1,1)},its cardinal number is1.Now suppose that the statement required is proved for S n−1.Letαi,j=(XX T+Y Y T)ij=x i x j+y i y jβi,j=(XY T−Y X T)ij=x i y j−x j y ithen we haveαj,k(x2i+y2i)=(αi,jαi,k+βi,jβi,k)βj,k(x2i+y2i)=(αi,jβi,k−αi,kβi,j)αi,j=αj,iβi,j=−βj,iαi,i≥0βi,i=0so it suﬃces to prove that the number of diﬀerent combinations of signs of(α1,n,β1,n,α2,n,β2,n,...,αn−1,n,βn−1,n)is4(n−1).We can assume that x2i+y2i=0for each i,otherwise,the elements in i-th row and i-th column are all zeros,which can be reduced to the case corresponds to S n−1.Thereforeαj,n=αi,jαi,n+βi,jβi,nx2i+y2iβj,n=αi,jβi,n−αi,nβi,jx2i+y2i(10)Further,we havesign(αj,n)=sign(αi,jαi,n),ifαi,jβi,jαi,nβi,n≥0sign(βj,n)=sign(αi,jβi,n),ifαi,jβi,jαi,nβi,n<0(11)Fix the signs ofα1,n andβ1,n,then it suﬃces to prove that the number of diﬀerent combinations of signs of(α2,n,β2,n,...,αn−1,n,βn−1,n)is n−1.Indeed,the number of diﬀerent combinations of signs of(α1,n,β1,n)is equal to4,it therefore deduces the statement required.From(11),we know that,when weﬁxα1,n andβ1,n,signs ofµ2,µ3,...,µn−1are deter-mined,whereµj=αj,n orβj,n,and we denote the other number in{αj,n,βj,n}exceptµj by ˜µj.Now we choose a sign of˜µ2,again,signs ofµ2,˜µ2determine signs of someµj,˜µj,without loss of generality,we assume that they determine signs ofµ3,...,µk+2,˜µk+3,...,˜µn−1.Weﬁrst proved that,ifα2,n,β2,n determine a sign of the same numberµj asα1,n,β1,n do, then they determine a same sign ofµj,that is,ifsign(α1,jβ1,jα1,nβ1,n)=sign(α2,jβ2,jα2,nβ2,n)(12) thensign(α1,jα1,n)=sign(α2,jα2,n)or sign(α1,jβ1,n)=sign(α2,jβ2,n)(13) Indeed,weﬁrst assume thatα1,jβ1,jα1,nβ1,n≥0.Ifα1,2β1,2α1,nβ1,n≥0,thenα1,2β1,2α1,jβ1,j≥0,sosign(α2,n)=sign(α1,2α1,n),sign(α2,j)=sign(α1,2α1,j).Thereforesign(α2,jα2,n)=sign(α1,2α1,jα1,2α1,n)=sign(α1,jα1,n).While ifα1,2β1,2α1,nβ1,n<0,thenα1,2β1,2α1,jβ1,j<0,sosign(β2,n)=sign(α1,2β1,n),sign(β2,j)=sign(α1,2β1,j).Thereforesign(β2,jβ2,n)=sign(α1,2β1,jα1,2β1,n)=sign(β1,jβ1,n).with the assumptionsign(α1,jβ1,jα1,nβ1,n)=sign(α2,jβ2,jα2,nβ2,n)=1,we can deduce thatsign(α1,jα1,n)=sign(α2,jα2,n).Similarly,ifα1,jβ1,jα1,nβ1,n<0,we can prove thatsign(α1,jβ1,n)=sign(α2,jβ2,n).We now use induction to prove that whenµ2,˜µ2determine signs of the same k numbers asα1,n,β1,n do,then number of diﬀerent combinations of signs of(α3,n,β3,n,...,αn−1,n,βn−1,n)after we chooseα1,n,β1,n,α2,n,β2,nis equal to k+1.When k equals1,the statement is obviously true,in the following,we suppose it has been proved for cases correspond numbers less then k.Now we choose˜µ3,thenµ3,˜µ3determine signs of p numbers in{µ4,...,µk+2},and k−1−p numbers in{˜µ4,...,˜µk+2}.From the induction hypothesis,we know this˜µ3corresponds p+1diﬀerent combinations of signs of(α4,n,β4,n,...,αn−1,n,βn−1,n),while when we choose the opposite sign for˜µ3with the former one we choose,µ3,˜µ3determine signs of k−1−p numbers in{µ4,...,µk+2},and p numbers in{˜µ4,...,˜µk+2},this˜µ3 corresponds k−1−p+1=k−p diﬀerent combinations of signs of(α4,n,β4,n,...,αn−1,n,βn−1,n).So the total number of diﬀerent combinations of signs of(α3,n,β3,n,...,αn−1,n,βn−1,n)is equal to(p+1)+(k−p)=k+1,this is what we require.Now return to the proof of the lemma,sinceµ2,˜µ2determine signs ofµ3,...,µk+2,˜µk+3,...,˜µn−1,this µ2,˜µ2corresponds k +1diﬀerent combinations of signs of them,while when we choose ˜µ2with opposite sign,then µ2,˜µ2determine signs of˜µ3,...,˜µk +2,µk +3,...,µn −1,this µ2,˜µ2corresponds n −k −3+1=n −k −2diﬀerent combinations of signs of them,so the total number of diﬀerent combinations of signs of(α2,n ,β2,n ,...,αn −1,n ,βn −1,n )is equal to (k +1)+(n −k −2)=n −1,as was required.Now we deﬁne a set V n for the Hermite interval matrix C I H=A I +iB I :Q =(u rs +iv rs )∈V n ⇔(M,N )∈S n ,and u rs = A rs ,M rs =1A rs ,M rs =−1,v rs = B rs ,N rs =1B rs ,N rs =−1(14)then from the proof of theorem 2,we can reduce the number of Hermite vertex matrices we used in it and its corollary.Theorem 2 .The minimum eigenvalues of Hermite matrices in a n ×n Hermite intervalmatrix C I H obtain the minimal value at a Hermite vertex matrix in V n corresponding to C I H.Corollary 1 .n ×n Hermite interval matrix C I H is total positive (semi)deﬁnite if and only if all Hermite vertex matrices in V n corresponding to C I Hare positive (semi)deﬁnite.References[1]Adjiman C S,Dallwig S,Floudas C A,et al.A Global Optimization Method,αBB,forGeneral Twice-diﬀerentiable Constrained NLPs–i.Theoretical Advances [J].Computers Chem.Engng,1998,Vol.22:1137-1158.[2]Adjiman C S,Androulakis I P,Floudas C A.A Global Optimization Method,αBB,forGeneral Twice-diﬀerentiable Constrained NLPs–ii.Implementation and Computational Results [J].Computers Chem.Engng,1998,Vol.22:1159-1179.[3]ShiZhicheng,GaoWeibin.A Necessary and Suﬃcient Condition for the Positive-deﬁniteness of Interval Symmetric Matrices [J].Internat.J.Control.1986,Vol.43:325-328.[4]Rohn Jiri.Positive Deﬁniteness and Stability of Interval Matrices [J].SIAM J.MatrixAnal.Appl.1994,Vol.15,No.1:175-184.[5]Rohn Jiri.Checking positive deﬁniteness or stability of symmetric interval matrices isNP-hard [J].Comment.Math.Univ.Carolinae.1994,Vol.35,No.4:795-797。

南京航空航天大学研究生课程《矩阵论》内容总结与习题选讲

《矩阵论》复习提纲与习题选讲Chapter1 线性空间和内积空间内容总结：z 线性空间的定义、基和维数；z 一个向量在一组基下的坐标；z 线性子空间的定义与判断；z 子空间的交z 内积的定义；z 内积空间的定义；z 向量的长度、距离和正交的概念；z Gram-Schmidt 标准正交化过程；z 标准正交基。

习题选讲：1、设表示实数域3]x [R R 上次数小于3的多项式再添上零多项式构成的线性空间（按通常多项式的加法和数与多项式的乘法）。

（1）求的维数；并写出的一组基；求在所取基下的坐标；3]x [R 3]x [R 221x x ++ （2）在中定义3]x [R ， ∫−=11)()(),(dx x g x f g f n x R x g x f ][)(),(∈ 证明：上述代数运算是内积；求出的一组标准正交基；3][x R （3）求与之间的距离；221x x ++2x 2x 1+−（4）证明：是的子空间；2][x R 3]x [R （5）写出2[][]3R x R x ∩的维数和一组基；二、设22R ×是实数域R 上全体22×实矩阵构成的线性空间（按通常矩阵的加法和数与矩阵的乘法）。

（1）求22R ×的维数，并写出其一组基；（2）在(1)所取基下的坐标； ⎥⎦⎤⎢⎣⎡−−3111（3）设W 是实数域R 上全体22×实对称矩阵构成的线性空间（按通常矩阵的加法和数与矩阵的乘法）。

证明：W 是22R ×的子空间；并写出W 的维数和一组基；（4）在W 中定义内积， )A B (tr )B ,A (T =W B ,A ∈求出W 的一组标准正交基；（5）求与之间的距离； ⎥⎦⎤⎢⎣⎡0331⎥⎦⎤⎢⎣⎡−1221 （6）设V 是实数域R 上全体22×实上三角矩阵构成的线性空间（按通常矩阵的加法和数与矩阵的乘法）。

证明：V 也是22R ×的子空间；并写出V 的维数和一组基；（7）写出子空间的一组基和维数。

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(3) A 0;
(4) tr ( A) i (i 1,2,, n) .
定理5.2.2 设 A是n 阶Hermite矩阵，则下列命题等价： (1) A是非负定矩阵； (2) 对任意n 阶可逆矩阵P, PHAP是Hermite非负定矩阵；
Ir （4）存在 n阶可逆矩阵 P使得P AP 0 r rank( A);
H
(3) A的n 个特征值均为非负数；
0 ,其中 0
(5) 存在秩为 r的矩阵 Q使得A Q Q ;
H
（6）存在 n阶Hermite矩阵S使得A S 2 .
为推论5.2.2 设A是n阶Hermite非负定矩阵(1) 如果Q是任一 n m矩阵，则 Q AQ 0;
H
(5.1.1)
其中1 , 2 ,, n均为实数。
定理5.1.4 设 A R nn ，则 A是实对称矩阵的充分必要条件是存在正交矩阵Q使得
QT AQ diag(1 , 2 ,, n )
(5.1.2)
其中1 , 2 ,, n均为实数。
5.1.2 矩阵的惯性
(4) 若A 0， A 0, 则A 0; (5) 若A 0，B 0, 则A B 0; (6) 若A B，B C , 则A C; (7) 若A B，A1 B1 , 则A A1 B B1; (8) 若A 0，B 0, 则A B 0; (9) 若A B，B C , 则A C;
(5.1.12)
称形如（5.1.12）的二次型为Hermite二次型的标准形。
定理5.1.7 对Hermite二次型 f (x) = xHAx，存在酉线性变换x = Uy（其中U是酉矩阵）使得Hermite 二次型f (x)变成标准形
1 y1 y1 2 y2 y2 n yn yn
H
(2) x H Ax半正定的充分必要条件为s r n； (3) x H Ax负定的充分必要条件为 s 0, r n； (4) x H Ax半负定的充分必要条件为s 0, r n； (5) x H Ax不定的充分必要条件为 0 s r n.
5.2
Hermite正定（非负定）矩阵
其中aij a ji，称为 Hermite二次型。记
a11 a 21 A a n1 a12 a 22 an 2
a1n x1 a2n x2 ,x x a nn n
则 A为Hermite矩阵。称矩阵A为Hermite二次型的矩阵，并且称 A的秩为Hermite二次型的秩。
其中 r = rank(A)，s = π(A).
Hermite二次型可分为五种情况
H
(1) 若s r n, 则规范形为 x Ax yi . 若x 0,
2 i 1
n
则y 0, x Ax 0.
H
( 2) 若s r n, 则规范形为 x Ax yi . 对任意
(3) 矩阵的“≥”是Hermite矩阵集合中的一种偏序关系。
定理5.3.1 设A, A1, B, B1, C均为n 阶Hermite矩阵，则 (1) A B( A B) A B( A B);
(2) A B( A B) 对任意 n阶可逆矩阵 P都有 P H AP P H BP( P H AP P H BP ) (3) 若A B( A B)，k为正数，则 kA kB(kA kB);
(4) 若A,B是Hermite矩阵，则 AB是Hermite矩阵的充分必要条件是AB = BA；
(5) A是Hermite矩阵的充分必要条件是对任意方阵 S， SH AS是Hermite矩阵。
定理5.1.1 设A (a jk ) C
nn
, 则A是Hermite 矩阵的充分
n H
必要条件是对任意x C , x Ax是实数.
H 2 i 1
r
x C n都有x H Ax 0.
H s 2 r
(5) 若0 s r n, 则规范形为 x Ax yi yi
i 1 i s 1
2
对不同的 x , x H Ax之值可以大于 0, 小于0或等于 0.
定义5.1.1 设f (x) = xHAx为Hermite二次型。
设A C
nn
, ( A)、 ( A)和 ( A)分别表示 A
的位于复平面上右半平开面、左半开平面和虚记 In( A) { ( A), ( A), ( A)} 则称In( A)为矩阵 A的惯性。
定理5.1.6(Sylvester惯性定律）设 A,B是n 阶Hermite 矩阵，则 A与B相合的充分必要条件是
正定（非负定）矩阵具有如下基本性质：
(1) 单位矩阵 I 0； (2) 若A 0, 数k 0, 则kA 0； (3) 若A 0, B 0, 则A B 0；
(4) 若A 0, B 0, 则A B 0.
定理5.2.1 设 A是n 阶Hermite矩阵，则下列命题等价： (1) A是正定矩阵； (2) 对任意n 阶可逆矩阵P,PHAP 都是Hermite正定矩阵； (3) A的n 个特征值均为正数； (4) 存在n 阶可逆矩阵P使得PHAP = I；
H
(2)
A 0;
(3) tr( A) i (i 1,2,, n) .
定理5.2.3 n 阶Hermite矩阵 A正定的充分必要条件是 A的顺序主子式均为正数，即
1 k k A 1 k 0
k 1,, n
定理5.2.4 n 阶Hermite矩阵 A正定的充分必要条件是 A的所有主子式全大于零。定理5.2.5 n 阶Hermite矩阵 A非负定的充分必要条件是A的所有主子式均非负。定理5.2.6 n 阶Hermite矩阵 A正定的充分必要条件是存在n 阶非奇异下三角矩阵 L 使得
x C n使得
P AP diag(1 ,, n ), P BP I
H H
其中1 , 2 ,, n是广义特征值问题 (5.2.5)的特征值。
5.3
矩阵不等式
定义5.3.1 设 A,B 都是n 阶Hermite矩阵，若A－B≥0，则称A大于或等于B（或称 B小于或等于 A），记作 A≥B（或B≤A）；若A－B＞0，则称A大于B（或称 B小于A），记作A>B或（B<A）。设 A,B 都是n 阶Hermite矩阵，由定义5.3.1得
轴上特征值的个数（特重征值按重数计算）。
In( A) In( B)
(5.1.6)
5.1.3
Hermite二次型
n n
由n个复变量 x1 ,, xn，系数为复数的二次齐式
f ( x1 ,, xn ) aij xi x j
i 1 j 1
(5.1.10)
(5) 存在n 阶可逆矩阵Q使得A = QHQ；
(6) 存在n 阶可逆Hermite矩阵S 使得A = S2.
推论5.2.1 设A是n阶Hermite正定矩阵，其特征值为
1 , 2 ,, n，则
(1) A1是正定矩阵； (2) 如果Q是任一 n m列满秩矩阵，则 Q H AQ 0;
H 2 i 1
r
x C 都有x Ax 0.
n H
( 3) 若s 0, r n, 则规范形为 x Ax yi .若x 0,
H 2 i 1
n
则y 0, x Ax 0.
H
(4) 若s 0, r n, 则规范形为 x Ax yi .对任意
第5章 Hermite矩阵与正定矩阵
5.1 5.2 Hermite矩阵与Hermite二次型 Hermite正定（非负定）矩阵
5.3
5.4
矩阵不等式
Hermite矩阵的特征值*
5.1
Hermite矩阵与Hermite二次型
Hermite矩阵矩阵的惯性
5.1.1 5.1.2
5.1.3
Hermite二次型
其中1 , 2 ,, n是Hermite矩阵A的特征值。
定理5.1.8 对Hermite二次型 f (x) = xHAx，存在可逆线性变换x = Py 使得Hermite二次型f (x)化为
f ( x) x H Ax y1 y1 ys ys ys1 ys1 yr yr
H
n 定义5.2.1 设A是n阶Hermite矩阵, 如果对任意 xC 且
x 0, 都有x Ax 0，则称 A为正定矩阵 , 记作A 0; 如果对任意 x C 都有x Ax 0, 则称A为非负定 (半正定 )
n H
矩阵, 记作A 0.
定理5.1.5 设 A是n 阶Hermite矩阵，则 A相合于矩阵
Is D0 0 0 0 Irs 0 0 0 On r ( 5 .1 .3 )
其中 r = rank(A)，s是 A的正特征值（重特征值按重数计算）的个数。
(5.1.3)中矩阵称为n 阶Hermite矩阵 A的相合标准形。
5.1.1 Hermite矩阵
Hermite矩阵具有如下简单性质： (1) 如果 A是Hermite矩阵，则对正整数 k，Ak 也是 Hermite矩阵；
(2) 如果 A是可逆Hermite矩阵，则A-1 是Hermite矩阵； (3) 如果 A,B是Hermite矩阵，则对实数k,p, kA+pB 是 Hermite矩阵；
利用Hermite二次型的矩阵，Hermite二次型可表示为 f ( x ) x H Ax