矩阵论第八章答案

习题8

1.解：若，则A 为单调矩阵，其充要条件是可从01≥=?B A 推出，这里X 是列向量.

0≥AX 0≥X 事实上，因为A 是单调矩阵，所以.若，则必有01≥?A 0≥AX ，从而.

01≥?AX A 0≥X

；再由0=AX A 非奇)(=?X A 异，即记A 01≥?j α2.为正矩阵0a ij >，故

∑==>==≥j j ij i j n i x a AX n j x 1)

,,1(0)(),0,,,1(0L L 所以至少有一个不为.

0>AX (2)，

0,0≥?>Y X A Q 如果由上题知；如果，,0≠?Y A AY AX Y X A >>?即,0)(0=?Y X 则，即.故有.

0)(=?Y X A AY AX =AY AX ≥（3）反证.若不然，对所有，即时，有

0≥X ),,1(0n j X j L =≥

，即，则当X 有一个分量时都有

0k X ，即AX ＜0

()()n i x a AX n j j

ij i ,,101L =<=∑=这与假设总有矛盾.故.

0≥AX 0≥A 3.证：因为，由Perron 定理知是A 的单特征值，且A 0>A )(A ρ的任一不等于的特征值λ必有，即只有1个模为)(A ρ)(A ρλ<)(A ρ的特征值，故A 为素矩阵.但反之不真，因为例如

????

??????=111001110A 就是素矩阵（不难验证），显然A 不是正矩阵.

04>A 4.证：因为，所以由非负矩阵谱半径的估计式

0≥A ∑∑=≤≤=≤≤≤≤n j ij n i n j ij n i a A a 1

111max )min （ρ又因为A 是素矩阵，即存在正整数m ，使得，因而，显然0>m A 0≠A ，故.∑=≤≤>n j ij n i a 1

10min 0)(>A ρ5.证：必要性.设A 是非负素矩阵，则是A 的正特征值，且)(A ρA 的任何一个其他特征值，都有，而的模λ)(A ρλ<)(A ρ?，显然就不是A 的特征值，故)()(A A ρρ=?)(A ρ?，即非奇异.

0)(()(≠??=+I A A I A A ρρI A A )(ρ+充分性.设，则不是A 的特值.由题设A 是n 阶

0)(≠+I A A ρ)(A ρ?

不可约非负对称矩阵，那么它的特征值全为实数，且由Perron-Frobenius 定理知为A 的（正的）单特征值，而又不是A )(A ρ)(A ρ?的特征值，所以模等于的实特征值只能是一个，再不会有第)(A ρ)(A ρ二个，即，故A 为素矩阵.

1=k 6.解：；

9)(=A ρ

则∞→([lim n ρ7.使得(+≥I r A 8.解：若是单调矩阵，则（非负）.

1?A 0≥A 9.证：注意到和A 有相同的特征值，可以转而考虑的谱半T A T A 径.根据随机矩阵A 应满足的条件，显然

∑===n

i ij n j a 1),,2,1(1L T

)1,,1,1(L =ζ是的特征向量，而且1是的特征值，是相应于特征值1的特

T A T A ζ

征向量.

另一方面，由于A 是正矩阵，也是正矩阵，又，依Perron-T A 0>ζFrobenius 定理知，是的优势特向量，即是相应于优势特征值ζT A ζ的特征向量，因此

)(T A ρ.

1)()(==T A A ρρ10.解：若A 和都是M 矩阵，则A 必为对角矩阵.

1?A 11.解：应当表述为矩阵满足，则A 为M 矩nxn ij a A )(=)(0j i a ij ≠≤阵的充要条件是，且矩阵满足，这),,2,1(0n i a ij L =>A D I B 1??=1)(

12.解：若矩阵A 和B 都是M 矩阵，但和A+B 不一定是M 矩阵，例如，设

，???????=5.0015.0A ??

?????=5.0105.0B 容易验证A 和B 都是M 矩阵，但是奇异的，??

??????=+1111B A 1?A 不存在，所以A+B 不是M 矩阵（不作特别声明时，M 矩阵指的是非奇异M 矩阵.

13.证：设，则由A 是M 矩阵满足推出

0≥=Y AX 01≥?A

.

01≥=?Y A X 14.证：因为A 为非奇异M 矩阵等价于A 的所有主子式为正数；而在A 是实对称的条件下，A 的所有主子式是正的等价于A 是正定的.

15.提示：证明很容易，考察????

?

?

?

???????=∑∑∑∑====n k kn nk n

k k nk b

k kn

k n

k k k b a b a b a b a AB 11111111L

M O M L 有

.

0)(111≥=???A B AB

南航矩阵论等价关系

Student’s Name: Student’s ID No.: College Name: The study of Equivalence Relations Abstract According to some relative definitions and properties, to proof that if B can be obtained from A by performing elementary row operations on A, ~ is an equivalence relation, and to find the properties that are shared by all the elements in the same equivalence class. To proof that if B is can be obtained from A by performing elementary operations, Matrix S A ∈ is said to be equivalent to matrix S B ∈, and ~A B means that matrix S A ∈ is similar to S B ∈, if let S be the set of m m ? real matrices. Introduction The equivalence relations are used in the matrix theory in a very wide field. An equivalence relation on a set S divides S into equivalence classes. Equivalence classes are pair-wise disjoint subsets of S . a ~ b if and only if a and b are in the same equivalence class.This paper will introduce some definitions and properties of equivalence relations and proof some discussions. Main Results Answers of Q1 (a) The process of the proof is as following,obviously IA=A,therefore ~ is reflexive;we know B can be obtained from A by performing elementary row operations on A,we assume P is a matrix which denote a series of elementary row operations on A.Then ,we have PA=B,(A~B),and P is inverse,obviously we have A=P -1B,(B~A).So ~ is symmetric.We have another matrix Q which denote a series of elementary row operations on B,and the result is C,so we have QB=C.And we can obtain QB=Q(PA)=QPA=C,so A~C.Therefore,~ is transitive. Hence, ~ is an equivalence relation on S . (b) The properties that are shared by all the elements in the same equivalence class are as followings: firstly,the rank is the same;secondly,the relation of column is not changed;thirdly,two random matrices are row equivalent;fourthly,all of the matrices

南航矩阵论2013研究生试卷及答案

南京航空航天大学2012级硕士研究生

二、(20分)设三阶矩阵，，. ????? ??--=201034011A ????? ??=300130013B ???? ? ??=3003003a a C (1) 求的行列式因子、不变因子、初等因子及Jordan 标准形； A (2) 利用矩阵的知识，判断矩阵和是否相似，并说明理由. λB C 解答: (1)的行列式因子为;…(3分)A 2121)1)(2()(,1)()(--===λλλλλD D D 不变因子为; …………………(3分)2121)1)(2()(,1)()(--===λλλλλd d d 初等因子为;……………………(2分) 2)1(,2--λλJordan 标准形为. ……………………(2分) 200011001J ?? ?= ? ??? (2) 不相似,理由是2阶行列式因子不同; …………………(5分) 0,a = 相似,理由是各阶行列式因子相同. …………………(5分) 0,a ≠共 6 页 第 4 页

三、(20分)已知线性方程组不相容. ?? ???=+=+++=++1,12,1434321421x x x x x x x x x (1) 求系数矩阵的满秩分解； A (2) 求广义逆矩阵； +A (3) 求该线性方程组的极小最小二乘解. 解答:(1) 矩阵,的满秩分解为 ???? ? ??=110021111011A A . …………………(5分)10110111001101A ??????=?????????? (2) . ……………………(10分)51-451-41-52715033A +?? ? ?= ? ??? (3) 方程组的极小最小二乘解为. …………(5分)2214156x ?? ? ?= ? ??? 共 6 页 第 5 页

南航双语矩阵论 matrix theory第三章部分题解

Solution Key to Some Exercises in Chapter 3 #5. Determine the kernel and range of each of the following linear transformations on 2P (a) (())'()p x xp x σ= (b) (())()'()p x p x p x σ=- (c) (())(0)(1)p x p x p σ=+ Solution (a) Let ()p x ax b =+. (())p x ax σ=. (())0p x σ= if and only if 0ax = if and only if 0a =. Thus, ker(){|}b b R σ=∈ The range of σis 2()P σ={|}ax a R ∈ (b) Let ()p x ax b =+. (())p x ax b a σ=+-. (())0p x σ= if and only if 0ax b a +-= if and only if 0a =and 0b =. Thus, ker(){0}σ= The range of σis 2()P σ=2{|,}P ax b a a b R +-∈= (c) Let ()p x ax b =+. (())p x bx a b σ=++. (())0p x σ= if and only if 0bx a b ++= if and only if 0a =and 0b =. Thus, ker(){0}σ= The range of σis 2()P σ=2{|,}P bx a b a b R ++∈= 备注: 映射的核以及映射的像都是集合,应该以集合的记号来表达或者用文字来叙述. #7. Let be the linear mapping that maps 2P into 2R defined by 10()(())(0)p x dx p x p σ?? ?= ??? ? Find a matrix A such that ()x A ασαββ??+= ??? . Solution 1(1)1σ??= ??? 1/2()0x σ?? = ??? 11/211/2()101 0x ασαβαββ????????+=+= ? ? ??????????? Hence, 11/210A ??= ??? #10. Let σ be the transformation on 3P defined by (())'()"()p x xp x p x σ=+ a) Find the matrix A representing σ with respect to 2[1,,]x x b) Find the matrix B representing σ with respect to 2[1,,1]x x + c) Find the matrix S such that 1B S AS -= d) If 2012()(1)p x a a x a x =+++, calculate (())n p x σ. Solution (a) (1)0σ=

南航矩阵论期中考试参考答案.doc

1) 一组基为q = .维数为3. 3) 南京航空航天大学双语矩阵论期中考试参考答案(有些答案可能有问题) Q1 1解矩阵A 的特征多项式为 A-2 3 -4 4I-A| =-4 2+6 -8 =A 2(/l-4) -6 7 A-8 所以矩阵A 的特征值为4 =0(二重)和/^=4. 人?2 3 由于(4-2,3)=1,所以D| (人)二1.又 彳 人+6=“2+4人=?(人) 4-2 3 、=7人+4=代(人)故(们3),代3))=1 ?其余的二阶子式(还有7个)都包含因子4, -6 7 所以 D? 3)=1 .最后 det (A (/L))=42(人.4),所以 D 3(A)=/l 2 (2-4). 因此矩阵A 的不变因子为d, (2) = d 2(2) = l, d 3 (2) = r (2-4). 矩阵A 的初等因子为人2, 2-4. 2解矩阵B 与矩阵C 是相似的.矩阵B 和矩阵C 的行列式因子相同且分别为9 3)=1 , D 2(/i)=A 2-/l-2 .根据定理：两矩阵相似的充分必要条件是他们有相同的行列式因子. 所以矩阵B 与矩阵c 相似. Q2 2)设k 是数域p 中任意数，a, 0, /是v 中任意元素.明显满足下而四项. (") = (",a) ； (a+月,/) = (",/) + (”,刃；(ka,/3) = k(a,/3) ； (a,a)>0, 当且仅当Q = 0时(a,a) = ().所以(。,/?)是线性空间V 上的内积. 利 用Gram-Schmidt 正交化方法，可以依次求出 ,p 2 =%-(％'5)与= 层=%-(％,弟与一(％,弓)役=

南航双语矩阵论matrixtheory第三章部分题解

Solution Key to Some Exercises in Chapter 3 #5. Determine the kernel and range of each of the following linear transformations on 2P (a) (())'()p x xp x σ= (b) (())()'()p x p x p x σ=- (c) (())(0)(1)p x p x p σ=+ Solution (a) Let ()p x ax b =+. (())p x ax σ=. (())0p x σ= if and only if 0ax = if and only if 0a =. Thus, ker(){|}b b R σ=∈ The range of σis 2()P σ={|}ax a R ∈ (b) Let ()p x ax b =+. (())p x ax b a σ=+-. (())0p x σ= if and only if 0ax b a +-= if and only if 0a =and 0b =. Thus, ker(){0}σ= The range of σis 2()P σ=2{|,}P ax b a a b R +-∈= (c) Let ()p x ax b =+. (())p x bx a b σ=++. (())0p x σ= if and only if 0bx a b ++= if and only if 0a =and 0b =. Thus, ker(){0}σ= The range of σis 2()P σ=2{|,}P bx a b a b R ++∈= 备注: 映射的核以及映射的像都是集合,应该以集合的记号来表达或者用文字来叙述. #7. Let be the linear mapping that maps 2P into 2R defined by 10 ()(())(0)p x dx p x p σ?? ?= ??? ? Find a matrix A such that ()x A ασαββ?? += ??? . Solution 1(1)1σ?? = ??? 1/2()0x σ?? = ??? 11/211/2()1010x ασαβαββ???? ???? +=+= ? ? ??????????? Hence, 11/21 0A ?? = ??? #10. Let σ be the transformation on 3P defined by (())'()"()p x xp x p x σ=+ a) Find the matrix A representing σ with respect to 2[1,,]x x b) Find the matrix B representing σ with respect to 2[1,,1]x x + c) Find the matrix S such that 1B S AS -= d) If 2012()(1)p x a a x a x =+++, calculate (())n p x σ. Solution (a) (1)0σ= ()x x σ=

南航07-14矩阵论试卷

南京航空航天大学07-14硕士研究生矩阵论试题 2007 ~ 2008学年《矩阵论》 课程考试A 卷 一、（20分）设矩阵 ?? ??? ??-----=111322211 A ， （1）求A 的特征多项式和A 的全部特征值； （2）求A 的行列式因子、不变因子和初等因子； （3）求A 的最小多项式，并计算I A A 236 -+； （4）写出A 的Jordan 标准形。 二、（20分）设2 2?R 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2 2?R 的维数，并写出其一组基； （2）设W 是全体22?实对称矩阵的集合， 证明：W 是2 2?R 的子空间，并写出W 的维数和一组基； （3）在W 中定义内积W B A BA tr B A ∈=,),(),(其中，求出W 的一组标准正交基； （4）给出22?R 上的线性变换T ： 22,)(?∈?+=R A A A A T T 写出线性变换T 在（1）中所取基下的矩阵，并求T 的核)(T Ker 和值域)(T R 。 三、（20分） （1）设 ? ??? ??-=121312A ，求1A ，2A ，∞A ，F A ； （2）设n n ij C a A ?∈=)(，令 ij j i a n A ,*max ?=， 证明： *是 n n C ?上的矩阵范数并说明具有相容性； （3）证明：*2*1 A A A n ≤≤。 四、（20分）已知矩阵 ?????? ? ??-=10010001111 1A ，向量 ??? ??? ? ??=2112b ， （1）求矩阵A 的QR 分解；

南航双语矩阵论 matrix theory第一章部分题解

Solution Key (chapter 1) #2. Take S , 2=. But 2S ?. If 2S ∈, then there are rational numbers a and b , such that 2=0a ≠ and 0b ≠.) This will lead to 22 423 2a b ab --= The right hand is a rational number and the left hand side is an irrational number. This is impossible. Thus, S is not closed under multiplication. Hence, S is not a field. #13. (a) Denote the set by S . Take 2()p x x x S =+∈, 2()q x x x S =-+∈. Then ()()2p x q x x S +=?. S is not closed under addition. Hence, S is not a subspace. (Or: The set S does not contain the zero polynomial, hence, is not a subspace.) (b) Denote the set by S . Take 3()1p x x S =+∈, 3()1p x x S =-+∈. Then ()()2p x q x S +=?. S is not closed under addition. Hence, S is not a subspace. (Or: The set S does not contain the zero polynomial, hence, is not a subspace.) (d) Denote the set by S . Take ()1p x x S =+∈, ()1p x x S =-+∈, ()()2p x q x S +=?. S is not closed under addition. Hence, S is not a subspace. #15. (c) Denote the set by S . Take ()p x x S =∈. But ()p x x S -=-?. Thus, the set S is not closed under scalar multiplication. Hence, S is not a subspace. (e) Denote the set by S . Take ()1p x x S =-∈ ()1q x x S =+∈. But ()()2p x q x x S +=?. S is not closed under addition. Hence, S is not a subspace. #17. Since 12{,,,}u v v v i s span ∈ for each i , all combinations of 12,,,u u u r are also in 12{,,,}v v v s span . Thus, 12{,,,}u u u r span is a subspace of 12{,,,}v v v s span . Therefore, 12dim({,,,})u u u r span ≤ 12dim({,,,})v v v s span . #25. (a) Let 12(,,,)b b b n B = . Then 12(,,,)b b b n AB A A A = . If AB O =, then b 0i A = for 1,2,,i n = . ()b i N A ∈ for 1,2,,i n = . All lineawr combinations of 12,,,b b b n are also in ()N A . Thus, ()()R B N A ?. ()R B is a subspace of ()N A .

西北工业大学矩阵论PPT课件

矩阵论讲稿 讲稿编者：张凯院 使用教材：《矩阵论》（第2版） 西北工业大学出版社 程云鹏等编 辅助教材：《矩阵论导教导学导考》 《矩阵论典型题解析及自测试题》 西北工业大学出版社 张凯院等编 课时分配：第一章 17学时第四章8学时第二章5学时第五章8学时 第三章8学时第六章8学时

第一章 线性空间与线性变换 §1.1 线性空间 一、集合与映射 1．集合：能够作为整体看待的一堆东西． 列举法：},,,{321L a a a S = 性质法：}{所具有的性质a a S = 相等(：指下面二式同时成立 )21S S =2121,S S S a S a ?∈?∈?即 1212,S S S b S b ?∈?∈?即 交：}{2121S a S a a S S ∈∈=且I 并：}{2121S a S a a S S ∈∈=或U 和：},{22112121S a S a a a a S S ∈∈+==+ 例1 R}0{2221111∈ ==j i a a a a A S R}0 {221211 2∈ ==j i a a a a A S ，21S S ≠ R},00{22112211 21∈ ==a a a a A S S I R},0{211222211211 21∈= ==j i a a a a a a a A S S U R}{2221 1211 21∈ ==+j i a a a a a A S S 2．数域：关于四则运算封闭的数的集合． 例如：实数域R ，复数域C ，有理数域，等等． Q 3．映射：设集合与，若对任意的1S 2S 1S a ∈，按照法则σ，对应唯一的

南京航空航天大学Matrix-Theory双语矩阵论期末考试2015

NUAA

Let 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 representing matrix 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 + -----------------------------------------------------------------------------------------------------------------------