Solution Key (chapter 1)
Exercise 2.
The show that this set is not closed under multiplication.Take
S ,2=.But 2S ?.If 2S ∈rational numbers a and b ,such that
2=It is clear that 0a ≠and
0b ≠.)This will 22423
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.Exercise 7.z
x y x +=+)
()()()(z x x y x x ++-=++-z x x y x x ++-=++-])[(])[(z 0y 0+=+z
y =Exercise 12It is a vector space.
A1:
A2:,Hence,A3:The existence of the zero element .The zero element must satisfy that for any ,
That is for any ,
.We obtain that the zero element is A4:The existence of additive inverse.For each ,its additive inverse is ,since
.(Note that is the zero element of )
M1:M2:M3:M4:Exercise 13.
(a)No,it is not a subspace.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 .
(b)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.)(c)Yes,it is a subspace.Check that this set is closed under addition and scalar multiplication.(d)No,it is not a subspace.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.
Exercise 15.
(a)Yes,it is a subspace.Check that this set is closed under addition and scalar multiplication.(b)Yes,it is a subspace.Check that this set is closed under addition and scalar multiplication.(c)Denote the set by S .Take ()p x x S =∈.But (1)()p x x S -=-?.
Thus,the set S is not closed under scalar multiplication.Hence,S is not a subspace.(d)Yes,it is a subspace.Check that this set is closed under addition and scalar multiplication.(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.
Exercise 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 .
Exercise 19
By Taylor expansion formula
()110(1)()32(1)!
j n n j j f f x x x j --==+=-∑22(1)(2)512(1)(1)(1)2!
n n n x x --=?+-?-+-+ 1
2(1)(2)()(1)2(1)!
j n n n n j x x j ----+-++- The coordinate vector is 2(1)(2)2(1)(2)()5,2(1),,,,,2)T
n n n n n j n ------ (Exercise 22
Use the definition of the transition matrix.
111011001?? ? ? ???
Exercise 25.
(b)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 linear 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 .
If ()R B is a subspace of ()N A ,then for each column b i of B ,we must have
b 0i A =.Hence,
12(,,,).
b b b n AB A A A O == (b)By part (a),we know that ()R B is a subspace of ()N A .Thus,
()dim(())dim(())r B R B N A =≤.
By the rank-nullity theorem,we obtain that ()()dim(())()r B r A N A r A n
+≤+=Exercise 26.
(a)Hint:First,show that each column vector of C is a linear combination of the column vectors of A.Then,linear combinations of the column vectors of C are linear combinations of the column vectors of A.(b)Hint:First,show that each row vector of C is a linear combination of the row vectors of B.Then,linear combinations of the row vectors of C are linear combinations of the row vectors of B.(c)By (a)and (b),()dim(())dim(()()rank C R C R A rank A =≤=And ()dim(())dim(())()
T rank C R C R B rank B =≤=Thus,()min{(),()}
rank C rank A rank B ≤Exercise 27.
(a)Hint:The column vectors of C are linearly independent if and only if the system 0x C =has only the trivial solution (the zero solution).If ()0x AB =,then ()0x A B =.x B must be zero since the column vectors of A are linearly independent.Thar is,0x B =.Since the column vectors of B are linearly independent,x must be zero.(b)Hint:T T T C B A =.Column vectors of T A are linearly independent.Column vectors of T B are linearly independent.The row vectors of C are linearly independent if and only if the column vectors of T C are linearly independent.Then apply part(a).
Exercise 29.
Let ,A B S ∈.Then ()T T T A B A B A B +=+=+,and ()T T kA kA kA ==.S is closed under addition and scalar multiplication.Thus,S is a subspace of n n
R ?Let ,A B K ∈.Then ()()T T T A B A B A B A B +=+=--=-+,and ()()T T kA kA kA ==-.K
is closed under addition and scalar multiplication.Thus,K is a subspace of n n R ?The proof of n n R S K ?=⊕
.Let .n n A R ?∈Then 11()()22T T A A A A A =++-.1()2T A A +is symmetric and 1()2T A A -is anti-symmetric.This show that n n R S K ?=+.Next,we show that the sum S K +is a direct sum.If A S K ∈?,then we have both T
A A =and T
A A =-.This will imply that A A =-.Thus,A must be the zero matrix.This proves that the sum S K +is a direct sum.
Exercise 32.
Let ij E denote the matrix whose (,)i j entry is 1,zero elsewhere.
ij F denote the matrix whose (,)i j entry is 1-,zero elsewhere.For
any ()m n ij ij A a C ?=+∈,where ,ij ij a b are real numbers,A can be written as
1111
n m n m
ij ij ij ij j i j i A a E b F =====+
∑∑∑∑.
This shows that the matrices {,|1,2,,,1,2,,} ij ij E F i m j n == forms a spanning set for m n C ?.If 1111
n m n m ij ij ij ij j i j i a E b F O ====+=∑∑∑∑,
then 0
ij ij a =for 1,2,,i m = ,1,2,,j n = .Thus,we must have 0ij ij a b ==for 1,2,,i m = ,1,2,,j n = .Therefore,{,|1,2,,,1,2,,} ij ij E F i m j n == forms a basis for m n
C ?.The
dimension
is 2mn .
Note that,all coefficients of linear combinations must be real numbers because the
underlying field is the real number field.
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
第 1 页 共 6 页 (A 卷) 学院 系 专业班级 姓名 学号 (密封线外不要写姓名、学号、班级、密封线内不准答题,违者按零分计) …………………………………………密…………………………封……………………………………线………………………………… 考试方式:闭卷 太原理工大学 矩阵分析 试卷(A ) 适用专业:2016级硕士研究生 考试日期:2017.1.09 时间:120 分钟 共 8页 一、填空选择题(每小题3分,共30分) 1-5题为填空题: 1. 已知??? ? ? ??--=304021101A ,则1||||A =。 2. 设线性变换1T ,2T 在基n ααα ,,21下的矩阵分别为A ,B ,则线性变换212T T +在基n ααα ,,21下的矩阵为_____________. 3.在3R 中,基T )2,1,3(1--=α,T )1,1,1(2-=α,T )1,3,2(3-=α到基T )1,1,1(1=β, T )3,2,1(2=β,T )1,0,2(3=β的过度矩阵为A = 4. 设矩阵??? ? ? ??--=304021101A ,则 5432333A A A A A -++-= . 5.??? ? ? ? ?-=λλλλλ0010 01)(2A 的Smith 标准形为 6-10题为单项选择题: 6.设A 是正规矩阵,则下列说法不正确的是 ( ). (A) A 一定可以对角化; (B )?=H A A A 的特征值全为实数; (C) 若E AA H =,则 1=A ; (D )?-=H A A A 的特征值全为零或纯虚数。 7.设矩阵A 的谱半径1)( 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 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= 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=南航矩阵论2013研究生试卷及答案
南航双语矩阵论 matrix theory第三章部分题解
南航矩阵论期中考试参考答案.doc
2016矩阵论试题A20170109 (1)