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=
22()22x x σ=+
002010002A ??
?
= ? ???
(b) (1)0σ=
()x x σ=
22(1)2(1)x x σ+=+
000010002B ??
?
= ? ???
(c)
2[1,,1]x x +2[1,,]x x =101010001??
?
? ???
The transition matrix from 2[1,,]x x to 2[1,,1]x x + is
101010001S ?? ?= ? ???
, 1
B S AS -= (d) 2201212((1))2(1)n n a a x a x a x a x σ+++=++
#11. Let A and B be n n ? matrices. Show that if A is similar to B then there exist n n ?
matrices S and T , with S nonsingular, such that A ST =and B TS =.
Proof There exists a nonsingular matrix P such that 1A P BP -=. Let 1S P -=, T BP =. Then A ST =and B TS =.
#12. Let σ be a linear transformation on the vector space V of dimension n . If there exist a vector v such that 1()v 0n σ-≠ and ()v 0n σ=, show that
(a) 1,(),,()v v v n σσ-L are linearly independent.
(b) there exists a basis E for V such that the matrix representing σ with respect to the basis E is
00001
0000
010??
?
?
? ???
L L M M M M L
Proof
(a) Suppose that
1011()()v v v 0n n k k k σσ--+++=L
Then 11011(()())v v v 0n n n k k k σσσ---+++=L
That is, 12210110()()())()v v v v 0n n n n n k k k k σσσσ----+++==L Thus, 0k must be zero since 1()v 0n σ-≠. 211111(()())()v v v 0n n n n k k k σσσσ----++==L
This will imply that 1k must be zero since 1()v 0n σ-≠.
By repeating the process above, we obtain that 011,,,n k k k -L must be all zero. This
proves that
1,(),,()v v v n σσ-L are linearly independent.
(b) Since 1,(),,()v v v n σσ-L are n linearly independent, they form a basis for V .
Denote 112,(),,()εv εv εv n n σσ-===L 12()εεσ= 23()εεσ= …….
1()εεn n σ-= ()ε0n σ=
12[(),(),,()]εεεn σσσL 121[,,,,]εεεεn n -=L 00001
0000
010??
?
?
? ???
L L M M M M L
#13. If A is a nonzero square matrix and k A O =for some positive integer k , show that A can not be similar to a diagonal matrix.
Proof Suppose that A is similar to a diagonal matrix 12diag(,,,)n λλλL . Then for each i , there exists a nonzero vector x i such that x x i i i A λ= x x x 0k k i i i i i A λλ=== since k A O =.
This will imply that 0i λ= for 1,2,,i n =L . Thus, matrix A is similar to the zero matrix. Therefore, A O =since a matrix that is similar to the zero matrix must be the zero matrix, which contradicts the assumption.
This contradiction shows that A can not be similar to a diagonal matrix. Or
If 112diag(,,,)n A P P λλλ-=L then 112diag(,,,)k k k k n A P P λλλ-=L .
k A O = implies that 0i λ= for 1,2,,i n =L . Hence, B O =. This will imply that A O =.
Contradiction!
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
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