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 +
-----------------------------------------------------------------------------------------------------------------------
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)
1
100
20012I 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 is
100021002J ?? ?
= ? ???
--------------------------------------------------------------------------------------------------------------------------
(3) For eigenvalue 1, 0
100
1001
1I A ??
?
-=- ? ?-?? , An eigenvector is 1(1,0,0)T p = For eigenvalue 2, 11020
0001
0I 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 that
3(1,1,0)T p =-
101001010P ?? ?
=- ? ???, 111000
1010P -??
?
= ? ?-?
? 1At J e Pe P -=222101001100010
00101000010t
t t t e e te e ??????
? ? ?=- ? ? ? ? ? ?-?????
?222200
00t t t t t t e e e e te
e ??
-
?= ? ?-?? --------------------------------------------------------------------------------------------------------------------
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 2
56x x --.
The minimal polynomial of A divides any annihilating polynomial of A. The possible minimal polynomials are
6x -, 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)2
05500
0A 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 less
than 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 3
1,()()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 from
n R to )(A R , then A is symmetric and the eigenvalues of
A 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 A
is Hermitian.
Let n n B A ?∈C , be Hermitian matrices, and A be
positive definite. Show that
AB 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 are
1’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 BA
Since 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.
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)