南京航空航天大学MatrixTheory双语矩阵论期末考试

南京航空航天大学MatrixTheory双语矩阵论期末考试
南京航空航天大学MatrixTheory双语矩阵论期末考试

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

矩阵论解题步骤-期末考试题

1. 广义逆(必考类型) 假设s x n 矩阵A 的广义逆为G ,且A 可以满秩分解为A = BC ,A 的秩r(A) = r ,则B 为s x r 矩阵,C 为r x n 矩阵。则G 可表示为: H 1 1 C (CC )(B B)B H H H G --= 例题: 步骤:显然,A 要分解为BC ,必须知道A 的秩,故先对A 进行行化简成最简式 ,r(A)=2,故A 满秩分解为A=(3x2) (2x4)=BC.根据A 的最简式来决定B 和C ,B 由A 最简式中只有1的原列组成,C 由A 的最简式的非零首元行组成。 B = , C = ,H 11C (CC )(B B)B H H H A --+=,通过计算即可 得到A 的广义逆。(若B 、C 中有单位矩阵,那么A 的广义逆表达式可去掉矩阵) 性质: 2. 证明r(ABC)r(B)r(AB)+r(BC)+>=

比较重要的性质 (1) ABX=0与BX=0同解 r(AB)=r(B) (2) r(A)=r(H A A ) (3) r(A+B)<=r(A)+r(B) (4) r(AB)<=min[r(A),r(B)] (5) r(AB)>=r(A)+r(B)-n ,其中A=s x n ,B=n x t 步骤: 设r(B)=r ,B 的满秩分解为B=HK ,所以ABC=AHKC , r(ABC)=r(AHKC)>=r(AH)+r(KC)-r (性质(5)) AB=AHK ,故r(AB)<=r(AH),同理得r(BC)<=r(KC),(性质(4)) 从而r(ABC)>=r(AB)+r(BC)-r(B),原式得证 知识点: A . 秩为r 的s x n 矩阵A 必可分解为A=BC ,其中B=s x r ,C=r x n 。该分解称为A 的 满秩分解。 3. nxn 2n n 2V {X |AX ,X C }n X ==∈,证明:12=V n C V ⊕ 证明包含两部分,1)证明12V V ⊕是直和 等价于 证明1 2V {0}V = 2)证明12V n C V ?⊕,12V n C V ?⊕ 步骤:

2016矩阵论试题

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=

矩阵理论2017-2018学年期末考试试题

矩阵理论2017-2018学年期末考试试题 ?、选择题 (每题5分,共25分) 1.下列命题错误的是(A)(B)若,且,则(C)设且,令,则的谱半径为1 (D)设为空间的任意?空间,则2.下列命题错误的是(A)若,则(B)若,则(C)若,则(D)设的奇异值分别为,,如果,则3.下列说法正确的是(A)若,则(B)若为收敛矩阵,则?定可逆 (C)矩阵函数对任何矩阵均有定义,?论A 为实矩阵还是复矩阵 (D)对任意?阵,均有4.下列选项中正确的是(A)且,则为收敛矩阵; (B)为正规矩阵,则(C),则(D)为的所有正奇异值,5.下列结论错误的是(A)若和分别是列满秩和?满秩矩阵,则(B)若矩阵为?满秩矩阵,则是正定矩阵(C)设为严格对?占优矩阵,,则的谱半径(D)任何可相似对?化的矩阵,皆可分解为幂等矩阵的加权和,即?、判断题(15分)(正确的打√,错误的打×) 1.若,且,,则 2.若且,则为到的值域上的正交投影 3.设都是可逆矩阵,且齐次线性?程组有?零解,为算?范数,则 4.,定义,则是上的范数 5.设矩阵的最?秩分解为,则当且仅当 ( ) (A ?B =?)H A H B H A ∈C n ×n =A A 2rank (A )=tr (A )μ∈C n μ=1μH H =E ?2μμH H ,V 1V 2V dim (+)=dim ()+dim () V 1V 2V 1V 2( ) =A ,=A A H A 2=A A +A =A A H A H (=(A m )+A +)m x ∈C n ∥x ≤∥x ≤∥x ∥∞∥2∥1 A , B ∈ C n ×n ≥≥?≥>0σ1σ2σn ≥≥?≥>0σ′1σ′2σ′ n >(i =1,2,?,n )σi σ′i ∥>∥A +∥2B +∥2 ( )A =????π000π001π????sinA =????0000000sin 10?? ??A E ?A e A A A ,B =e A e B e A +B ( )A ∈C n ×n ∥A <1∥m A A ∈C n ×n r (A )=∥A ∥2A ∈(r >0)C m ×n r ∥A =A +∥F r √≥≥?≥σ1σ2σr A ∥=A +∥21σ1 ( ) A B (AB =)+ B +A + A A A H Hermite A =()∈(n >1)a ij C n ×n D =diag (,,?,)a 11a 22a nn E ?A D ?1r (E ?A )≥1 D ?1(i =1,2,?,n )A i A =∑n i =1λi A i A ∈C m ×n A ≠0(A =A A ?)H A ?∥A =n A ?∥2 ( ) A ∈,G ∈C m ×n C n ×m AGA =A y =AGx ,?x ∈C m C m A ( ) A , B ∈ C n ×n (A +B )x =0∥?∥∥A ∥≥1B ?1 ( )?(x ,y )∈R 2f (x ,y )=2+3?4xy x 2y 2 ̄  ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄√f (x ,y )R 2 ( )A A =BD Ax =0Dx =0 ( )

2016矩阵论试题A20170109 (1)

第 1 页 共 4 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 _______ 3332345=-++-A A A A A . 5.??? ? ? ? ?-=λλλλλ0010 1)(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)(

南航双语矩阵论matrix theory第7章部分习题参考答案

第七章部分习题参考答案 Exercise 1 Show that a normal matrix A is Hermitian if its eigenvalues are all real. Proof If A is a normal matrix, then there is a unitary matrix that diagonalizes A . That is, there is a unitary matrix U such that H A UDU = where D is a diagonal matrix and the diagonal elements of D are eigenvalues of A . If eigenvalues of A are all real, then ()H H H H H H A UDU UD U UDU A ==== Therefore, A is Hermitian. Exercise 2 Let A and B be Hermitian matrices of the same order. Show that AB is Hermitian if and only if AB BA =. Proof If AB BA =, then ()()H H H H AB BA A B AB ===. Hence, AB is Hermitian. Conversely, if AB is Hermitian, then ()H AB AB =. Therefore, H H AB B A BA ==. Exercise 3 Let A and B be Hermitian matrices of the same order. Show that A and B are similar if they have the same characteristic polynomial. Proof Since matrix A and B have the same characteristic polynomial, they have the same eigenvalues 12,,,n λλλ . There exist unitary matrices U and V such that 12diag(,,,)H n U AU λλλ= , 12diag(,,,)H n V BV μμμ= . Thus, H H U AU V BV =. (11,H H U U V V --==) That is 1()H H UV AUV B -=. Hence, A and B are similar. Exercise 4 Let A be a skew-Hermitian matrix, i.e., H A A =-, show that (a) I A - and I A + are invertible. (b) 1()()I A I A --+ is a unitary matrix with eigenvalues not equal to 1-. Proof of Part (a) Method 1: (a) since H A A =-, it follows that ()()H I A I A I AA I A A -+=-=+ For any x 0≠ ()()0x x x x x x x x x x H H H H H H H I A A A A A A +=+=+> Hence, ()()I A I A -+ is positive definite. It follows that ()()I A I A -+ is invertible. Hence, both I A - and I A + are invertible. Method 2:

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=

2016北京邮电大学《矩阵分析与应用》期末试题

北京邮电大学 《矩阵分析与应用》期末考试试题(A 卷) 2015/2016学年第一学期(2016年1月17日) 注意:每题十分,按中间过程给分,只有最终结果无过程的不给分。 一、 已知22 R ?的两组基: 111000E ??=? ??? ,120100E ??=????,210010E ??=????,220001E ??=????; 11100 0F ??=? ???,121100F ??=????,211110F ??=????,221111F ??=????。 求由基1112212,,,E E E E 到11122122,,,F F F F 的过渡矩阵,并求矩阵 3542A -?? =?? ?? 在基11122122,,,F F F F 下的坐标。 二、 假定123x x x ,,是3 R 的一组基,试求由112323y x x x =-+, 2123232y x x x =++,312413y x x =+;生成的子空间()123,,L y y y 的基。 三、 求下列矩阵的Jordan 标准型 (1)1 0002 10013202 31 1A ???? ? ?=??????(2)310 0-4-1007121-7-6-10B ?? ????=?????? 四、 设()()123123,,,,,x y ξξξηηη==是3 R 的任意两个向量, 矩阵 210=120001A ?? ???????? ,定义(),T x y xAy = (1) 证明在该定义下n R 构成欧氏空间; (2) 求3 R 中由基向量()()()1231,0,0,1,1,0,1,1,1x x x ===的度量矩阵; 五、 设y 是欧氏空间V 中的单位向量,x V ∈,定义变换 2(,)Tx x y x y =- 证明:T 是正交变换。

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;

南航矩阵论英文小论文_最小二乘法

Anti-frequency Offset Method for Precision Synchronization on Highly Dynamic Condition Based on the Least Squares Abstract: How to achieve precise synchronization is a technical problem that must be addressed in today’s military operations in the highly dynamic conditions. In the conditions, the bit rate offset and carrier frequency offset will affect the precise synchronization accuracy. An improved Pseudo Noise (PN) code phase measurement method based on the least squares method is proposed, and the precise synchronization ability of this method in high-dynamic conditions is proved by theoretical analysis and simulation. The theoretical analysis shows that the method can eliminate the impact of carrier frequency offset and be of strong anti-noise ability. While the numerical simulation manifests that this method is not sensitive to bit rate offset when the length of the selected PN sequence is not very long, and proves that it is of strong anti-frequency offset. The measurement results demonstrate that this method not only achieves a high measurement accuracy, but also possesses good anti-noise and anti-frequency offset ability. Key words: Wireless communication; Pseudo Noise (PN) phase estimate; Precision synchronization;Anti-frequency offset; Highly dynamic condition; Least Squares (LS) 1Introduction With the continuous development of wireless communication and network, the precise synchronization requirement of the system is higher and higher, such as the American GPS , Chinese Beidou Positioning System, American Tactical Tatgeting Network Technology and Intra Flight Data Link Network Technology put forward new requirements on precise synchronization technology, and often need to work under the condition of high dynamic and low noise ratio. Direct Sequence Spread Spectrum, because of its good correlation and good anti-noise ability, is often used in the process of synchronous communication system, especially in the field of aerospace and military communications, Direct Spread Spectrum signal usually applied to complete the range and positioning work [1, 2]. Both synchronous and ranging, the phase measuring accuracy of PN code sequence is the underlying factors of affecting the performance, so a lot of literatures do many researches on the phase measurement of the PN sequence [3-6].For example, literature[3] analysis of the effect of deviation of phase measurement in the time domain, and subsection cross-correlation method is proposed to improve the ability to resist frequency deviation of the system, but the article didn't mention the error range of system accurate synchronization in the end.The literature[4], with fast Fourier transform method, by the process of calculating reference signal and the input signal correlation spectrum achieves DSSS signal PN code phase measurement, this method is of high efficiency and widely used. Defects of the method is, however, that it must increase

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 .

矩阵论考试试题(含答案)

矩阵论试题 一、(10分)设函数矩阵 ()??? ? ??-=t t t t t A sin cos cos sin 求:()?t dt t A 0和(()?2 0t dt t A )'。 解:()?t dt t A 0=()???? ? ??-????t t t t tdt tdt dt t dt t 0 sin cos cos sin =??? ? ??---t t t t cos 1sin sin cos 1 (()?2 t dt t A )'=()??? ? ? ?-=?22 22 2sin cos cos sin 22t t t t t t t A 二、(15分)在3R 中线性变换σ将基 ????? ??-=1111α,????? ??-=1202α,??? ?? ??-=1013α 变为基 ????? ??-=0111β,????? ??-=1102β,??? ? ? ??-=2303β (1)求σ在基321,,ααα下的矩阵表示A ; (2)求向量()T 3,2,1=ξ及()ξσ在基321,,ααα下的坐标; (3)求向量()()ξσξ及T 3,2,1=在基321,,βββ下的坐标。 解:(1)不难求得: ()2111ααβασ-== ()32122αααβασ++-== ()321332αααβασ++-==

因此σ在321,,ααα下矩阵表示为 ??? ? ? ??---=110211111A (2)设()??? ?? ??=321321,,k k k αααξ,即 ??? ? ? ??????? ??---=????? ??321111021101 321k k k 解之得:9,4,10321-=-==k k k 所以ξ在321,,ααα下坐标为()T 9,4,10--。 ()ξσ在321,,ααα下坐标可得 ???? ? ??--=????? ??--????? ??---=????? ??1332239410110211111321y y y (3)ξ在基321,,βββ下坐标为 ??? ? ? ??-=????? ??--????? ??--=????? ??---61519410011111101 94101A ()ξσ在基321,,βββ下坐标为 ????? ??--=????? ??--????? ??--=????? ??---94101332230111111011332231A 三、(20分)设??? ? ? ??-=301010200A ,求At e 。 解:容易算得 ()()()()212--=-=λλλλ?A I

相关文档
最新文档