南航矩阵论期中考试参考答案.doc
南京航空航天大学矩阵论07-08A试卷及答案.doc
南京航空航天大学研究生考试试卷r 1 1 -2'一、(20 分)设矩阵4= —2 —2 3 ,<-1 -1 1 >(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A6+3A —2/;(4)写出A的Jordan标准型二、(20分)设Z?2"2是实数域上的全体2x2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)求尺2"2的维数，并写山其一组基；(2)设W是全体2x2实对称矩阵的集合，证明：W是/?2x2的子空间，并写出W的维数和一组基；(3)在W中定义闪积G4，B) = Zr(&4),其中人BeW,求出W的一组标准正交基；(4)给出尺〜2上的线性变换7\ T(A) = A+A r, VA G R^2写出线性变换T在(1)中所取基下的矩阵，并求7的核/^r(r)和值域/?(r)。
三、(20分)证明： 是C'w 上的矩阵范数并说明具有相容性（1）求矩阵A 的07?分解；（3）用广义逆判断方程组Av = 6是否相界？若相界，求其通解；若不相容，求其极小最小二乘解 五、（20分）证明：A,, >0, Ar-AgAjAuSO 。
(I-1 1、’2' 1 11，向量/?=11 、0 0b<2>四、（20分）已知矩阵4 =，5 3 2>12、 1）设矩阵汲二3 2 t ，B = 1 1 0.5/t 2； /<2 0.5/ 1 ,，其中f 为实数问当Z 满足什么条件时，A 〉B 成立？Ai A 2 A2 ^22>0,其巾 A u eCkxkau(1)设乂 =2 13 -1 21 ，喇"K, ML, h(2)设4 =(〜)e C ,IX \ 令p=n • max 騸⑶证明：-||<<||<<(2)设 n 阶 Hermite 矩阵 A =(3)己知Hermite 矩陈A =(七)€ (?■ , a ij〉工a ij (= l，2，".，n )，证明：A 正定一、（20 分）(2) VA ，fielV ，V 々e/?，贝ij v (A +B)7= A 7+ B 7= A+B , /. A + B G W ；v (M)7 =kA T = kA ； /.MeW 。
南航矩阵论期中考试试题2013可修改全文
Mid-term Exam of Matrix TheoryPart IRequired Questions (4×15′=60′).Q1.1)Let A = 2−344−686−78,1.Calculate the characteristics polynomial and eigenvalues of A .2.Find the determinant divisors,invariant divisors and elementary divisors of A .2)Given B =(17−645−16)and C =(14−603−13),please determine if B and C are similar or not.And prove your conclusion.Q2.Denote V ={(a 11a 12a 21a 22)∈R 2×2|a 11=a 22}.For any X ∈V ,let T (X )=P X +XP ,where P =(0110).1.Find a basis of V and show the dimension.2.Arbitrarily given A =(a 11a 12a 21a 22)and B =(b 11b 12b 21b 22)in V ,define (A,B )=a 11b 11+2a 12b 12+a 21b 21.Please show that (A,B )is an inner product on V .3.Given an orthonormal basis of V under the inner product of 2.4.Prove that T is a linear transformation on V ,and show the matrix representation of T with respect to the basis given in item 1.Q3.Consider the inner product space C [−1,1]with inner product defined as(f,g )=∫1−1f (x )g (x )dx,∀f (x ),g (x )∈C [−1,1].11.Show that1and3x2−1are orthogonal.2.Determine∥1∥and∥3x2−1∥.3.Let S=L{1,3x2−1}be a subspace of R[x]3,find the optimal approximation of xover S.Q8.Denote R[x]3to be the vector space of zero and polynomials with degree less than3.1.Determine the dimension of R[x]3and give a basis of R[x]3.2.Define the linear transformation D on R[x]3,D(f(x))=f′(x),∀f(x)∈R[x]3.Please give the matrix representation of D with respect to the basis given in the above item.Show R(D)and ker(D).3.Prove that D is not diagonalizable.4.Define the inner product on R[x]3,(f,g)=∫1−1f(x)g(x)dx,∀f(x),g(x)∈R[x]3,please Gram-Schmidt orthogonalize the basis given in item1.Part IIPreferential Questions(2×20′=40′).Q5.For any x∈R n,several definitions are given as follows,∥x∥0=∑x i=0|x i|0,∥x∥p=(m∑i=1|x i|p)1p(0<p<1),∥x∥1=m∑i=1|x i|.(1)1.Please determine if∥x∥0,∥x∥p and∥x∥1are valid vector norms or not.And try todefense your decision.2.Especially when n=2,plot the curves of∥x∥0=1,∥x∥p=1and∥x∥1=1respectively.2Q6.Given A∈R n×n,summarize the necessary and sufficient conditions of A to be di-agonalizable,and prove at least one of them.Determine if the matrix A given in Q1is diagonalizable or not.If yes,please explain why,if not,please give the Jordan canonical form of A.Q7.Given A∈R n×n,denote W={X∈R n×n|AX=XA}.1.Show that W is a subspace of R n×n.2.DenoteD=λ10 00λ2 0............00···λn,whereλ1,λ2,···,λn are different from each other.If A=D,please determine the dimension of W.3.If A is similar to D defined as in item2,please prove that any X∈W is diagonalizable.4.Given some X∈W,if X and A are both diagonalizable,then there exists a nonsingularmatrix P∈R n×n such that P−1XP and P−1AP are diagonal simultaneously.3。
南航双语矩阵论matrixtheory第五章部分习题参考答案
第五章部分习题参考答案#2. Find determinant divisors and elementary divisors of each of the following matrices.(a) 1000100015432λλλλ-⎛⎫ ⎪-⎪ ⎪- ⎪+⎝⎭ (b)001010100000λλλλ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭Solution（a ） 100010()0015432A λλλλλ-⎛⎫ ⎪- ⎪= ⎪- ⎪+⎝⎭det (())A λ4322345λλλλ=++++100det 10101λλ-⎛⎫⎪-=- ⎪ ⎪-⎝⎭. Hence, the determinant divisors are 123()()()1D D D λλλ===,4324()2345D λλλλλ=++++. Invariant divisor are 123()()()1d d d λλλ===,4324()2345d λλλλλ=++++Unfortunately, it is not easy to factorize 4324()2345d λλλλλ=++++ by hand. With the help of Maple or Matlab, we can see that ()A λ has four distinct linear elementary divisors. (b) 44()D λλ=, 123()()()1D D D λλλ===. There is a unique elementary divisor 4λ #3. Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , a a B a εε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭ be n n ⨯ matrices, where 0ε≠. Show that A and B are similar.Proof The Smith normal forms of both I A λ- and I B λ-are11()n a λ⎛⎫ ⎪⎪ ⎪ ⎪-⎝⎭. A and B have the same set of elementary divisors. Hence they are similar to each other. #4. Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , 11a a B a ε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭be n n ⨯ matrices, where 0ε≠. Show that A and B are NOT similar. ProofThe determinant of I A λ- is ()n a λ- . The determinant of I B λ- is ()n a λε--. A and B have distinct characteristic polynomials. Hence, they are not similar.#11. How many possible Jordan forms are there for a 66⨯ complex matrix with characteristic polynomial 42(2)(1)x x +-?Solution The possibilities for the sets of elementary divisors are { 42(2),(1)x x +-}, {4(2),(1),(1)x x x +--}{32(2),(2),(1)x x x ++-}, {3(2),(2),(1),(1)x x x x ++--} {222(2),(2),(1)x x x ++-}, {22(2),(2),(1),(1)x x x x ++--},{22(2),(2),(2),(1)x x x x +++-}, {2(2),(2),(2),(1),(1)x x x x x +++--}{2(2),(2),(2),(2),(1)x x x x x ++++-}, {(2),(2),(2),(2),(1),(1)x x x x x x ++++--}. For each set of elementary divisors, there is a Jordan canonical form up to similarity. There are 10 Jordan canonical forms up to similarity.#12. Classify up to similarity all 33⨯ complex matrices A such that 3A I =. Solution An annihilating polynomial of A is 321(1)()()x x x x ωω-=---, where ω A is diagonalizable.The possibilities for the minimal polynomial of A are1x -, x ω-, 2x ω-;(1x -)(x ω-), (x ω-)(2x ω-), (1x -)(2x ω-);2(1)()()x x x ωω---Up to similarity, all 33⨯ complex matrices A are100010001⎛⎫ ⎪ ⎪ ⎪⎝⎭, 000000ωωω⎛⎫⎪ ⎪ ⎪⎝⎭, 222000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 10001000ω⎛⎫⎪ ⎪ ⎪⎝⎭, 1000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 22000000ωωω⎛⎫ ⎪⎪ ⎪⎝⎭, 2000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭;221000000ωω⎛⎫⎪ ⎪ ⎪⎝⎭,210001000ω⎛⎫⎪ ⎪ ⎪⎝⎭21000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭#14. If N is a nilpotent (幂零的) 33⨯ matrix over C , prove that 21128A I N N =+- satisfies2A I N =+, i.e., A is a square root of I N +. Use the binomial series for 1/2(1)t + to obtain asimilar formula for a square root of I N +, where N is any nilpotent n n ⨯ matrix over C .Use the result above to prove that if c is a non-zero complex number and N is a nilpotent complex matrix, then cI N +has a square root. Now use the Jordan form to prove that every non-singular complex n n ⨯ matrix has a square root.Solution If N is an n n ⨯ matrix and k N O =, then k x is an annihilating polynomial for N . The minimal polynomial of N must be of the form p x , where p n ≤ and p k ≤ since the minimal polynomial of a matrix divides its characteristic polynomial. Thus, n N O =.(1) If N is a nilpotent 33⨯ matrix, then 3N O =. By straightforward computation, we can verify that 2A I N =+.(2) If N is an n n ⨯ nilpotent matrix, n N O =.1/22111111(1)(1)((1)1)122222(1)122!(1)!n n t t t t n -----++=+++++- 1/22111111(1)(1)((1)1)122222()22!(1)!n n I N I N N N n -----++=++++-(3) Since1N c is a nilpotent matrix, 1I N c + has a square root 1/21()I N c+. cI N + has a square root 1/21/21()c I N c+.(4) Suppose that 12121()0()000()r d d d r J J P AP J J λλλ-⎛⎫ ⎪⎪==⎪ ⎪ ⎪⎝⎭. Then each ()k d k J λ has asquare root 1/2()k d k J λ since ()k d k J λ is of the form k I N λ+, where 0k λ≠ because A is nonsingular and N is nilpotent.Let 121/211/2211/2()000()000()r d d d r J J B P P J λλλ-⎛⎫⎪⎪=⎪ ⎪⎪⎝⎭, then 2B A =. Hence, A has a squareroot.#20. Prove that the minimal polynomial of a matrix is equal to the characteristic polynomial if andonly if the elementary divisors are relatively prime in pairs.Proof Suppose that a Jordan canonical form of A is1212()000()000()r d d d r J J J J λλλ⎛⎫⎪ ⎪=⎪ ⎪ ⎪⎝⎭(where 12,,,r λλλ are not necessarily distinct. Each ()i d i J λ is a Jordan block.)The minimal polynomial of A is the same as that of J . The characteristic polynomial of A is the same as that of J . The elementary divisors of A are 11()d λλ-, , ()rd r λλ-The minimal polynomial of ()i d i J λ is ()i d i λλ-. The minimal polynomial of J is the least common multiple (最小公倍式) of 11()d λλ-, , ()rd r λλ-. The characteristicpolynomial of J is 1212()()()()rd d d r p λλλλλλλ=--- .The least common divisor of 11()d λλ-, , ()rd r λλ- is equal to the product of11()d λλ-, , ()r d r λλ- if and only if ()j dj λλ-and ()k d k λλ-are relatively prime forj k ≠. Thus the minimal polynomial of a matrix is equal to the characteristic polynomial ifand only if the elementary divisors are relatively prime in pairs.。
南航矩阵论课后习题答案
南航矩阵论课后习题答案南航矩阵论课后习题答案矩阵论是数学中的一个重要分支，广泛应用于各个领域，包括物理学、工程学、计算机科学等等。
南航的矩阵论课程是培养学生数学思维和解决实际问题的重要环节。
在课后习题中，学生需要运用所学的矩阵理论知识，解答各种问题。
下面是南航矩阵论课后习题的一些答案和解析。
1. 已知矩阵A = [1 2 3; 4 5 6; 7 8 9]，求A的逆矩阵。
解析：要求一个矩阵的逆矩阵，需要先判断该矩阵是否可逆。
一个矩阵可逆的充要条件是其行列式不为零。
计算矩阵A的行列式，得到det(A) = -3。
因此，矩阵A可逆。
接下来，我们可以使用伴随矩阵法求解逆矩阵。
首先，计算矩阵A的伴随矩阵Adj(A)，然后将其除以行列式的值，即可得到逆矩阵。
计算得到A的伴随矩阵为Adj(A) = [-3 6 -3; 6 -12 6; -3 6 -3]。
最后，将伴随矩阵除以行列式的值，即可得到矩阵A的逆矩阵A^-1 = [-1 2 -1; 2 -4 2; -1 2 -1]。
2. 已知矩阵A = [2 1; 3 4]，求A的特征值和特征向量。
解析：要求一个矩阵的特征值和特征向量，需要先求解其特征方程。
特征方程的形式为|A - λI| = 0，其中A为给定矩阵，λ为特征值，I为单位矩阵。
计算得到特征方程为|(2-λ) 1; 3 (4-λ)| = (2-λ)(4-λ) - 3 = λ^2 - 6λ + 5 = 0。
解这个二次方程，得到特征值λ1 = 1，λ2 = 5。
接下来，我们可以求解对应于每个特征值的特征向量。
将特征值代入(A - λI)x = 0，即可求解出特征向量。
对于特征值λ1 = 1，解得特征向量x1 = [1; -1]；对于特征值λ2 = 5，解得特征向量x2 = [1; 3]。
3. 已知矩阵A = [1 2; 3 4]，求A的奇异值分解。
解析：奇异值分解是将一个矩阵分解为三个矩阵的乘积：A = UΣV^T，其中U和V是正交矩阵，Σ是对角矩阵。
南航双语矩阵论-matrix-theory第三章部分题解精选全文
可编辑修改精选全文完整版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 by10()(())(0)p x dx p x p σ⎛⎫⎪= ⎪⎝⎭⎰ Find a matrix A such that()x A ασαββ⎛⎫+= ⎪⎝⎭.Solution1(1)1σ⎛⎫= ⎪⎝⎭ 1/2()0x σ⎛⎫= ⎪⎝⎭11/211/2()1010x 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= ()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 + is101010001S ⎛⎫ ⎪= ⎪ ⎪⎝⎭, 1B 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 thatA ST =andB 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- are linearly independent.(b) there exists a basis E for V such that the matrix representing σ with respect to the basis E is000010000010⎛⎫⎪⎪⎪⎪⎝⎭Proof(a) Suppose that1011()()v v v 0n n k k k σσ--+++= Then 11011(()())v v v 0n n n k k k σσσ---+++=That is, 12210110()()())()v v v v 0n n n n n k k k k σσσσ----+++==Thus, 0k must be zero since 1()v 0n σ-≠. 211111(()())()v v v 0n n n n k k k σσσσ----++==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 - must be all zero. Thisproves that1,(),,()v v v n σσ- are linearly independent.(b) Since 1,(),,()v v v n σσ- are n linearly independent, they form a basis for V .Denote 112,(),,()εv εv εv n n σσ-=== 12()εεσ= 23()εεσ= …….1()εεn n σ-= ()ε0n σ=12[(),(),,()]εεεn σσσ121[,,,,]εεεεn n -=000010000010⎛⎫⎪⎪⎪⎪⎝⎭#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 λλλ. 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 =. 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, whichcontradicts the assumption.This contradiction shows that A can not be similar to a diagonal matrix. OrIf 112diag(,,,)n A P P λλλ-= then 112diag(,,,)k k k k n A P P λλλ-=. k A O = implies that 0i λ= for 1,2,,i n =. Hence, B O =. This will imply that A O =.Contradiction!。
南京航空航天大学2007-2014硕士研究生矩阵论matrixTheory试题
2 3 4 A 4 6 8 6 7 8 。 一（20 分） （1）设
2010 ~ 2011 学年《矩阵论》 课程考试 A 卷
（i）求 A 的特征多项式和 A 的全部特征值； （ii）求 A 的行列式因子，不变因子和初等因子； （iii）写出 A 的 Jordan 标准形；
1 A* A2 A* （3）证明： n 。
1 1 1 1 A 0 0 0 0 四、 （20 分）已知矩阵
（1）求矩阵 A 的 QR 分解；
1 2 0 1 b 1 1 2 1 ，向量 ，
（2）计算 A ；
17 6 14 60 A , B 45 16 3 13 ，试问 A 和 B 是否相似？并说明 （2）设
原因。
2 1 A 1 2 3 1 ，求 A 1 ， A 2 ， A ， A F ； 二（20 分） （1）设
（3）用广义逆判断方程组 Ax b 是否相容？若相容,求其通解;若不相容,求其极小最小二乘解。
五、 （20 分）
（1）设矩阵
问当 t 满足什么条件时, A B 成立？
5 3 2 0 1 A 3 2 t , B 1 1 2 t 2 2 0 .5 t
五（20 分）设
A ( a ij )
为 n 阶 Hermite 矩阵，证明：
3
存在唯一 Hermite 矩阵 B 使得 A B ；
2
（2）
（3） 如果 A 0 ，则 tr ( A)tr ( A ) n 。
1
如果 A 0 ，则 tr ( A ) (tr ( A)) ；
2
南京航空航天大学2009_矩阵论考试考题及答案
(3)，判断方程组 Ax b 是否相容？若相容，求其最小范数解；若不相容，求其极小最小二乘 解。(4 分)
解：
2 0 0 8 1 0 0 4 行 (1): A 0 2 8 0 0 1 4 0 ,故矩阵 A 的满秩分解为： 2 2 8 8 0 0 0 0 2 0 2 0 1 0 0 4 1 0 0 4 A 0 2 CD, C 0 2 , D 。 0 1 4 0 0 1 4 0 2 2 2 2
k
k
k 1
A |||| A k 1 |||| A || || A || k . （5 分）
k
2. || A || 1 lim || A || k 0 lim || A k || 0. （5 分） 3. lim || A k || 0 lim || A k 0 || 0 lim A k 0. （5 分）
学院 ------------------------------ 线 ----------------------------------------------------------------
年级 ----------
从而其极小最小二乘解为：
南航双语矩阵论matrix theory第4章部分习题参考答案
)
If i is a root of p( ) 0 , then p(i ) 0 . We obtain that eigenvalue of C T with eigenvector x (1, i ,, in 2 , in 1 )T .
Exercise 16
Let be an orthogonal transformation on a Euclidean space V (an inner product space over the real number field). If W is a -invariant subspace of V, show that the orthogonal complement of W is also -invariant. Proof Let V W W , where W is -invariant. Let {u1 , u2 ,, uk } be an orthonormal basis for
0 1 T C x 0 0 0 0 1 0 0 0 0 0 0 an 0 an 1 0 an 2 1 a1
T
i i 1 2 2 i i i n2 n 1 n 1 i i i n 1 n n 1 a a a p ( i n i n 1 i 1 i i
C T x i x . Then i is an
(b) If p( ) has n distinct roots, then all roots of p( ) are eigenvalues of C T . We obtain that the characteristic polynomial of C T and p( ) have the same n distinct roots. And also they have the same degree and the same leading coefficient. Hence, the characteristic polynomial of C T is the same as p( ) . Since C and C T have the same characteristic polynomial, we know that p( ) is the characteristic polynomial of C.
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⎪⎪⎪⎭⎫ ⎝⎛-----=111322211A ， （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22⨯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22⨯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22⨯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nn ij C a A ⨯∈=)(，令ijji 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*1A A A n ≤≤。
四、（20分）已知矩阵⎪⎪⎪⎪⎪⎭⎫⎝⎛-=100100011111A ，向量⎪⎪⎪⎪⎪⎭⎫⎝⎛=2112b ， （1）求矩阵A 的QR 分解；（2）计算+A ；（3）用广义逆判断方程组b Ax =是否相容？若相容,求其通解;若不相容,求其极小最小二乘解。
五、（20分）（1）设矩阵⎪⎪⎪⎭⎫⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛=15.025.011210,2223235t t B t t A ，其中t 为实数，问当t 满足什么条件时, B A >成立？（2）设n 阶Hermite 矩阵022121211>⎪⎪⎭⎫⎝⎛=A A A A A H，其中k k C A ⨯∈11，证明：0,012111122211>->-A A A A A H。
南航双语矩阵论matrix theory第5章部分习题参考答案
第五章部分习题参考答案Exercise 2Find determinant divisors and elementary divisors of each of the following matrices.(a) 1000100016733λλλλ-⎛⎫ ⎪- ⎪ ⎪- ⎪--+⎝⎭ (b)001010100000λλλλ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭ Solution（a ） 100010()0016733A λλλλλ-⎛⎫ ⎪-⎪= ⎪- ⎪--+⎝⎭det (())A λ432323376(1)(46)λλλλλλλλ=+--+=-++-222(1)(56)(1)(2)(3)λλλλλλ=-++=-++There is a 3x3 submatrix whose determinant is100det 10101λλ-⎛⎫ ⎪-=- ⎪ ⎪-⎝⎭. Hence, the determinant divisors are 123()()()1D D D λλλ===,4324()3376D λλλλλ=+--+. Invariant divisor are 123()()()1d d d λλλ===,4324()3376d λλλλλ=+--+ The elementary divisors of ()A λ are 2(1)λ-, 2λ+, 3λ+(b) 44()D λλ=, 123()()()1D D D λλλ===. There is a unique elementary divisor 4λExercise 3Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , a a B a εε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭be n n ⨯ matrices, where 0ε≠. Show that A and B are similar.Proof The Smith normal forms of both I A λ- and I B λ-are11()n a λ⎛⎫ ⎪⎪ ⎪ ⎪-⎝⎭.A andB have the same set of elementary divisors. Hence they are similar to each other.Exercise 4Let11a a A a ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ , 11a a B a ε⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭be n n ⨯ matrices, where 0ε≠. Show that A and B are NOT similar.ProofThe determinant of I A λ- is ()n a λ- . The determinant of I B λ- is ()n a λε--. A and B have distinct characteristic polynomials. Hence, they are not similar.Exercise 6For each of the following matrices, find the Jordan Canonical Form J and matrix P such that 1P AP J -=(a) 040140122-⎛⎫ ⎪- ⎪ ⎪--⎝⎭(b)134478677-⎛⎫⎪- ⎪ ⎪-⎝⎭Solution(a)40140122I A λλλλ⎛⎫ ⎪-=-+ ⎪ ⎪-+⎝⎭Determinant divisors are 33()det()(2)D I A λλλ=-=+, 2()det()(2)D I A λλλ=-=+,1()1D λ= Invariant divisors are 23()(2)d λλ=+, 2()(2)d λλ=+, 1()1d λ= Elementary divisors are 2(2)λ+, (2)λ+A Jordan canonical form is 210020002-⎛⎫⎪- ⎪ ⎪-⎝⎭. Let123(,,)p p p P =, then1121233222p p p p p p p A A A =-=-=-Solving (2)x 0A I +=, that is, 123240012001200x x x -⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪-= ⎪⎪ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭, we obtain that {1p ,3p } form abasis for (2)((0,0,1),(2,1,0))T T N I A Span -= . Let 1(2,1,0)(0,0,1),p T T a b =+ To obtain 2p , we solve the system212(2)p p a A I a b ⎛⎫⎪-== ⎪ ⎪⎝⎭, that is,1232402120120y a y a y b -⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪-= ⎪⎪ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭. This system is consistent only if a b =. Let 1a b ==. Then 1(2,1,1)p T =, We solve the system above for vector2(1,0,0)p T =. Take 3(0,0,1)p T =.210100101P ⎛⎫ ⎪= ⎪ ⎪⎝⎭(b) 134478677I A λλλλ--⎛⎫ ⎪-=-+- ⎪ ⎪--⎝⎭23()det()(1)(3)D I A λλλλ=-=+-, 2()1D λ= , 1()1D λ=Invariant divisors are 23()(1)(3)d λλλ=+-, 2()1d λ=, 1()1d λ= Elementary divisors are 2(1)λ+, (3)λ-A Jordan canonical form is 110010003-⎛⎫⎪- ⎪ ⎪⎝⎭. Let123(,,)p p p P =, then11212333p p p p p p p A A A =-=-= 3(1,2,2)p T =, 1(1,2,1)p T =, 2(1,1,0)p T =--111212102P -⎛⎫⎪=- ⎪ ⎪⎝⎭Exercise 8Show that if p A I = for some positive integer p , then A is similar to a diagonal matrix over the complex number field.Proof Since p A I =, 1p x - is an annihilating polynomial. The minimal polynomial ()m x of A must divide 1p x -. Since the polynomial 1p x - has only single roots(单根)，()m x has only single roots. Therefore, by Theorem 5.2.7 (see lecture notes p124), matrix A is diagonalizable.Exercise 9Prove that if an n n ⨯ matrix satisfies 256A A I -=then A is diagonalizable.Proof Since 256A A I -=, 256x x -- is an annihilating polynomial of A . . The minimal polynomial ()m x of A must divide 256x x --. Since 256(6)(1)x x x x --=-+, the minimal polynomial must be a product of distinct linear factors. By Theorem 5.2.7 (see lecture notes p124), matrix A is diagonalizable.Exercise 10Show that if A is nonsingular, then 1A - can be written as a polynomial of A .Proof Let 1110()n n n p x c c c λλλ--=++++ be the characteristic polynomial of A . The constant term of ()p x must not be zero since A is nonsingular. By Cayley-Hamilton Theorem,()p A O =. That is, 1110n n n A c A c A c I O --++++= . Thus,1121101()n n n A A c A c I c ----=-+++ , which is a polynomial of A .Exercise 11How many possible Jordan forms are there for a 66⨯ complex matrix with characteristic polynomial 42(2)(1)x x +-?Solution The possibilities for the sets of elementary divisors are { 42(2),(1)x x +-}, {4(2),(1),(1)x x x +--}{32(2),(2),(1)x x x ++-}, {3(2),(2),(1),(1)x x x x ++--} {222(2),(2),(1)x x x ++-}, {22(2),(2),(1),(1)x x x x ++--},{22(2),(2),(2),(1)x x x x +++-}, {2(2),(2),(2),(1),(1)x x x x x +++--}{2(2),(2),(2),(2),(1)x x x x x ++++-}, {(2),(2),(2),(2),(1),(1)x x x x x x ++++--}. For each set of elementary divisors, there is a Jordan canonical form up to similarity. There are 10 Jordan canonical forms up to similarity.Exercise 12Classify up to similarity all 33⨯ complex matrices A such that 3A I =.Solution An annihilating polynomial of A is 321(1)()()x x x x ωω-=---, where ω= A is diagonalizable.The possibilities for the minimal polynomial of A are1x -, x ω-, 2x ω-;(1x -)(x ω-), (x ω-)(2x ω-), (1x -)(2x ω-);2(1)()()x x x ωω---Up to similarity, all 33⨯ complex matrices A are100010001⎛⎫ ⎪ ⎪ ⎪⎝⎭, 000000ωωω⎛⎫⎪ ⎪ ⎪⎝⎭, 222000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 10001000ω⎛⎫⎪ ⎪ ⎪⎝⎭, 1000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭; 22000000ωωω⎛⎫ ⎪⎪ ⎪⎝⎭, 2000000ωωω⎛⎫ ⎪ ⎪ ⎪⎝⎭;221000000ωω⎛⎫⎪ ⎪ ⎪⎝⎭,210001000ω⎛⎫⎪ ⎪ ⎪⎝⎭21000000ωω⎛⎫ ⎪ ⎪ ⎪⎝⎭Exercise 14If N is a nilpotent (幂零的) 33⨯ matrix over C , prove that 21128A I N N =+- satisfies2A I N =+, i.e., A is a square root of I N +. Use the binomial series for 1/2(1)t + to obtain asimilar formula for a square root of I N +, where N is any nilpotent n n ⨯ matrix over C .Use the result above to prove that if c is a non-zero complex number and N is a nilpotent complex matrix, then cI N +has a square root. Now use the Jordan form to prove that every non-singular complex n n ⨯ matrix has a square root.Solution If N is an n n ⨯ matrix and k N O =, then k x is an annihilating polynomial for N . The minimal polynomial of N must be of the form p x , where p n ≤ and p k ≤ since the minimal polynomial of a matrix divides its characteristic polynomial. Thus, n N O =.(1) If N is a nilpotent 33⨯ matrix, then 3N O =. By straightforward computation, we canverify that 2A I N =+.(2) If N is an n n ⨯ nilpotent matrix, n N O =.1/22111111(1)(1)((1)1)122222(1)122!(1)!n n t t t t n -----++=+++++- 1/22111111(1)(1)((1)1)122222()22!(1)!n n I N I N N N n -----++=++++-(3) Since1N c is a nilpotent matrix, 1I N c + has a square root 1/21()I N c+. cI N + has a square root 1/21/21()c I N c+.(4) Suppose that 12121()00()000()r d d d r J J P AP J J λλλ-⎛⎫ ⎪⎪==⎪ ⎪ ⎪⎝⎭. Then each ()k d k J λ has asquare root 1/2()k d k J λ since ()kd k J λ is of the form k I N λ+, where 0k λ≠ because A is nonsingular and N is nilpotent.Let 121/211/2211/2()000()000()r d d d r J J B P P J λλλ-⎛⎫⎪⎪=⎪ ⎪⎪⎝⎭, then 2B A =. Hence, A has a squareroot.Exercise 20Prove that the minimal polynomial of a matrix is equal to the characteristic polynomial if andonly if the elementary divisors are relatively prime in pairs.Proof Suppose that a Jordan canonical form of A is1212()000()000()r d d d r J J J J λλλ⎛⎫ ⎪⎪= ⎪ ⎪⎪⎝⎭(where 12,,,r λλλ are not necessarily distinct. Each ()id i J λ is a Jordan block.)The minimal polynomial of A is the same as that of J . The characteristic polynomial of A is the same as that of J . The elementary divisors of A are 11()d λλ-, , ()rd r λλ-The minimal polynomial of ()id i J λ is ()id i λλ-. The minimal polynomial of J is theleast common multiple (最小公倍式) of 11()d λλ-, , ()rd r λλ-. The characteristicpolynomial of J is 1212()()()()rd d d r p λλλλλλλ=--- .The least common divisor of 11()d λλ-, , ()rd r λλ- is equal to the product of11()d λλ-, , ()r d r λλ- if and only if ()j dj λλ-and ()k d k λλ-are relatively prime for j k ≠. Thus the minimal polynomial of a matrix is equal to the characteristic polynomial ifand only if the elementary divisors are relatively prime in pairs.。
矩阵论试卷(2012A)答案(1[1].5)
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⎛1 0⎞ ⎛0 0⎞ 5分 W 的一组基为 X1 = ⎜ ⎟ , X = ⎜ 2 ⎜0 1⎟ ⎜1 1 ⎟ ⎟, 维数是 2. ⎝ ⎠ ⎝ ⎠ （ 2 ） 对 任 意 X1 , X 2 ∈ W , k ∈ R, 都 有 T ( X 1 + X 2 ) = T ( X1 ) + T ( X 2 ) ， 9分 T ( kX1 ) = kT ( X1 ) ，所以， T 为线性变换 ⎛1 0⎞ ⎛0 0⎞ （3）对于 W 的一组基为 X1 = ⎜ ⎜0 1⎟ ⎟, X 2 = ⎜ ⎜1 1 ⎟ ⎟, 有： ⎝ ⎠ ⎝ ⎠ ⎛ 0 0⎞ ⎛ − 1 0⎞ T ( X 1) = ⎜ ⎜ 0 0⎟ ⎟ = 0X1 + 0 X2 ， T ( X 2 ) = ⎜ ⎜ 2 1⎟ ⎟ = −1 X1 + 2 X 2 ， ⎝ ⎠ ⎝ ⎠ ⎛ 0 − 1⎞ ⎛ 0 −1⎞ T ( X 1 , X 2 ) = ( X 1 , X 2 )⎜ ⎜0 2 ⎟ ⎟ ， T 在（1）中所取基下的矩阵是 A = ⎜ ⎜0 2 ⎟ ⎟ 14 分 ⎝ ⎠ ⎝ ⎠ ⎛1 0⎞ ⎛0 0⎞ （4）对于 W 的一组基为 X1 = ⎜ ⎜0 1⎟ ⎟, X 2 = ⎜ ⎜1 1 ⎟ ⎟, ⎝ ⎠ ⎝ ⎠ ⎛ − b 0⎞ ⎛ 0 0⎞ 若 T ( aX 1 + bX 2 ) = ⎜ ⎟ ⎜ 2b b ⎟ ⎟=⎜ ⎜ ⎟ ，则有： b = 0 ， ⎝ ⎠ ⎝ 0 0⎠ 所以， Ker (T ) = {kI2 : k ∈ R )} ，维数为 1, 17 分 ⎛ −1 0⎞ R (T ) = {T ( X ) : X ∈ W } = span{T ( X 1 ), T ( X 2 )} = {k ⎜ ⎜ 2 1⎟ ⎟ : k ∈ R} ， ⎝ ⎠
南航双语矩阵论matrixtheory第7章部分习题参考答案
南航双语矩阵论matrixtheory第7章部分习题参考答案第七章部分习题参考答案Exercise 1Show 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 2Let A and B be Hermitian matrices of the same order. Show that AB is Hermitian if and only if AB BA =. ProofIf 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 3Let 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 that12diag(,,,)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 4Let 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:If I A - is singular, then there exists a nonzero vector x such that()x 0I A -=. Thus, x x A =,x x x x H H A =. (1)Since x x H is real, it follows that()x x x x H H H A =.That is x x x x H H H A =. Since H A A =-, it follows thatx x x x H H A -= (2)Equation (1) and (2) implies that 0x x H =. This contradicts the assumption that x is nonzero. Therefore, I A - is invertible. Method 3:Let λ be an eigenvalue of A and x be an associated eigenvector. x x A λ=x x x x H H A λ=. ()x x x x x x x x x x x xH H H H H H H H A A A λλ===-=-Hence, λ is either zero or pure imaginary. 1 and 1- can not be eigenvalues of A . Hence, I A -and I A + are invertible.Method 4: Since H A A =-, A is normal. There exists a unitary matrix U such that 12diag(,,,)H n U AU λλλ=12()()diag(,,,)H H H H H H n U AU U A U U AU ==-= 12diag(,,,)n λλλ= 12diag(,,,)n λλλ- Each j λ is pure imaginary or zero. 12(diag(,,,))H n I A U I U λλλ-=-12diag(1,1,,1))H n I A U U λλλ-=---Since 10i λ-≠ for 1,2,,j n = , det ()0I A -≠. Hence, I A - is invertible. Similarly, we can prove that I A + is invertible.Proof of Part (b) Method 1:Since ()()()()I A I A I A I A +-=-+, it follows that11[()()]()()H I A I A I A I A ---+-+11()()()()H H I A I A I A I A --=+--+ ( Note that 11()()H H P P --= if P is nonsingular.)11()()()()I A I A I A I A --=-+-+ 11()()()()I A I A I A I A I --=--++=Hence, 1()()I A I A --+ is a unitary matrix. Denote 1()()B I A I A -=-+.Since 111(1)(1)()()()()2()I B I I A I A I A I A I A I A -----=---+=-++-+=-+,1det()(2)det[()]0n I B I A ---=-+≠Hence, 1- can not be an eigenvalue of 1()()I A I A --+. Method 2:By method 4 of the Proof of Part (a),12diag(1,1,,1))H n I A U U λλλ-=---12diag(1,1,,1))H n I A U U λλλ+=+++1()()I A I A --+1212111diag(,,,))111H n nU U λλλλλλ---=+++ The eigenvalues of 1()()I A I A --+ are1212111,,,111n nλλλλλλ---+++ , which are all not equal to 1-.Method 3: Since ()()()()I A I A I A I A +-=-+, it follows that11()()()()I A I A I A I A ---+=+-If 1- is an eigenvalue of 1()()I A I A --+, then there is a nonzero vector x , such that1()()x x I A I A --+=-. That is 1()()x x I A I A -+-=-.It follows that()()x x I A I A -=-+.This implies that x 0=. This contradiction shows that 1- can not be an eigenvalue of1()()I A I A --+.Exercise 6If H is Hermitian, show that i I H - is invertible, and 1(i )(i )U I H I H -=+- is unitary. Proof Let i A H =-. Then A is skew-Hermitian. By Exercises #4, I A - and I A + are invertible, and 1()()U I A I A -=-+ is unitary. This finishes the proof.Exercise 7Find the Hermitian matrix for each of the following quadratic forms. And reduce each quadratic form to its canonical form by a unitary transformation (a) 12312131213(,,)i i f x x x x x x x x x x x =+-+ Solution()1123123230i 1(,,)i 00100x f x x x x x x x x ???? ???=- ??? , 0i 1i 00100A ?? ?=- ? ???3d e t ()2I A λλλ-=-. Eigenvalues of Aare 1λ=2λ=30λ=.Associated unit eigenvectors are1i 1,)22u T =-, 2i 1,)22u T =-, and3u T =, respectively. 123,,u u u form an orthonormal set.Let 123(,,)u u u U =, and x y U =. Then we obtain the canonical form1122y yExercise 9Let A and B be Hermitian matrices of order n , and A be positive definite. Show that AB issimilar to a real diagonal matrix.Proof Since A is positive definite, there exists an nonsingular Hermitian matrix P such that H A PP = 1()H H AB PP B P P BP P -==AB is similar to H P BP . Since H P BP is Hermitian, it is similar to a real diagonal matrix. Hence, AB is similar to a real diagonal matrix.Exercise 10Let A be an Hermitian matrix of order n . Show that there exists a real number 0t such that t I A +is positive definite.Proof 1: The matrix t I A + is Hermitian for real values of t . If the eigenvalues of A are12,n λλλ，，, then the eigenvalues of t I A +are 12,,n t t t λλλ+++ ，. Let 12max{,,}n t λλλ> ，Then the eigenvalues of t I A + are all positive. And hence, tI A +is positive definite.Proof 2: The matrix t I A + is Hermitian for real values of t . Let r A be the leading principle minor of A of order r .d e t ()r r r I A t +=+terms involving lower powers in t . Hence, det()r r t I A + is positive for sufficiently large t .Thus, if t is sufficiently large, all leading principal minors of t I A + will be positive.That is, there exists a real number 0t such that det()r r t I A + is positive for 0t t > and for each r . Thus t I A + is positive definite for 0t t >.Exercise 11 Let11121222H A A A A A ??=be an Hermitian positive definite matrix. Show that 1122det()det()det()A A A ≤Proof We first prove that if A is Hermitian positive definite and B is Hermitian semi-positivedefinite, then det()det()A B A +≥. Since A is positive definite, there exists a nonsingular hermitian matrix P such thatHA P P =11(())H H A B P I P B P P --+=+ 11det()det()det(())H A B A I P B P --+=+11()H I P B P --+ is positive semi- definite. Its eigenvalues are all greater than or equal to 1.Thus11det(())1H I P B P --+≥111121112112111222H H I O A A I A A A A I A A O I --??-?-11112111112112212111222121112H H A A A O I A A O A A A A O A A A A O I ---??-== ? ? ?--?122121112H A A A A -- is positive definite, and 1121112H A A A - is positive semi-definite, and11122121112det()det()det()H A A A A A A -=- Hence, 111222212111212111222121112det()det()det(H H H A A A A A A A A A A A A ---=-+≥-）This finishes the proof.Exercise 12Let A be a positive definite Hermitian matrix of order n . Show that the element in A with the largest norm must be in the main diagonal.Proof Let ()ij A a =. Suppose that 00i j a is of the largest norm, where 00i j ≠. Consider theprincipal minor 00000000i i i j i j j j a a a a ??. It must be positive definite since A is positive definite. (Recall that an Hermitian matrix is positive definite iff all its principal minors are positive.) Thus, 00000000det 0i i i j i j j j a a a a ??>. On the other hand, 000000000000002det 0i i i j i i j j i j i j j j a a a a a a a ??=-≤since 00i j a is of the largest norm.(Remark: The diagonal elements in an Hermitian matrix must be real.)This contradiction implies that the element in A with the largest norm must be in the main diagonal.。
南京航空航天大学矩阵论11-12试卷及答案.doc
南京航空航天大学2011级硕士研宄生共5页第1页2011〜2012学年第1学期《矩阵论》课程考试A卷考试日期：2012年1月9日，：学院专业学号姓名成绩、2 1、二(20 分)(1)设-1 23>(2)设A =(力)eC，，证明：(i)对m阶酉矩阵f/和n阶酉矩阵V,有||[MV||F=||<.;(ii)若胭A(A) = r，…，<7,.为A的全部正奇异值，则玄〜f。
A-I /=1 7=1AM 14 00 5Vl4；11<=3;11<=初.r 4’r r⑵(i)||[MV||F =[zr((f/AV)H[/AV)]2 = [tr(y H A H U H UAV}}=[tr(y n A n AV)y = [tr(V-l A,J AV)y =[tr(A n A)y =||A||(ii)因为md(A) = r,则l+l奇异值分解定理知，存在m阶酉矩阵(7和zi阶酉矩阵V,其中，…，C7,.),从而Z 0 04 10 1、三（20分）设焱= 0 110，b = 0J 21 1,<4；（1） 计算A 的满秩分解；（2） 计算广义逆矩阵4+;（3） 用广义逆矩阵判定线性方程组Ar = /7是否相容。
若相容，求其通解; 若不相容，求其极小最小二乘解。
22 A += c T(cc Tr\B TBy 1B T z 5 -4 r 03 3 15 -57 2 5-4 169，，1 51"3<55>0、，1 1 0 1、 0 10 1 1 o y J b \ 71) A715 本b，所以Ax = 6不相容,Ax = b 的极小最小二乘解为x = /V7?1 15 ，19、12-7r 2 -1 0 '四(20分)(1)设4= -13-3,判断A是否是正定或半正定矩阵，并<0-3 2〉说明理由；(2)设A是H阶Hermite正定矩阵，5是阶Hermite矩阵，证明：相似于实对角矩阵；(3)设B均为《阶Hermite矩阵，并且= /I是AB的特征值，证明：存在A的特征值汉和S的特征值A，使得A =(1) 因为A的顺序主子式A1=2〉0，A2=5〉0,A3=-8<0,所以A不是正定的。
南航矩阵论考试试题
南航矩阵论考试试题南航矩阵论考试试题南航矩阵论考试是一门重要的数学课程，旨在培养学生的逻辑思维和解决问题的能力。
本文将介绍一些典型的南航矩阵论考试试题，帮助读者更好地理解这门课程的内容和要求。
一、基础知识部分1. 请解释矩阵的定义和基本性质。
矩阵是由数个数按矩形排列而成的表格。
它的定义包括行数和列数两个维度，记作m×n。
矩阵有很多基本性质，如加法、数乘、转置等。
矩阵的加法满足交换律和结合律，数乘满足分配律。
矩阵的转置是将矩阵的行和列互换得到的新矩阵。
2. 什么是方阵和单位矩阵？方阵是行数等于列数的矩阵。
单位矩阵是一个对角线上全为1，其余元素全为0的方阵。
单位矩阵在矩阵运算中起到了重要的作用，类似于数学中的“1”。
二、矩阵运算部分1. 请计算以下矩阵的和：A = [1 2 3; 4 5 6]，B = [7 8 9; 10 11 12]。
矩阵的和等于对应位置元素相加得到的新矩阵。
根据题目给出的矩阵，可以计算得到A + B = [8 10 12; 14 16 18]。
2. 请计算以下矩阵的积：C = [1 2; 3 4]，D = [5 6; 7 8]。
矩阵的乘法需要注意行列对应元素的乘积。
根据题目给出的矩阵，可以计算得到C × D = [19 22; 43 50]。
三、线性方程组部分1. 请解以下线性方程组：2x + 3y = 8，4x - 5y = 7。
线性方程组可以转化为矩阵的形式，即AX = B，其中A为系数矩阵，X为未知数矩阵，B为常数矩阵。
根据题目给出的线性方程组，可以得到矩阵形式为：[2 3] [x] [8][4 -5] [y] = [7]通过矩阵的逆运算，可以解得x = 3，y = 2。
2. 请解以下线性方程组：x + 2y + 3z = 6，2x - y + z = 1，3x + 4y + 5z = 10。
同样地，将线性方程组转化为矩阵形式：[1 2 3] [x] [6][2 -1 1] [y] = [1][3 4 5] [z] [10]通过矩阵的逆运算，可以解得x = 1，y = 2，z = 1。
矩阵论试题及答案可编辑全文
2006矩阵论试题答案一．填空（每题4分，共40分）1． 设−−=41311221222832A ，则A 的值域4(){,R }R A y y Ax x ==∈的维数=)(dim A R 2 .2． 设A 的若当标准型−−−=10000011000001100000020000012000002J ，则A 的最小多项式=)(λψm 32(1)(2)λλ+−.3． 设110430102A −=−，则()5432333h A A A A A A =−++−=110430102−− −−. 4． 设埃尔米特阵为 −−+=2005111i i i i A ， 则矩阵A 为 正定的 埃尔米特阵.5． 在3R 中有下列两组向量：()13,1,2Tα=−−，()21,1,1Tα=−，()32,3,1Tα=−； ()11,1,1Tβ=，()21,2,3Tβ=，()32,0,1Tβ=，则由321,,ααα到321,,βββ的过渡矩阵=P 619113421270−−−−−− −− .6．设33CA ×∈，21332211{}ij m j i A a ===∑∑，H AA 的非零特征值分别为15 ,5 ,3，则=2mA.7． 设12102101, 11111137A B −== −−，12,V V 分别为齐次线性方程组 0Ax =，0Bx =的解空间，则=)dim(21V V ∩ 1 .8． 设1(1)1(1)121()321nn n n n n n A n n n n +−−=++ −，则lim n n A →∞=1311e .9． 设213121202A −=，则A 的 LDU 分解为 A =100121012/51 2001123205200115004/5001− − − 10．设 −=5221A ，=0242B ，则2448204048102040100A B−−−⊗=. 二．（10分）设T 为n 维欧氏空间V 中的线性变换，且满足：),(),(Ty x y Tx −=，试证明：T 在标准正交基下的矩阵A 为反对称阵（T A A −=）证明：设n ααα,,,21 为V 的标准正交基，n n ij a A ×=}{，下证：ji ij a a −=： 由=),,,(21n T ααα A n ),,,(21ααα 知n ni i i i a a a T αααα+++= 2211，n nj j j j a a a T αααα+++= 2211， ),(),(j i j i T T αααα−=；=),(j i T ααji j n ni i i a a a a =+++),(2211αααα ， =),(j i T ααij n nj j j i a a a a =+++),(2211αααα , 所以：ji ij a a −=.三．（10分）在复数域上求矩阵−−−=7137341024A 的若当标准形J ，并求出可逆矩阵P 使得J AP P =−1.解： A 的若当标准形210021002J=. 令123(,,)P p p p =，则有112123232,2,2Ap p Ap p p Ap p p ==+=+;1213262100621062104170,417,4173150315315p p p p p −−−−=−=−= −−−解得：123(2,1,1),(0,1,0),(1,2,1)T T Tp p p ===− , 201112101P=−.四. （10分）已知 =654321x x x x x xX ，162534()sin()x x f X e x x x x =++，求dXdf . 解答：16161234652543225516cos()cos()x x x x ff f x x x df dX ff f x x x x e x x x x x x x x x e ∂∂∂∂∂∂== ∂∂∂ ∂∂∂. 五.（10分）已知311202113A −=−−−，求4sin()A π，Ae .解：3||(2)E A λλ−=−，A 的最小多项式2)2()(−=λλϕ .待定系数一：令24sin ()(2)q a b πλλλλ=−++，则21,0a b b +==,4sin()A E π=;令2()(2)e q a b λλλλ=−++，则222,a b e b e +==.222211212112A e e e E e A −−=−+=− −−.待定系数二：令324sin ()(2)q a b c πλλλλλ=−+++，则22222414018,8,32216a b c b c a b c c ππππ ++=+=⇒=−==− =− ; 224sin()(44)32A E E A A E ππ=−−+=.令32()(2)e q a b c λλλλλ=−+++，则2222222414,,22a b c e b c e a e b e c e c e++= +=⇒==−== ; 2221()2211212112A e e E A A e −−− =− +−−= .六.（10分）设−=01200110A ，求A 的奇异值分解. 解答一：=5002A A H ，A 的奇异值为5,2; 00Σ= , 25H HV A AV = ，1001V =; 1100100100200100U AV −−− =Σ==; 00000000U− =; 0000010001 0 000 0 000A=.解答二：=5002A A H ，那么A 的奇异值为5,2，A A H对应于特征值5，2的标准特征向量为 = =01,1021x x ，=0110V ; 再计算H AA 的标准正交特征向量，解得分别与5，2，0，0对应的四个标准正交特征向量=0520511υ， −=2102102υ，−=0510523υ，=2102104υ，−−=210210051052210210052051U ; 所以=∆=HV UA 0000000000000110.七．（10分）设n n i A ×∈≠C 0，2rank rank i i A A =),,2,1(n i =，且当i j ≠时),,2,1,(0n j i A A j i ==.试用归纳法证明存在同一个可逆阵n n P ×∈C 使 得对所有的i ),,2,1(n i =有1−=P PE a A ii i i ，其中C ∈i a . 证明：1n =时，命题显然.假设n k ≤时，命题成立. 当1n k =+时，设1rank A r =.由若当分解11111000D A P P − =，其中1C r rD ×∈可逆; 当2,,j n = 时，由110j j A A A A ==可得1(1)(1)1100, C 0n n j jj A P P B B −−×− =∈(直接推出的j B 为()()n r n r −×−的) 再由0i j A A =得0i j B B =(,,2,,)i j i j n ≠= ;0j B ≠,2rank rank j j B B =也是明显的.由假设知存在可逆阵(1)(1)C n n Q −×−∈使得1j j jj B a QE Q −=，其中C j a ∈，2,,j n = .此时,再由110j j A A A A ==得到11111111110101010000000a A P P a P P Q Q −−− == ; 记1100P P Q =，则 11111111100000000 (2,,).0 j j j jj j j jj jj A P P P P B a QE Q a P P a P E P j n E −−−−− =====由归纳原理知命题为真.。
南京航空航天大学07-08矩阵论答案(B)
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南航矩阵论第一章作业答案与提示
矩阵论作业答案与提示第一章（P41-P44）8提示：设044332211=+++ααααx x x x ，解得04321====x x x x ，因此4321,,,αααα线性无关．10（1）提示：考虑n 阶反对称矩阵构成的线性空间V ．设ij α是处的元素为1，处的元素为-1，而其余元素均为零的n 阶反对称矩阵（），则),(j i ),(i j j i <n n n 2n ,123112,,,,,,,−ααααL L L A ij 线性无关．又若V a α∈=)(，则有∑≤<≤=nj i ijija A 1α，即A 可以由n n n n ,1223112,,,,,,,−αααααL L L 线性表示，因此.2)1(12)2()1()dim(−=+++−+−=n n n n V L 同理，若V 是n 阶对称矩阵构成的线性空间，则.2)1()dim(+=n n V 12提示：设A x x x x =+++44332211αααα，解得1,1,3,24321−===−=x x x x ，因此A 在基4321,,,αααα下的坐标是.)1,1,3,2(T −−18提示：（1）对任意P ,,∈∈k W Y X ，直接验证W kX Y X ∈+,．（2）在中取向量W )(i i e diag =α，其中表示第i 个分量为1，其余分量为零的n 维行向量，，则i e n i ,,2,1L =n ααα,,,21L 线性无关．又若，则由W x X n n ij ∈=×)(XA AX =得到，即)(0j i x ij ≠=),,,(2211nn x x x diag X L =．于是∑==ni i ii x X 1α，即X 可由n ααα,,,21L 线性表示．因此n ααα,,,21L 是的一组基，而．W n W =)dim(19（1）提示：设},{},,{212211ββααspan V span V ==，则},,,{212121ββααspan V V =+．由于121,,βαα是向量组2121,,,ββαα的极大线性无关组，所以，而3)dim(21=+V V 121,,βαα是21V V +的一组基．接下来，求的维数和基．设21V V I 21V V I ∈α，则有24132211ββαααk k k k −−=+=，从而024132211=+++ββααk k k k ．解这个向量方程得到：,,3,4,4321k k k k k k k k =−=−==其中k 是任意常数．此时,)4,3,2,5()3()4(2121T k k k −−−=−=−=ββααα即．于是})4,3,2,5{(21T span V V −−−=I 1)dim(21=V V I ，而是的一组基． T )4,3,2,5(−−−21V V I 21提示：设,)1,,1,1(,)1,1,0,,0(,)0,,1,1,0(,)0,,0,1,1(121Tn T n T T L L L L =−=−=−=−αααα则},,,{},{},,,,{112121211n n n n span V V span V span V ααααααα−−=+==L L ．由于n n ααα,,,11−L 线性无关，所以它们构成n R 的一组基，从而．注意到，于是． n R V V =+21}0{21=V V I n R V V =+⋅2124提示：在V 中取向量,1100,0010,0011321⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=ααα 则321,,ααα线性无关，且321αααc b a c c b a a ++=⎟⎟⎠⎞⎜⎜⎝⎛+， 从而，而3)dim(=V 321,,ααα是V 的一组基．定义映射如下：3:R V →σ,)(⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛+c b a c c b a a σ 由于是向量在基T c b a ),,(⎟⎟⎠⎞⎜⎜⎝⎛+=c c b a a α321,,ααα下的坐标，所以σ是V 到3R 的同构映射．27（1）提示：首先将321,,ααα化为标准正交向量组，得到.)2,1,1,2(101,)2,3,3,2(261,)1,2,2,1(101321T T T −−−=−=−=εεε其次，解方程组，求得基础解系，将其单位化，得0321===x x x TT T εεεT )3,2,2,3(4−=αT )3,2,2,3(2614−=ε，则4321,,,εεεε是V 的标准正交基．最后，直接计算，得到311010εεα+=． （2）解答：212321362),31(4103,26,21εεαεεε+=−===x x ．。
矩阵论考试题和答案(详细)
页
第 1 页
课程编号: A000003 考试日期: 2009 年 1 月 13 日
λ I − A = λ (λ + 1)2
---------------3 ----------------3 -------------6 --------------2 ---------------2
= P −1 AP 满足相容矩阵范数的四个条件。
共 4 页
第 2 页
三、 （20 分）
（1） A 的满秩分解为 1 0 − 1 0 1 A = 0 1 0 1 0 − 1 0
A + = C T ( CC
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(tr ( A))2 = (λ1 + L + λn ) 2 ≥ λ12 + L + λn2 = tr ( A2 ) 。 ---------------4
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n = tr ( I ) = tr ( AA−1 ) = tr ( AH A−1 ) ≤ tr ( AH A)tr ( A− H A−1 ) 2 = tr ( A2 )tr ( A−2 ) 2
由（2）知 tr ( A2 ) ≤ tr ( A), tr ( A−2 ) ≤ tr ( A−1 ) ，因此n ≤ tr ( A)tr ( A−1 ) 。 -则存在与 . 相容的向量范数 . a ，从而
| λ | x a = λ x a = Ax a ≤ A x a , | λ −1 | x a ≤ A−1 x
南航双语矩阵论第1章部分习题参考答案
Solution Key (chapter 1)Exercise 2.The show that this set is not closed under multiplication.TakeS ,2=.But 2S ∉.If 2S ∈rational numbers a and b ,such that2=It is clear that 0a ≠and0b ≠.)This will 224232a 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.zx y x +=+)()()()(z x x y x x ++-=++-z x x y x x ++-=++-])[(])[(z 0y 0+=+zy =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 in12{,,,}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 19By 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 --=⋅+-⋅-+-+ 12(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)Tn n n n n j n ------ （Exercise 22Use 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 haveb 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 nR ⨯Let ,A B K ∈.Then ()()T T T A B A B A B A B +=+=--=-+,and ()()T T kA kA kA ==-.Kis 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 TA A =and TA 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.Forany ()m n ij ij A a C ⨯=+∈,where ,ij ij a b are real numbers,A can be written as1111n m n mij 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 1111n m n m ij ij ij ij j i j i a E b F O ====+=∑∑∑∑,then 0ij 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 nC ⨯.Thedimensionis 2mn .Note that,all coefficients of linear combinations must be real numbers because theunderlying field is the real number field.。
南航双语矩阵论matrix theory第3章部分习题参考答案
1
Exercise 8
Let S be the subspace of C[a, b] spanned by e x , xe x , and x 2 e x . Let D be the differentiation operation of S, i.e., D( f ) f ' . Find the matrix representing D with respect to [e x , xe x , x2 e x ] Solution
3
This will imply that k1 must be zero since n 1 ( v) 0 . By repeating the process above, we obtain that k0 , k1 , , kn 1 must be all zero. This proves that
Solution
(1) 1
1
1
( x)
1 / 2 0
( x) 1 0 1
Hence, A
1 1/ 2 1 0
1/ 2 1 1/ 2 0
4
If B is nonsingular, then AB B1 ( BA) B . AB and BA are similar.
A P diБайду номын сангаасg(1 , 2 , , n ) P1 then Ak P diag(1k , 2k ,
, n . Hence, diag(1 , 2 ,
, n k ) P1 .
Ak O implies that i 0 for i 1, 2,
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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=
T "） = %2+"12 _1
_
02+"12 -o r
%2 +Sn _ i
%1+”21
%2 +腐2 _ 1 0 %1 +021 +。
12 +"12
+811 + %2 +012 %2 +§T1 + G 1 +J1
%2 +0\2 +%1 +目21 T （a ）+T （/?）=T （a+"）.T （ka ） = R 任1 +。
12 。
》+名] 0] +0>7 +4] 灯（a ）.所以T 是V 上的线性变换.T （/） = T （%） =
1 1
0 0矩阵A 称为线性变换T 在
0 0
Q3 1）
2）
l|| = 7（n ）=V2 , |3X 2
-1|| = ^
（3X 2-1,3X
2
3）
设S 中的任意向量为K I +K2（3X 2-
-（^ + ^（3\2-1））|| = ^ j （x-3K 2x 2-K 1 + K2）2</% =
则£|,旦，§是标准正交基•
4）设Q, ”CV, k 为数域P 上的任意数.
0 丁（。
|,%,。
3）二（。
1，%，％）人. A= 2
2
QIZ 2 7
—+2K
《+—.明显看 5 I
3
出当K!=K 2=O 时上式值
最小。
所以x 在S 上的最佳逼近为0向量.
Q4
1）维数为 3, 一组基为 0=1, a 2=x , a y = x 1.
D （l ）=0-l+0x + 0-x 2
2） D （x ）=l-l+0 x + 0-x 2 或表示成D （l, x, x 2） = （l, x, x?）A.则线性变换在
D （X 2
）=0-1+2-X + 0-X
2
I
（1,3亍一 1）= j （3x 2
-l ）^/x = 0所以 1 和 3X 2
-1 是正交的.
-1
1 %+坊 0 %1+”21 8
3)
4) 0 1 0
基1, X, x2下的矩阵为A= 0 0 2
0 0 0
A -1 |AE-A|= 0 2 0 0
-2 =23 4.当4=0的时候，秩r(AE-A )= 2,所以D 不可对角化.
A
£\ -ii^ii ->
腐=%一(%,£|)弓=X ，£2
~
-
二。
3 - ( ％，*1 )与 一 (0，£ ?)《2 = JV~ - I / 3 ,《
Q5
1) Ikllp ,||域0为无效范数，kill 为有效范数.
对于||圳°,当x={i ，i ，i …}时，|HHHII %P G P
不满足齐次性，所以为无效 范数. 对于 x p,当a={0,a,0,(),•••}, 0={a,0,0,(),•••}, (x,/3e. R n , p =—时.
2)
Q6
R + "||p =(2网)2 =4|d ,而妆|| +1加|| = 2同，所以不满足三角不等式.||x||p 为 无效范数，另外根据Minkowski 不等式，只有当P21才满足三角不等式的条件, 也可也得到lk||p 为无效范数.对于凤，容易证明满足非负性，齐次性和三角不等 式。
故为有
效范数.
1) 证明存在可逆矩阵P,使P 'AP=A ,其中A 为对角矩阵.
证明
A 的特征值互异.
A* X】…，* X,…，X n]PA
对于Q1中的矩阵，是不可对角矩阵.因为当特征值为0.
.
. .
因为X], X2,---, Xn线性无关，故P为满秩矩阵，所以AP=PA ,即存在可逆矩阵使得P"AP=A,即A可对角化.
必要性：已知存在可逆矩阵P使P J A P=A,设P=[P|, P2,…，P n],则有
[A P p AP2,..-, APJ = [/11P1, /l2P2,---, /l n P n]，可知九为A 的特征值，％为A 的特
征向量.即A有n个线性无关的特征向量.
秩为2壬〃一§,S,・=2/ = 3.根据Q1的初等因子可得Jordan形矩阵为
3) 证明对于A的每一个特征值W都有秩(4义- A) = 〃 - S’，S,.是九的重数.
4) 矩阵A可对角化的充要条件是有n个线性无关的特征向量.
证明：充分性：设A的n个线性无关的特征向量为XI, X2, X3,…，Xn.则, i = 1,2,…，〃。