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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。
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。
《矩阵论》复习提纲与习题选讲Chapter1 线性空间和内积空间内容总结：z 线性空间的定义、基和维数；z 一个向量在一组基下的坐标；z 线性子空间的定义与判断；z 子空间的交z 内积的定义；z 内积空间的定义；z 向量的长度、距离和正交的概念；z Gram-Schmidt 标准正交化过程；z 标准正交基。
习题选讲：1、设表示实数域3]x [R R 上次数小于3的多项式再添上零多项式构成 的线性空间（按通常多项式的加法和数与多项式的乘法）。
（1） 求的维数；并写出的一组基；求在所取基下的坐标；3]x [R 3]x [R 221x x ++ （2） 在中定义3]x [R ， ∫−=11)()(),(dx x g x f g f n x R x g x f ][)(),(∈ 证明：上述代数运算是内积；求出的一组标准正交基；3][x R （3）求与之间的距离；221x x ++2x 2x 1+−（4）证明：是的子空间；2][x R 3]x [R （5）写出2[][]3R x R x ∩的维数和一组基；二、 设22R ×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22R ×的维数，并写出其一组基；（2） 在(1)所取基下的坐标； ⎥⎦⎤⎢⎣⎡−−3111（3） 设W 是实数域R 上全体22×实对称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）。
证明：W 是22R ×的子空间；并写出W 的维数和一组基；（4） 在W 中定义内积， )A B (tr )B ,A (T =W B ,A ∈求出W 的一组标准正交基；（5）求与之间的距离； ⎥⎦⎤⎢⎣⎡0331⎥⎦⎤⎢⎣⎡−1221 （6）设V 是实数域R 上全体22×实上三角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）。
证明：V 也是22R ×的子空间；并写出V 的维数和一组基；（7）写出子空间的一组基和维数。
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。
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!。
