南京航空航天大学Matrix-Theory双语矩阵论期末考试2015

NUAALet 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 representingmatrix 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)110020012I 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 is100021002J ⎛⎫ ⎪= ⎪ ⎪⎝⎭--------------------------------------------------------------------------------------------------------------------------(3) For eigenvalue 1, 010010011I A ⎛⎫⎪-=- ⎪ ⎪-⎝⎭ ， An eigenvector is 1(1,0,0)T p = For eigenvalue 2, 1102000010I 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 that3(1,1,0)T p =-101001010P ⎛⎫ ⎪=- ⎪ ⎪⎝⎭, 1110001010P -⎛⎫⎪= ⎪ ⎪-⎝⎭ 1At J e Pe P -=22210100110001000101000010tt t t e e te e ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭22220000t t t t t t e e e e tee ⎛⎫-⎪= ⎪ ⎪-⎝⎭ --------------------------------------------------------------------------------------------------------------------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 256x x --.The minimal polynomial of A divides any annihilating polynomial of A. The possible minimal polynomials are6x -, 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)2055000A 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 lessthan 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 31,()()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 fromn R to )(A R , then A is symmetric and the eigenvalues ofA 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 Ais Hermitian.Let n n B A ⨯∈C ， be Hermitian matrices, and A bepositive definite. Show thatAB 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 are1’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 BASince 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.。

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.。

南京航空航天大学研究生课程《矩阵论》内容总结与习题选讲

《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。

南京航空航天大学矩阵论08-09A试卷及答案

（1）求的维数，并写出的一组基；
（2）在中定义线性变换：
求在（1）中所取基下的矩阵表示；
（3）求（2）中线性变换的值域和核，并确定他们的维数；
（4）在中能否取一组基使得（2）中线性变换在所取基下的矩阵为对角矩阵？如果能，则取一组基；如果不能，则说明理由。
五（20分）设为阶Hermite矩阵，证明：
（1）存在唯一Hermite矩阵使得；
（2）如果，则；
（3）如果，则。
三20分）设，，
（1）作出的满秩分解，计算；
（2）应用广义逆矩阵判断线性方程组是否相容。若相容，求其通解；
若不相容，求其极小最小二乘解。
（3）设是实矩阵，是维实向量，证明：不相容线性方程组的最小二乘解唯一当且仅当列满秩
四（20分）设表示实数域上全体上三角矩阵作成的线性空间（对矩阵的加法和数量乘法）。
一、（20分）设，
（1）求的特等因子和最小多项式；
（3）写出
的Jordan标准形
。
二（20分）（1）设，求，，，；
（2）设是上的相容矩阵范数，证明：
（i）如果是阶可逆矩阵，是的任一特征值，则；
（ii）如果是可逆矩阵，令，则是上的相容矩阵范数。

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.

南京航空航天大学MatrixTheory双语矩阵论期末考试

(2) The Jordan canonical form is
--------------------------------------------------------------------------------------------------------------------------
(2) Find a basis for such that with respect to this basis,thematrixBrepresenting is diagonal.
(3) Find thekernel（核）andrange（值域）of this transformation.
Solution:
南京航空航天大学Matrix-Theory双语矩阵论期末考试
———————————————————————————————— 作者：
———————————————————————————————— 日期：
Part I (必做题，共5题，70分)
第1题（15分）
得分
Let denote the set of all real polynomials of degree less than 3 withdomain（定义域） .The addition and scalar multiplication are defined intheusual way.Definean inner product on by
第2题（15分）
得分
Let be the linear transformation on (the vector space of real polynomials of degree less than 3) defined by

南航矩阵论试卷

1．作出的一个满秩分解；
2．求的加号逆；
3．求方程组的极小范数解〔要求解中不含有参数t〕．
答案及评分标准:
1. 的一种满秩分解为；
(注意：满秩分解不唯一，需要检验)．
2.因为 ,所以
.
3.由相容性,解得 ,从而极小范数解为 .
共5页第4页
四、(20分)设矩阵 .
1．求；
2．证明对于中的任意矩阵，有；
案及评分标准：
1.特征多项式为；初等因子为 .
2. 的最小多项式是，Jordan标准形为 .
3.因为的初等因子为，与的初等因子不同,所以与不相似.
〔5分〕
共5页第2页
二、(20分)设，映射使得
.
1．证明是的一个子空间，并求它的维数和基；
2．证明是的线性变换，并求在题1所取基下的矩阵；
3．求的核与值域的维数和基；
4．证明： .
答案及评分标准:
1.直接验证，知是线性子空间. 的维数是3，一组基是
.
2.直接验证,知是线性变换.
在题1所取基下的矩阵是 .
3.由于，所以，为的一组基；由于，所以，是的一组基.
4.由于，所以 .
共5页第3页
三、(20分)设非齐大学2015级硕士研究生
共5页第1页
2015~ 2016学年第1学期《矩阵论》课程考试A卷
考试日期：2015年12月28日课程编号：A080001命题教师：阅卷教师：
学院专业学号成绩
一、(20分)设阶矩阵．
1．求的特征多项式和初等因子；
2．求的最小多项式和Jordan标准形；
3．问：与矩阵是否相似？并说明理由．

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Matrix-Theory双语矩阵论期末考试2015

NUAALet 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 representingmatrix 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)110020012I 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 is100021002J ⎛⎫ ⎪= ⎪ ⎪⎝⎭--------------------------------------------------------------------------------------------------------------------------(3) For eigenvalue 1, 010010011I A ⎛⎫⎪-=- ⎪ ⎪-⎝⎭ ， An eigenvector is 1(1,0,0)T p = For eigenvalue 2, 1102000010I 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 that3(1,1,0)T p =-101001010P ⎛⎫ ⎪=- ⎪ ⎪⎝⎭, 1110001010P -⎛⎫⎪= ⎪ ⎪-⎝⎭ 1At J e Pe P -=22210100110001000101000010tt t t e e te e ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭22220000t t t t t t e e e e tee ⎛⎫-⎪= ⎪ ⎪-⎝⎭ --------------------------------------------------------------------------------------------------------------------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 256x x --.The minimal polynomial of A divides any annihilating polynomial of A. The possible minimal polynomials are6x -, 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)2055000A 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 lessthan 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 31,()()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 fromn R to )(A R , then A is symmetric and the eigenvalues ofA 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 Ais Hermitian.Let n n B A ⨯∈C ， be Hermitian matrices, and A bepositive definite. Show thatAB 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 are1’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 BASince 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.。

南京航空航天大学2009_矩阵论考试考题及答案

二、（1பைடு நூலகம் 分）设矩阵
考试试卷 A
(考试时间：2009 年 11 月？日晚 7：00-9：00 考试方式：闭卷 A)
成绩：
一、（15 分）在 R 4 中有两组基，
1 0 2 A 0 1 1 ， 0 1 0
计算： 2 A8 3 A5 A4 A2 4E 。

1 1 4 4
(3) ，因容易验证 AA b b ，故方程组 Ax b 相容，最小范数解为
1 1 3 3 y1 0 1 1 x A b E2 A A y 3 1 3 y 2 34 0 3 3 3 1 y3 4
个基有相同坐标的非零向量为 k x1 x2 x3 x4 ， k 非零常数。
（5 分）
共 4 页，第 1 页
学院年级班学号姓名 ------------------------------线--------------------------------- ---------- -----------------------封--------------------------------------- --------------------------------------密--------------------------------
(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

历年矩阵论试题

南京航空航天大学矩阵论历年试题整理者：王正华2007.1.28一设2615115126A −=− −（1）求A 的特征多项式和A 的全部特征值；（2）求A 的行列式、不变因子，初等因子；（3）求A 的最小多项式；（4）写出A 的Jordan 标准形二（1）设210121A= −，1）求12,,,F A A A A ∞；（2）设A 为n阶矩阵，证明21,max ij i j na A∞≤≤≤≤三（1）111111112A − =− −，作出A 的满秩分解并求出A +；（2）利用该矩阵判断如下方程组1231231231121x x x x x x x x x −+=−++=− −+= ，是否相容？若相容求通解；若不相容，求极小最小二乘解四设V 是数域P 上全体3阶实对称矩阵作成的线性结构（1）求V 的维数，并写出一组基（2）在V 中定义变换100100()011010001011T X X=，证明T 是线性变换，并求T 在（1）中所取基下的矩阵五（1）设2010252,022024220t A t B −==，其中t 是实数，t 满足什么条件时A B >成立？（2）设,A B 均为Hermite 半正定矩阵，证明：○1若A >0, 则AB 相似于半正定对角阵； ○2若A >0, 则()00tr AB B =⇒=； ○3若()0,tr AB = 则0AB =一（20分）已知 A =1001225i i −，其中i（1）求12,,,F A A A A ∞（2）证明：A ≥0 （3）设,,nH c B αβαβ∈=，证明22FBαβ=二（20分）设A ＝110101101211 ，b ＝314(1)作出A 的满秩分解 (2) 计算A +(3)利用广义逆矩阵方法判断线性方程组A x ＝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三（20分）设A ＝110430211− − −（1）求A 的特征多项式和A 的全部特征值（2）求A 的不变因子、初等因子和最小多项式（3）写出A 的Jordan 标准形（4）设A 为n 阶矩阵，证明：A 非奇异的充要条件是存在常数项不为零的多项式()f x ，使()f x ＝0 四（20分）（1）设A 、B 均为Hermite 矩阵（n 阶），且A B ＝B A ，证明：（a ）如果A >0,且A B >0 , 则B >0（b ）如果A >0, B >0,且33A B >，则A B >（2）若A 是2阶实正规矩阵，且i αβ±是A 的一对共轭实特征值，证明：存在正交矩阵Q ，使得Q AQ αββα+ =−五（20分）设实数域上线性空间32R ×的子集W ＝22{,()0}A R tr A ×∈=（1）W 是22R×的子空间（2）给出W 的变换T （A ）＝A A ++，A W ∀∈，证明：T 是W 上的线性变换（3）求Ker （T ）及其维数（4）求W 的一组基和维数，并写出线性变换T 在所取基下的矩阵一（20分）设[]n R X 表示实数域R 上次数小于n 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[]n R X 的维数并写出[]n R X 的一组基；（2）在[]n R X 中定义线性变换D ：(())'(),()[]n D f x f x f x R x =∈，求D 在（1）中所取基下的矩阵表示，并求R （D ）和Ker （D ）（3）证明D 在任何一组基下的矩阵都不可能是对角矩阵（4）在[]n R X 中定义内积11(,)()(),f g f x g x dx −=(),()[]n f x g x R X ∈,求出3[]R X 的一组标准正交基二（20分）设A ＝3615125125− −−三（16分）（1）设A ＝11121013 − −，求12,,,F A A A A ∞ （2）设A 为n 阶矩阵，证明：()1A ρ<的充要条件是存在某种相容矩阵范数.，使得1A <四（14分）设111021111021A − −−=（1）作出A 的一个QR 分解，即求满足T Q Q I =的4×3矩阵和3阶上三角矩阵R ，使得A QR ＝（2）计算A +五（16分）（1）设311120102A − = − ，121211111B =，问A ≥B 是否成立（2）设A 为n 阶Hermite 正定矩阵，B 为n 阶Hermite 半正定矩阵，并且AB BA =，证明（i ）AB 为Hermite 半正定矩阵（ii ）如果A ≥B ，则2A ≥2B六（14分）（1）设222i i A i i i i =− −−，其中i =，证明A 是正定矩阵；（2）若n n A C ×∈，且21A<，则A >B ≥0(3)设,n n A B C ×∈是Hermite 矩阵，证明如果A >B ≥0，则A B −≤A ，且等号成立一（20分）（1）设A 为n 阶非奇异复矩阵，试述矩阵A 的QR 分解定理；（2）设110101111010A= −（i ）作出A 的一个满秩分解（ii ）计算广义逆矩阵A +二（18分）（1）设210123032A=− −，求12,,,F A A A A ∞；（2）设A 为n 阶可逆矩阵，.是满足1I =的矩阵范数，证明11AA −−≥，21A ≤三（22分）设3117937100480024A −−−−−= − −（1）求A 的特征多项式和A 的全部特征值；（2）求A 的不变因子、初等因子和最小多项式；（3）写出A 的Jordan 标准形；（4）求lim k k A →∞；（5）计算Ae 四（20分）（1）设622250207A −=−，证明A 为正定矩阵；（2）设A ，B 均为Hermite 矩阵，证明：（i ）如果A >0, 则A B 相似于对角矩阵；（ii ）如果A >0, B >0, 则A B 的特征值均为正数；（iii ）如果A >0, B >0,且A B ＝B A ，则A B 是Hermite 正定矩阵五（20分）设V 是实数域R 上全部3阶实反对称矩阵作成的线性空间（按矩阵的加法和数量乘法）（1）求V 的维数，并写出V 的一组基；（2）证明：若A 是3阶实对称矩阵，且X V ∈，则必有AX XA V +∈；（3）作映射T 如下：011011()101101,110110T X X X X V −−=+∈ −−证明：T 是V 上的线性变换；（4）求T 在（1）中所取基下的矩阵表示。

2015年硕士研究生矩阵论期末考试试题卷

硕士研究生考试试卷 2014— 2015学年第一学期考试科目: 矩阵论考试时间：120分钟出卷教师: 出卷时间: 阅卷负责人签名：姓名、学号必须写在指定位置 1 填空题（18分）（1）123654789A ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦，111X ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦，则1A = m A ∞= 1AX = AX ∞= （2）1821A -⎡⎤=⎢⎥-⎣⎦，幂级数06k k k k z ∞=∑的收敛半径是，矩阵幂级数06k k k k A ∞=∑是（收敛、发散），理由（3）1210A -⎡⎤=⎢⎥-⎣⎦，0110B ⎡⎤=⎢⎥⎣⎦，则A B ⊗的全部特征值是（4）123b b b α⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦是给定的向量，123X ξξξ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦是向量变量，且112233()f X b b b ξξξ=++，则dfdX =专业：姓名：学号：2 设110430102A-⎡⎤⎢⎥=-⎢⎥⎢⎥⎣⎦，（16分）（1）求A的Jordan标准形J （2）求变换矩阵P，使1P AP J-=3 设221022212A⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦，求A的QR分解（12分）4 设101102221453A-⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦（16分）（1）求A的满秩分解（2）求A的More-Penrose逆A+5设20312102810A⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦（14分）（1）写出A的行盖尔圆，并在复平面上画图表示（2）隔离A的特征值（3）利用实矩阵特征值性质，改进得到的结果6 设()m nij m n A a C ⨯⨯=∈，规定,ijG i j A a = （12分）（1）证明G ⋅是m n C ⨯上的一种范数（2）证明矩阵范数与向量2-范数相容7 求解微分方程初值问题（12分）11221221431dx x x dt dx x x dt ⎧=++⎪⎪⎨⎪=+-⎪⎩，12(0)1,(0)2x 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,