2022年最新南京航空航天大学Matrix-Theory双语矩阵论期末考试 .pdf

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1、精品文档精品文档Part I (必做题，共5 题， 70 分) 第 1 题（15 分） 得分Let 1,1Pdenote the set of all real polynomials of degree less than 3 with domain （定义域） 1,1. The addition and scalar multiplication are defined in the usual way. Define an inner product on 1,1Pby 11,( ) ( )p qp t q t dt . (1) Construct an orthonormal basi

2、s for 1,1Pfrom the basis 21,x xby using the Gram-Schmidt orthogonalization process. (2) Let2( )1f xx1,1P. Find the projection of fonto the subspace spanned by1, x . Solution: (1)1111 , 12dx, 112u, 1111111,02222pxxdx, 12132xpuxxp, 22211331,22322pxxxx, 22232210(31)4xpuxxp- (2) 2211221,1,projxuuxuu2211

3、331,1,2222xxxx22120332- 第 2 题（15 分）得分Let be the linear transformation on 3P(the vector space of real polynomials of degree less than 3) defined by ( )( )( )p xxp xpx . (1) Find the matrix Arepresenting with respect to the ordered basis 21,x x for 3P . (2) Find a basis for 3Psuch that with respect to

4、 this basis, the matrix B representing is diagonal. (3) Find the kernel （核）and range （值域） of this transformation. Solution: 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 5 页 - - - - - - - - - 精品文档精品文档(1) 221022xxxx（）（ ）（）002010002A- (2) 101010001T(The column v

5、ectors of T are the eigenvectors of A) The corresponding eigenvectors in 3Pare 1000010002TAT(T diagonalizes A) 221, ,11, ,x xx xT. With respect to this new basis 21, ,1x x, the representing matrix of is diagonal. - (3) The kernel is the subspace consisting of all constant polynomials. The range is t

6、he subspace spanned by the vectors 2,1x x- 第 3 题（20 分）得分Let 110020012A. (1) Find all determinant divisors and elementary divisors ofA. (2) Find a Jordan canonical form of A. (3) Compute Ate. (Give the details of your computations.) Solution: (1) 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - -

7、 - 名师精心整理 - - - - - - - 第 2 页，共 5 页 - - - - - - - - - 精品文档精品文档110020012IA,(特征多项式2( )(1)(2)p. Eigenvalues are 1, 2, 2.) Determinant divisor of order 1( )1D, 2( )1D, 23( )( )(1)(2)DpElementary divisors are 2(1) and (2). - (2) The Jordan canonical form is 100021002J- (3) For eigenvalue 1, 010010011IA，

8、An eigenvector is 1(1,0,0)TpFor eigenvalue 2, 1102000010IA, An eigenvector is 2(0,0,1)TpSolve 32(2 )AI pp , 331100(2 )00000101AI ppwe obtain that 3(1, 1,0)Tp101001010P, 1110001010P1AtJePe P22210100110001000101000010tttteetee22220000tttttteeeetee- 第 4题 （10 分） 得分Suppose that 33RAand OIAA652. (1) What

9、are the possible minimal polynomials ofA? Explain. (2) In each case of part (1), what are the possible characteristic polynomials ofA? Explain. Solution: (1) An annihilating polynomial of A is 256xx. The minimal polynomial of A divides any annihilating polynomial of A. The possible minimal polynomia

10、ls are 6x, 1x, and 256xx. 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 3 页，共 5 页 - - - - - - - - - 精品文档精品文档- (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

11、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)xCase 1x, the characteristic polynomial is 3(1)xCase 256xx, the characteristic polynomial is 2(1) (6)xxor 2(6) (1)xx- 第 5 题（10 分）得分Let 120000A. Find the M

12、oore-Penrose inverse Aof A. Solution: 12011200000AP G1()(1, 0 )TTPP PP, 111()250TTGGGG110112 (1,0)2055000AG 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. 第 6 题Let 4Pbe the vector space consisting of all real polynomials of degr

13、ee less than 4 with usual addition and scalar multiplication. Let 123,x xxbe three distinct real numbers. For each pair of polynomials fand gin4P , define 31,() ()iiif gf x g x. Determine whether ,f gdefines an inner product on 4Por not. Explain. 第 7 题LetnnAR. Show that if xxA)(is the orthogonal pro

14、jection from nRto)(AR, then A is symmetric and the eigenvalues of A are all 1 s and 0 s. 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 5 页 - - - - - - - - - 精品文档精品文档第 8 题LetnnAC. Show thatxxAHis real-valued for allnCxif and only if A is Hermitian. 第 9 题Let nnB

15、AC，be Hermitian matrices, and A be positive definite. Show that AB is similar to BA , and is similar to a real diagonal matrix. 若正面不够书写，请写在反面. 第 6 题解答Let 123( )()()()f xxxxxxx. Then ,0ff. But 0f. This does not define an inner product. 第 7 题解答For any x, ( )()xx TAR AN A, ()xx0TAA. Hence, TTA AA. Thus

16、. TAA. From above, we have 2AA. This will imply that 2is an annihilating polynomial of A. The eigenvalue of A must be the roots of 02. Thus, the eigenvalues of A are 1 s and 0 s. 第 8 题解答See Thm 7.1.1, page 182. 也可以用其它方法. 第 9 题解答Since A is nonsingular, 1()ABA BA A. Hence, A is similar to BASince A is

17、 positive definite, there is a nonsingular hermitian matrix P such that HAP P. 1()HHABPP BP P BP PSince HP BPis Hermitian, it is similar to a real diagonal matrix. is similar to HABP BP, HP BPis similar to a real diagonal matrix. Thus AB is similar to a real diagonal matrix. 选做题得分名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 5 页，共 5 页 - - - - - - - - -