南京航空航天大学Matrix-Theory双语矩阵论期末考试

1、南京航空航天大学 Matrix-Theory 双语矩阵论期末 考试2015NUAAMatrixTheory,Final,Test Date : 2015 年 12 月 28 日矩阵论班号：学号姓名必做题（70分）选做题 （30 分）总分1234512页)第2页（共Part I （必做题，共5题，70分）第1题 （15 分）得分Let P I,” denote the set of all real polynomials of degree less than 3 with domain (定义域)i,u. Theaddition and scalar multiplication are d

2、efined in the usual way. Define an inner product on p by1 p,q 1P(t)q(t)dt.(1) Construct an orthonormal basis for 耳 from the basis 1,x,x2 by using the Gram-Schmidt orthogonalization process.(2) Let f(x) x2 1 PE. Find the projection of f onto the subspace spanned by1,x.第4页（共12页）Solution:11dx 2,pi11x 2

3、.2U2xP1xP1I3 _xP2x2, 3x . 3x 1- 223,U32 x2 xP2P210 (3x2 1)43x第2题 （15 分）得分Let be the linear transformation onP3 (thevector space of real polynomials of degree less than 3) defined by(p(x) xp(x) p”(x).(1) Find the matrix a representing with respect to the ordered basis 1,x,x1 0 1t 0 1 0 (The column ve

4、ctors of T are0 0 1the eigenvectors of A)The corresponding eigenvectors in P3 for P3.(2) Find a basis for P3 such that with respect tothis basis, the matrix B representing is diagonal.(3) Find the kernel (核)and range (值域)this transformation.Solution:(1)(10(x) x (x2) 2 2x2第5页(共12页)are000T 1AT 0 10 (T

5、 diagonalizes A) 0 0 2i,x,x2 i i,x,x2T . With respect to this new basis 1,x,x2 1, the representing matrix of is diagonal.(3) The kernel is the subspace consisting of all constant polynomials.The range is the subspace spanned by the vectors x,x2 1第8页（共12页）第3题 （20 分）得分11 0Let a 0 2 0 .01 2(1) Find all

6、 determinant divisors and elementary divisors of a.(2) Find a Jordan canonical form of a.(3) Compute e computations.) Solution:(1)111A 0201(Give the details of00 ,（特征多项式p（）（2your21)(2).Eigenvalues are 1, 2, 2.)Determinant divisor of order Di( ) 1, d2( ) 1, D3( ) p( ) (1)(2)2Elementary divisors are (

7、1) and ( 2)2 .(2) The Jordan canonical form is010(3) For eigenvalue 1, i a 0 1 0011Aneigenvector is o。”For eigenvalue 2,1102I A 0 0 0010Aneigenvector is P2 (o,o,1)TSolve (a 2i)p3 访(A 2I)p311 00000 P3001 01weobtain thatpp3(1,1,o)T1 011100 01p 1001,0 1001 01 0 1et00110eAtPeJ P 10 010 e2t te2t 0 01t t

8、2t -e e e 02t0 e 02t 2t0 te e0 1000e2t01 0整页(共12页)-第4题|得分 (10加|Suppose that a r33 and a 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 polyno

9、mial of A have the same linear factors. 5A 6i o .(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 x2 5x 6 .The minimal polynomial of A divides any

10、annihilating polynomial of A.The possible minimal polynomials arex 6, x 1, and x2 5x 6.第9页(共12页)Case x i, the characteristic polynomial is (x i)3Case x2 5x 6, the characteristic polynomial is (x 1)2(x 6) or (x 6)2(x 1)第5题（10 分）得分Let a ； 2 0 . Find the Moore-Penrose inverse aof a.Solution:1 A011 2 0

11、PG 01_T _ 1 _T_T T 1 1 _P (P P) P (1,0), G G (GG )2(1,0)1 125 05 01 125 0也可以用SVD求.Part II（选做题,每第12页（共12页）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 P4 be the vector space consisting of all real polynomials of degree less than 4 with usual addition and

12、 scalar multiplication. Let 入 be three distinct real numbers. For each pair of polynomials f and g in P4, define3 f ,gf(x)g(x).i 1 Determine whether f,g defines an inner produc on p, or not. Explain.第 7 题 Let a Rnn. Show that if (x) Axis the orthogonal projection from Rn to R(A), then A is symmetric

13、 and the eigenvalues of A are all 1 s and 0s.第 8 题| Let a cnn. Show that xH ax is real-valued for allx cnif apd only if A is Hermitian.第 9 题| Let a, b cn n be Hermitian matrices, and AbeLpositiveLdefinite.ShowthatABis第11页（共12页）similar to BA , and is similar to a real diagonal matrix.m 若正面不够书写，请写在反面.

14、第6题解答|Let f（x）（x xi）（x X2）（x X3）. Then f,f o. But f o.This_does_notdefine an inner product.第7题解答|For any x, A x r（a） n（at）, at（ax x） 0. Hence, ATA AT .Thus. a AT.From above, we have a2 a. This will imply that2 is an annihilating polynomial of A. The eigenvalue of A must be the roots of 2 0. Thus, the eigenvalues of A are 1s and 0s.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, AB a（ba）a1. Hence, A is第12页(共12页)similar to BASince A is positive definite, there is a. nonsingular hermitian matrix P such thati_iHH1A PPH .AB PP B P(P BP)PSince phbp is Hermitian,