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Đề thi Olympic sinh viên thế giới năm 1998

" Đề thi Olympic sinh viên thế giới năm 1998 " . Đây là một sân chơi lớn để sinh viên thế giới có dịp gặp gỡ, trao đổi, giao lưu và thể hiện khả năng học toán, làm toán của mình. Từ đó đến nay, các kỳ thi Olympic sinh viênthế giới đã liên tục được mở rộng quy mô rất lớn. Kỳ thi này là một sự kiện quan. | 5th INTERNATIONAL MATHEMATICS COMPETITION FOR UNIVERSITY STUDENTS July 29 - August 3 1998 Blagoevgrad Bulgaria First day PROBLEMS AND SOLUTIONS Problem 1. 20 points Let V be a 10-dimensional real vector space and Ui and U2 two linear subspaces such that Ui C U2 dimR U1 3 and dimR U2 6. Let E be the set of all linear maps T V V which have Ui and U2 as invariant subspaces i.e. T U1 C U1 and T U2 C U2 . Calculate the dimension of E as a real vector space. Solution First choose a basis 1 v2 3 of U1. It is possible to extend this basis with vectors v4 v5 and 6 to get a basis of U2. In the same way we can extend a basis of U2 with vectors v7 . 10 to get as basis of V. Let T 2 E be an endomorphism which has U1 and U2 as invariant subspaces. Then its matrix relative to the basis v1 . v10 is of the form 0 0 0 . 000 000 000000 000000 000000 000000 So dimRE 9 18 40 67. Problem 2. Prove that the following proposition holds for n 3 5 points and n 5 7 points and does not hold for n 4 8 points . For any permutation 1 of 1 2 . . n different from the identity there is a permutation fl2 such that any permutation fl can be obtained from fl1 and fl2 using only compositions for example fl fl1 O fl-1 o 2 o fl-1 . Solution Let Sn be the group of permutations of 1 2 . n . 1 When n 3 the proposition is obvious if x 12 we choose y 123 if x 123 we choose y 12 . 2 n 4. Let x 12 34 . Assume that there exists y 2 Sn such that S4 x y . Denote by K the invariant subgroup K id 12 34 13 24 14 23 . By the fact that x and y generate the whole group S4 it follows that the factor group S4 K contains only powers of y yK i.e. S4 K is cyclic. It is easy to see that this factor-group is not comutative something more this group is not isomorphic to S3 . 3 n 5 a If x 12 then for y we can take y 12345 . b If x 123 we set y 124 35 . Then y3xy3 125 and y4 124 . Therefore 123 124 125 2 x y - the subgroup generated by x and y. From the fact that 123 124 125 generate the alternating subgroup A 5 it follows that A