tailieunhanh - Linear Agebra Prolem Book Part 10

Tham khảo tài liệu 'linear agebra prolem book part 10', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 256 UNEAR ALGEBRA PROBLEM BOOK linear combination u 31U1 32 2 H---F 3nun applied to one of the Xj s with j m yields 0 because Xj is in M and at the same time yields dj because Ui xj 0 when i j it follows that the coefficients of the early Uj s are all 0. Consequence u is a linear combination of the latter Ui s or in other words u is in N or better still M c N. The conclusions of the preceding two paragraphs imply that M N and hence since N has a basis of n - m elements that dimM n m. 80 Solution 80. If the spaces V and V are identified as suggested by Problem 77 then by definition M00 consists of the set of all those vectors x in M such that u x 0 for all u in V . Since by the definition of V the equation u x 0 holds for all X in M and all u in M so that every X in M satisfies the condition just stated for belonging to M00 it follows that M c M00. If the dimension of V is n and the dimension of M is m then see Problem 78 the dimension of M is n - m and therefore by the same result the dimension of M00 is n n - m . In other words M is an m-dimensional subspace of an m-dimensional space M00 and that implies that M and M00 must be the same. 81 Solution 81. Suppose that A is a linear transformation on a finite-dimensional vector space V and A is its adjoint on V . If u is an arbitrary vector in ker A so that A u 0 then of course A ù x 0 for every X in V and consequently u Ax 0 for every X in V. The latter equation says exactly that u takes the value 0 at every vector in the range of A or simpler said that u belongs to ran A 0. The argument is reversible if u belongs to ran A 0 so that u Ax 0 for every X then A u x 0 for every X and therefore A u 0 or simpler said u belongs to ker A . Conclusion keryV ran A 0. It should not come as too much of a surprise that annihilators enter. The range and the kernel of A are subspaces of V and the range and the kernel of A are subspaces of V what possible relations can there be between SOLUTIONS Chapter 5 257 subspaces of V and .

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