CHAPTER II DIMENSION In the present chapter we investigate ...

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Appendix A*: quotient spaces Appendices for Chapter II Let V be a vector space, not necessarily finite dimensional. For subsets S and T in V , and a scalar a, we define S + T (aS respectively) to be the set of vectors which can be expressed in the form u + v (au respectively) with u in S and v in T , that is, S + T = {u + v | u ∈ S and v ∈ T }, aS = {au | u ∈ S}. When S consists of a single point u, we write u + T for S + T . Let M be a linear subspace of V . We call a subset of V of the form u + M (where u is some vector in V ) a coset (of M). Notice that u in generally is not uniquely determined. In fact, it is easy to check that two cosets u + M and v + M are equal if and only if u − v is in M. In case V is a 2–dimensional space represented by a plane with a point representing the zero vector, called the origen, then a 1–dimensional subspace M is a line through the origin of this plane and the cosets of M are lines parallel the to line representing M. It is easy to show that if A and B are cosets of M, then so are A + B and aA, where a is any scalar. In fact, when A = u + M and B = v + M, we have A + B = u + v + M and aA = au + M. Furthermore, with addition and scalar multiplication of cosets defined in this manner, the collection of all cosets form a vector space. This vector space of all cosets of M in V is called the quotient space of V over M, denoted by V/M: V/M = {u + M| u ∈ V }. There is a natural linear map Q from V to the quotient space V/M given by Qu = u+M. The map Q : V → V/M is called the quotient map. It is easy to check that the range of Q is V/M and the kernel of Q is M. Hence it follows from Theorem 2.3.1 of the present chapter that, when V is finite dimensional, dim V/M = dim V − dim M. Now suppose that N is another subspace of V (assumed to be finite dimensional). Let R be the restriction of Q to N. Thus R: N → V/M is the linear map given by Ru = u+M for all u in N. Then we can check that the kernel of R is M ∩ N and the range of R is (N + M)/M. So, by Theorem 2.3.1 again, we get dim N − dim(M ∩ N) = dim (M + N)/M = dim(M + N) − dim N. 28
Hence dim(M + N) + dim M ∩ N = dim M + dim N. Thus we have given a conceptual proof of Theorem 2.3.2. Let T : V → W be a linear map between vector spaces. Let M = ker T , the kernel of T and let Q : V → V/M be the quotient map. Then there is a unique isomorphism ˜T : V/M → T (V ) such that T = j ˜ T Q, where j : T (V ) → W is the inclusion map for the subspace T (V ) of W . This fact roughly says that a linear map is essentially a quotient map, followed by an injection. (It is analogous to the first isomorphism theorem in group theory or ring theory.) Appendix B*: Duality Recall that the dual V ′ of a vector space V over F is the space of all linear functionals on V : V ′ = L (V, F). The dual of V ′ , naturally denoted by V ′′ , is called the bidual of V . There is a natural map J : V → V ′′ defined by (Ju)(φ) = φ(u) for all u ∈ V and φ ∈ V ′′ . Now we prove: when V is finite dimensional, J is an isomorphism between V and V ′′ . As we know, dim V ′ = dim V . Applying this identity to V ′ , we have dim V ′′ = dim V ′ . Consequently dim V ′′ = dim V . So it is enough to show that ker J = {0}. To this end, we show that, given a nonzero vector u in V , Ju = 0, there is, there exists φ ∈ V ′ such that (Ju)(φ) = φ(u) = 0. Since u = 0, there is a basis of vectors b1, . . . , bn in V such that b1 = u. Now define a linear functional φ by putting φ(x1b1 + · · · + xnbn) = x1 for any vector x = x1b1 + · · · + xnbn in V . Then φ(u) = 1 = 0. Now we can conclude that J : V → V ′′ is an isomorphism. Using this natural isomorphism J, we can treat V ′′ and V as the same and hence V as the dual space of V ′ . Let M be a subspace of V . The annihilator M o of M is defined as follows: M o = {φ ∈ V ′ | φ(x) = 0 for all x ∈ M}. Let φ be an element in M o . Then φ(v + M) = φ(v) for all v ∈ V and hence φ can be treated as a linear functional on V/M. Conversely, if ψ is a linear functional on V/M, then ψ ◦ Q is a linear functional on V belonging to M o . This shows that the dual of V/M is naturally isomorphic to M o : (V/M) ′ ∼ = M o . By a more sophisticated argument, we can show M ′ ∼ = V ′ /M o . Appendix C*: Projective Space Let V be a vector space over F of dimension n + 1. For any positive integer k with k ≤ n, denote by Gk(V ) the set of all k–dimensional subspaces of V ; (G here stands 29
Page 1 and 2: CHAPTER II DIMENSION In the present
Page 3 and 4: ♠ Aside. What is supposed to be p
Page 5 and 6: The set L of all linear combination
Page 7 and 8: combination of them, that is, v can
Page 9 and 10: Applying elementary row operations
Page 11 and 12: EXERCISE SET II. 1. Review Question
Page 13 and 14: Exercises 1. Let T be a linear map
Page 15 and 16: has n + 1 elements. Now we determin
Page 17 and 18: first, the other letting j run firs
Page 19 and 20: Here, M + N stands for the subspace
Page 21 and 22: 2.5. Consider the following general
Page 23 and 24: q(x) be in ker T . Then aq(x + 1) =
Page 25 and 26: (h) If S is a set of 123 vectors sp
Page 27: 6. Suppose that T is a linear opera
Page 31: (x, y) with x = x1/x0, y = x2/x0 on

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