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

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They are among the most important operators used in differential equations and difference equations. We can check that each of the following (finite dimensional) subspaces is invariant for both D and Ta: V1 = span {1, x, x 2 }, V2 = span {e kx , xe kx , x 2 e ax }, V3 = span {e kx cos bx, e kx sin bx}, V4 = span {cos bx, sin bx, x cos bx, x sin bx, x 2 cos bx, x 2 sin bx}. Here k and b are any real constants. Here we only check that V3 is invariant for both D and Ta and we leave the rest of verification to the reader as an exericse; (see Drill 7). Indeed, by the product rule in differentiation, we have D(e kx cos bx) = D(e kx ) cos bx + e kx D(cos bx) = ke kx cos bx − be kx sin bx D(e kx sin bx) = D(e kx ) sin bx + e kx D(sin bx) = ke kx sin bx + be kx cos bx which are linear combinations of e kx cos bx and e kx sin bx and hence are in V3. This proves D(V3) ⊆ V3. Next, T (e kx cos bx) = e k(x+ a) cos(b(x + a)) = e kx+ ka cos(ab + bx) = e ak e kx (cos ab cos bx − sin ab sin kx) = e ak cos ab(e kx cos bx) − e ak sin ab(e kx sin kx) which is a linear combination of e kx cos bx and e kx sin bx and hence are in V3. Similarly we can check that T (e kx sin bx) is in V3. Hence Ta(V3) ⊆ V3. (We recommend the reader to review Examples 3.5.1 to 3.5.4 in the last chapter. In each of these examples the choice of the invariant subspace is suggested by our present example.) If v is a linear combination of a set of linear independnt vectors v1, v2, . . . , vr, then this linear combination must be unique. Indeed, suppose we have both v = α1v1 + α2v2 + · · · + αnvr and v = β1v1 + β2v2 + · · · + βnvr. Subtract these identities: 0 = v − v = (α1 − β1)v1 + (α2 − β2)v2 + · · · + (αr − βr)vr. Since, by assumption, v1, v2, . . . , vr are linearly independent, we must have α1 − β1 = 0, α2 − β2 = 0, etc. and hence α1 = β1, α2 = β2, etc. Using the argument presented here, we can prove: a set of vectors form a basis of V if and only if: 1. this set is linearly independent, and 2. it spans V . Recall from Definition 2.6.1 that vectors v1, v2, . . . , vn in V form a basis of V if and only if each vector v can be uniquely written as a linear 6
combination of them, that is, v can be written as a1v1 + a2v2 + · · · + anvn, where the scalars a1, a2, . . . , an are uniquely determined by v. 1.4. Let us take an arbitrary finite set of vectors v1, v2, . . . , vr in a linear space V . For each positive integer k with k ≤ r, let Vk be the span of v1, v2, . . . , vk, that is, Vk = span {v1, v2, . . . , vk}. For convenience, let V0 = {0}, the trivial subspace. So V0 = {0}, V1 = span {v1}, V2 = span {v1, v2}, . . . Vr = span {v1, v2, . . . , vr}. Clearly the family of subspaces Vk is increasing: {0} = V0 ⊆ V1 ⊆ V2 ⊆ · · · ⊆ Vr−1 ⊆ Vr. (That Vk ⊆ Vk+ 1 means that Vk is contained in Vk+ 1, that is, every vector in Vk is also a vector in Vk+ 1.) For each positive integer k with 1 ≤ k ≤ r, there are two possibilites Case 1: Vk−1 = Vk. This happens exactly when vk is in Vk−1. Case 2: Vk−1 = Vk. This happens exactly when vk is not in Vk−1. When the second case occurs, we call vk a leading vector. Thus vk is a leading vector if and only if vk is not in the span of the preceding vectors, namely v1, v2, . . . , vk−1. Now we claim Theorem 1.4.1. W ith the notation and terminology as above, the we have (a) leading vectors are linearly independent, and (b) any non-leading vector is a unique linear combination of preceding leading vectors. We postpone the proof of this theorem to the end of the present section. Here let us draw some important conclusions from it. First, let us take a spanning set of vectors v1, v2, . . . , vr in V . From part (a) of the above theorem, we see that the leading vectors in this set are linearly independent. From part (b) we see that all vectors vk (1 ≤ k ≤ r) in this spanning set are linear combinations of leading vectors. This shows that leading vectors also span V . Now the leading vectors are both linearly independent and spanning. Hence the leading vectors form a basis of V . We have proved Corollary 1.4.2. Any finite set of vectors spanning a linear space V contains a subset of vectors that form a basis of V . 7
Page 1 and 2: CHAPTER II DIMENSION In the present
Page 3 and 4: ♠ Aside. What is supposed to be p
Page 5: The set L of all linear combination
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 and 28: 6. Suppose that T is a linear opera
Page 29 and 30: Hence dim(M + N) + dim M ∩ N = di
Page 31: (x, y) with x = x1/x0, y = x2/x0 on

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

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