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

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In order to prove that a given set of vectors v1, v2, . . . , vr is linearly dependent, we may write down a1v1 + a2v2 + · · · · · · + arvr = 0 and treat it as an equation with a1, a2, . . . , ar as unknowns and try to find a nontrivial solution of it. To give a quick example, let us check if the vectors v1 = (i, 1), v2 = (1, −i) in C 2 are linearly dependent. We begin by writing down a1v1 + a2v2 = 0, or a1(i, 1) + a2(1, −i) = (0, 0). which gives ia1 + a2 = 0 and a1 − ia2 = 0. This system of equations in a1, a2 has a nontrivial solution, such as a1 = i and a2 = 1. So the given vectors are linearly dependent. The opposite of linear dependence is linear independence. A set of vectors is linearly independent if this set is not linearly dependent. Recall that v1, v2, . . . , vr are linearly dependent if there is a nontrivial linear relation among them. So “vectors v1, v2, . . . , vr are linearly independent” means that there is no nontrivial linear relation among them, in other words, the only possible linear relation among them is the trivial one. Thus, if we have a linear relation a1v1 + a2v2 + · · · + arvr = 0 among linearly independent vectors v1, v2, . . . , vr, then all ak (0 ≤ k ≤ r) must be zeroes. Thus we have arrived at Definition. We say that vectors v1, v2, . . . , vr are linearly independent if any linear relation among them, say a1v1 + a2v2 + · · · · · · + arvr = 0, is necessarily trival, that is, a1 = a2 = · · · = ar = 0. To prove that given vectors v1, v2, . . . , vr are linearly independent, we almost always start with something like “Suppose we have a1v1 + a2v2 + · · · + arvr = 0.” After that, we try to figure out why a1, a2, . . . , ar all must be equal to zero. In the next subsection, we give a few examples to see how it works. 1.2. First we give a quick example to show the steps. We are asked to prove that vectors v1 = (1, 0), v2 = (1, 2) in R 2 are linearly independent. We begin with the sentence “Suppose that a1v1 +a2v2 = 0”. Then we proceed by rewriting a1v1 +a2v2 = 0 as a1(1, 0) + a2(1, 2) = (0, 0), or (a1 + a2, 2a2) = (0, 0), which gives us a1 + a2 = 0 and 2a2 = 0, from which we deduce a1 = a2 = 0. Here we emphasize that the first sentence in our proof must be correct, no matter how hard or how easy the given problem is. Example 1.2.1. Prove that vectors u and v in a vector space V are linearly independent if u + v and u − v are linearly independent. 2
♠ Aside. What is supposed to be proved? Well, the linear independence of u and v. So we should start with something like “Assume au + bv = 0”. Then we try to derive a = 0 and b = 0. A common error is to start with “Assume a(u + v) + b(u − v) = 0” which results in receiving no credit for the rest of the work. ♠ Solution. Assume au + bv = 0. Let p = u + v and q = u − v. Then, by assumption, p and q are linearly independent. Notice that 2u = p + q and 2v = p − q. Hence 0 = 2(au + bv) = a(2u) + b(2v) = a(p+q) + b(p−q) = (a+b)p + (a−b)q. Since p, q are linearly independent, a + b = 0 and a − b = 0. So a = b = 0. Done. Example 1.2.2. Let T be a linear operator on V and let v be in V . Assume that T 3 v = 0 but T 2 v = 0. Prove that v, T v, T 2 v are linearly independent. Solution. Assume a1v + a2T v + a3T 2 v = 0. (1.2.1) Applying T to this identity, we have T (a1v + a2T v + a3T 2 v) = T 0 = 0. Rewrite the last identity as a1T v + a2T 2 v + a3T 3 v = 0. Since T 3 v = 0, we have a1T v + a2T 2 v = 0. (1.2.2) Applying T to the last identity again, we have a1T 2 v + a2T 3 v = 0, or a1T 2 v = 0. Since T 2 v = 0, we must have a1 = 0. Thus (1.2.2) becomes a2T 2 v = 0. Since T 2 v = 0, we must have a2 = 0. Now (1.2.1) becomes a3T 2 v = 0. Since T 2 v = 0, we must have a3 = 0. We have shown that a1 = a2 = a3 = 0. Hence v, T v, T 2 v are linearly dependent. Remark. More generally, if T m v = 0 and T m+ 1 v = 0, then the vectors vm = v, vm−1 = T v, vm−2 = T 2 v, . . . , v1 = T m−1 v, v0 = T m v are linearly independent. This remark will be refered to in the future and the labeling of vectors here is competible to the discussion there. Example 1.2.3. Let T be a linear operator on a vector space V and let u and v be nonzero vectors in V such that T u = 2u and T v = 3v. Prove that u, v are linearly independent. Solution. Assume au + bv = 0. (1.2.3) 3
Page 1: CHAPTER II DIMENSION In the present
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 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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