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

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§2. Dimension 2.1. The main purpose of the present subsection is to establish the theorem stated below. The key role played by this theorem in linear algebra is to justify the definition of the dimension of a vector space. Theorem 2.1.1. Any two bases of a vector space have the same number of vectors. Let V be a finite dimensional vector space. Take any basis in V ; (we can do this because V has a basis according to Corollary 1.4.3 in the previous section. This theorem tells us that the number of vectors in it does not depend on which basis we pick. So we can define the dimension of V to be the number of vectors in any basis of V , denoted by dim V . We provide two proofs of this theorem. The first proof given here is based on the material about leading vectors described in the last section. Let us take two bases of V , say E consisting of vectors e1, e2, . . . , em and F consisting of f1, f2, . . . , fn. Here of course m and n are number of vectors in the bases E and F respectively. We need to prove m = n. Without loss of generality, let us assume n ≤ m. We use F as a coordinate system to convert vectors in E into column vectors in F n . Let Ck = [ek]F (1 ≤ k ≤ m) and C = [C1 C2 · · · Cm]. Notice that C is an n × m matrix, with n ≤ m. Since the vectors e1, e2, . . . , em are linearly independent, all of them are leading vectors. This fact carries over to the column vectors of C: all of C1, C2, . . . , Cm are leading vectors. Hence the reduced row echelon form of C must be ⎡ 1 0 ⎤ 0 ⎢ 0 ⎢ ⎣ 1 . .. 0 ⎥ ⎦ 0 0 1 which is necessarily a square matrix. So C is a square matrix as well. Thus m = n. Example 2.1.2. Now we determine the dimensions of familiar spaces. First, the dimension of F n is n, since the standard basis given as follows has exactly n vectors: e1 = (1, 0, . . . , 0, 0), e2 = (0, 1, . . . , 0, 0), . . . , en = (0, 0, . . . , 0, 1). The linear space Mmn(F) of all m × n matrices over F can be identified with F mn (as long as the linear structure is concerned and hence its dimension is mn. Next, the space Pn of all polynomials of degree not exceeding n is of dimension n + 1, since its standard basis consisting of 1, x, x 2 , . . . , x n−1 , x n 14
has n + 1 elements. Now we determine the dimension of the space L (V, W ) of all linear mappings from V into W , where dim V = n and dim W = m. Take a basis E in V and F in W , Then each linear map T ∈ L (V, W ) corresponds to the m × n matrix [T ] E F and vice versa. Thus the dimension of L (V, W ) is the same as the dimension of the space of all m × n matrices, which is clear mn. Finally, th dual V ′ of a vector space is the special case of L (V, W ) with W = F and hence the dimension of V ′ is 1 × n = n. In summary, dim F n = n, dim Mmn = mn, dim Pn = n + 1, dim L (V, W ) = (dim V )(dim W ), dim V ′ = dim V. Example 2.1.3. Determine dimension of the subspace M of C 3 consisting of vectors of the form (x, x + iy, 2y), where x, y are arbitrary complex numbers. Solution. We can rewrite (x, x+iy, 2y) as x(1, 1, 0)+y(0, i, 2). Let b1 = (1, 1, 0), b2 = y(0, i, 2). Then b1, b2 span M. Also, we can easily check that b1, b2 are linearly independent. Hence b1, b2 form a basis of M. Therefore dim M = 2. Example 2.1.4. In Example 1.2.4 from the last section, we took n + 1 distinct complex numbers c1, c2, . . . , cn+ 1 and consider n+1 vectors vk = (c k 1 , ck 2 , . . . , ck n , ck n+ 1 ) (k = 0, 1, 2, . . . , n) in C n+ 1 . We showed that these vectors are linearly independent. Now the number of vectors here is the same as the dimension of the space C n+ 1 , namely n + 1. From the following corollary we see that these vectors form a basis of C n+ 1 . Corollary 2.1.1. If E is a linearly independent set of m vectors in a linear space V with dim V = n, then m ≤ n. If, furthermore, m = n, then E forms a basis of V . Indeed, by Corollary 1.4.4 from the last section, we know that E can be extended to a basis F of V . Since dim V = n, F is a set of n vectors. Now the corollary is clear. Corollary 2.1.2. If E is a spanning set of m vectors in a linear space V and with dim V = n, then m ≤ n. If, furthermore, m = n, then E forms a basis of V . To see this, we use Corollary 1.4.2 from the last section to infer that E contains a basis of V , say B. Since dim V = n, there are n vectors in B. The rest of argument is clear. 2.2. Now we give a second proof of Theorem 2.1.1 in a more direct way, not relying on the material about leading vectors in the previous section and the fact that every matrix can be row reduced to a reduced row echelon form. However, basic skill in handling the summation symbol is required and this proof can be skipped or postponed. §2 of 15
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: Exercises 1. Let T be a linear map
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

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