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

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(Don’t forget that the theorem we are proving now mainly concerns a linear transformation, namely T . Naturally we apply T to both sides of the identity and see what happens.) Apply T to (2.3.1) above and write down T (a1v1 + · · · + akvk + b1u1 + · · · + brur) = T 0. Because T is linear, we can rewrite this as a1T v1 + · · · + akT vk + b1T u1 + · · · + brT ur = 0. Recall what happens when we apply T to these v’s and u’s: T vj = 0, T uk = wk. So b1w1 + b2w2 + · · · + brwr = 0. (2.3.2) But w1, w2, . . . , wr form a basis of T (V ). In particular, they are linearly independent. So (2.3.2) entails b1 = b2 = · · · = br = 0. Now go back to (2.3.1). It becomes a1v1 + a2v2 + · · · + akvk = 0. Since v1, v2, . . . , vk are linearly independent, (because they form a basis of ker T ), we have a1 = a2 = · · · = ak = 0. Hence B ≡ {v1, . . . , vk, u1, . . . , ur} are linearly independent. Next we prove that B spans V . To this end, take an arbitrary vector v in V . Our goal is to write v as a linear combination of vectors from B. Applying T to v, we get T v, a vector in the range T (V ). In T (V ) we have already picked a basis, namely {w1, w2, . . . , wr}, which is waiting to be used. Thus T v as a linear combination of them, say T (v) = β1w1 + β2w2 + · · · + βrwr. Recall that w1 = T u1, w2 = T u2 etc. So This tells us T z = 0, where T v = β1T u1 + · · · + βrT ur = T (β1u1 + · · · + βrur). z = v − (β1u1 + · · · + βrur). (2.3.3) In other words, z is in ker T . Now, in ker T there is a basis of vectors waiting for us, namely v1, v2, . . . , vk. Hence z is a linear combination of these basis vectors, say z = α1v1 + α2v2 + · · · + αkvk. (2.3.4) Combining (2.3.3) and (2.3.4), we end up with v = α1v1 + · · · + αkvk + β1u1 + · · · + βrur, which is exactly what we want. Theorem 2.3.2. If M and N are subspaces of a (finite dimensional) vector space V , then dim(M + N) + dim(M ∩ N) = dim M + dim N. 18
Here, M + N stands for the subspace consisting of vectors which can be written in the form u + v, where u is in M and v is in N, and M ∩ N, called the intersection of M and N, stands for the subspace consisting of vectors in both M and N. Thus M + N = {u + v: u ∈ M and v ∈ N} M ∩ N = {w ∈ V : w ∈ M and w ∈ N}. The proof of this Theorem can be done by using similar argument as the previous theorem. It is left as an exercise for the reader. (See Exercise 13.) 2.4. Now we draw some consequences of Theorem 2.3.1. Given vector spaces V and W , assume there is an invertible linear mapping T from V to W (Recall that “T is invertible” means that there is a linear map S from W to V such that ST = IV and T S = IW .) The invertibility of T tells us that ker T = {0} and T (V ) = W . Thus the identity dim ker T + dim T (V ) = dim V becomes dim W = dim V and hence dim V = dim W . We have shown Corollary 2.4.1. If there is an invertible linear mapping from V to W , then V and W has the same dimension. ♠ Remark: There is a subject in mathematics called category theory (which is considered as “abstract general nonsense” by some people) that puts every branch of mathematics under its umbrella. According to that theory (perhaps “philosophy” is a more appropriate word here) every branch of mathematics deals with a family of “objects” and “mophisms” between objects. Invertible morphisms are called isomorphisms. If there is an isomorphisms between two objects, say X and Y , then we say that X and Y are isomorphic. Two isomorphic objects are indistinguishable in their behavior and often considered as the same. In linear algebra, the objects are finite dimensional vector spaces and morphisms are linear mappings between them. The above corollary tells us that isomorphic vector spaces have the same dimension. The converse of this is also true: if V and W are vector spaces over F having the same dimension, say dim V = dim W = n, then they are isomorphic. Indeed, according to the last paragraph of subsection 2.6 in Chapter I, both V and W are isomorphic to F n and hence they are isomorphic to each other. Thus, dimension is the only thing we need to tell if two vector spaces are “the same” or “different”. Comparing to other branches of mathematics, objects in linear algebra are considered to be “easy”. ♠ Next, let T be a linear operator on a finite dimensional space, (that is, T is a linear map from V into V itself). Assume that ker T is the zero space, that is, ker T = {0}. The identity dim ker T + dim T (V ) = dim V becomes dim T (V ) = dim V . Now T (V ) is a subspace of V , and it has the same dimension as V . So T (V ) must coincide with V . This shows that all vectors in V are in the range of T , in other words, T is surjective. 19
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: first, the other letting j run firs
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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