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Chapter Two Vector Spaces Chapter Two Vector Spaces The first chapter began by introducing Gauss’ method and finished with a fair understanding, keyed on the Linear Combination Lemma, of how it finds the solution set of a linear system. Gauss’ method systematically takes linear com- binations of the rows. With that insight, we now move to a general study of linear combinations. We need a setting for this study. At times in the first chapter, we’ve com- bined vectors from R2, at other times vectors from R3, and at other times vectors from even higher-dimensional spaces. Thus, our first impulse might be to work in Rn, leaving n unspecified. This would have the advantage that any of the results would hold for R2 and for R3 and for many other spaces, simultaneously. But, if having the results apply to many spaces at once is advantageous then sticking only to Rn’s is overly restrictive. We’d like the results to also apply to combinations of row vectors, as in the final section of the first chapter. We’ve even seen some spaces that are not just a collection of all of the same-sized column vectors or row vectors. For instance, we’ve seen a solution set of a homogeneous system that is a plane, inside of R3. This solution set is a closed system in the sense that a linear combination of these solutions is also a solution. But it is not just a collection of all of the three-tall column vectors; only some of them are in this solution set. We want the results about linear combinations to apply anywhere that linear combinations are sensible. We shall call any such set a vector space. Our results, instead of being phrased as “Whenever we have a collection in which we can sensibly take linear combinations . ”, will be stated as “In any vector space . ”. Such a statement describes at once what happens in many spaces. The step up in abstraction from studying a single space at a time to studying a class of spaces can be hard to make. To understand its advantages, consider this analogy. Imagine that the government made laws one person at a time: “Leslie Jones can’t jay walk.” That would be a bad idea; statements have the virtue of economy when they apply to many cases at once. Or, suppose that they ruled, “Kim Ke must stop when passing the scene of an accident.” Contrast that with, “Any doctor must stop when passing the scene of an accident.” More general statements, in some ways, are clearer. 79 80 Chapter Two. Vector Spaces I Definition of Vector Space We shall study structures with two operations, an addition and a scalar multi- plication, that are subject to some simple conditions. We will reflect more on the conditions later, but on first reading notice how reasonable they are. For instance, surely any operation that can be called an addition (e.g., column vec- tor addition, row vector addition, or real number addition) will satisfy all the conditions in (1) below. I.1 Definition and Examples 1.1 Definition A vector space (over R) consists of a set V along with two operations ‘+’ and ‘¢’ such that (1) if ~v; ~w 2 V then their vector sum ~v + ~w is in V and ² ~v + ~w = ~w + ~v ² (~v + ~w) + ~u = ~v + (~w + ~u) (where ~u 2 V ) ² there is a zero vector ~0 2 V such that ~v + ~0 = ~v for all ~v 2 V ² each ~v 2 V has an additive inverse ~w 2 V such that ~w + ~v = ~0 (2) if r; s are scalars (members of R) and ~v; ~w 2 V then each scalar multiple r ¢ ~v is in V and ² (r + s) ¢ ~v = r ¢ ~v + s ¢ ~v ² r ¢ (~v + ~w) = r ¢ ~v + r ¢ ~w ² (rs) ¢ ~v = r ¢ (s ¢ ~v) ² 1 ¢ ~v = ~v. 1.2 Remark Because it involves two kinds of addition and two kinds of mul- tiplication, that definition may seem confused. For instance, in ‘(r + s) ¢ ~v = r ¢ ~v + s ¢ ~v ’, the first ‘+’ is the real number addition operator while the ‘+’ to the right of the equals sign represents vector addition in the structure V . These expressions aren’t ambiguous because, e.g., r and s are real numbers so ‘r + s’ can only mean real number addition. The best way to go through the examples below is to check all of the con- ditions in the definition. That check is written out in the first example. Use it as a model for the others. Especially important are the two: ‘~v + ~w is in V ’ and ‘r ¢ ~v is in V ’. These are the closure conditions. They specify that the addition and scalar multiplication operations are always sensible — they must be defined for every pair of vectors, and every scalar and vector, and the result of the operation must be a member of the set (see Example 1.4). Edited by Foxit PDF Editor Copyright (c) by Foxit Software Company, 2004 For Evaluation Only. Section I. Definition of Vector Space 81 1.3 Example The set R2 is a vector space if the operations ‘+’ and ‘¢’ have their usual meaning. µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ x y x + y x rx 1 + 1 = 1 1 r ¢ 1 = 1 x2 y2 x2 + y2 x2 rx2 We shall check all of the conditions in the definition. There are five conditions in item (1). First, for closure of addition, note that for any v1; v2; w1; w2 2 R the result of the sum µ ¶ µ ¶ µ ¶ v w v + w 1 + 1 = 1 1 v2 w2 v2 + w2 is a column array with two real entries, and so is in R2. Second, to show that addition of vectors commutes, take all entries to be real numbers and compute µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ v w v + w w + v w v 1 + 1 = 1 1 = 1 1 = 1 + 1 v2 w2 v2 + w2 w2 + v2 w2 v2 (the second equality follows from the fact that the components of the vectors are real numbers, and the addition of real numbers is commutative). The third condition, associativity of vector addition, is similar. µ ¶ µ ¶ µ ¶ µ ¶ v w u (v + w ) + u ( 1 + 1 ) + 1 = 1 1 1 v2 w2 u2 (v2 + w2) + u2 µ ¶ v + (w + u ) = 1 1 1 v2 + (w2 + u2) µ ¶ µ ¶ µ ¶ v w u = 1 + ( 1 + 1 ) v2 w2 u2 For the fourth we must produce a zero element — the vector of zeroes is it. µ ¶ µ ¶ µ ¶ v 0 v 1 + = 1 v2 0 v2 Fifth, to produce an additive inverse, note that for any v1; v2 2 R we have µ ¶ µ ¶ µ ¶ ¡v v 0 1 + 1 = ¡v2 v2 0 so the first vector is the desired additive inverse of the second. The checks for the five conditions in item (2) are just as routine. First, for closure under scalar multiplication, where r; v1; v2 2 R, µ ¶ µ ¶ v rv r ¢ 1 = 1 v2 rv2 is a column array with two real entries, and so is in R2. This checks the second condition. µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ v (r + s)v rv + sv v v (r + s) ¢ 1 = 1 = 1 1 = r ¢ 1 + s ¢ 1 v2 (r + s)v2 rv2 + sv2 v2 v2 Edited by Foxit PDF Editor Copyright (c) by Foxit Software Company, 2004 For Evaluation Only. 82 Chapter Two. Vector Spaces For the third condition, that scalar multiplication distributes from the left over vector addition, the check is also straightforward. µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ v w r(v + w ) rv + rw v w r ¢ ( 1 + 1 ) = 1 1 = 1 1 = r ¢ 1 + r ¢ 1 v2 w2 r(v2 + w2) rv2 + rw2 v2 w2 The fourth µ ¶ µ ¶ µ ¶ µ ¶ v (rs)v r(sv ) v (rs) ¢ 1 = 1 = 1 = r ¢ (s ¢ 1 ) v2 (rs)v2 r(sv2) v2 and fifth conditions are also easy. µ ¶ µ ¶ µ ¶ v 1v v 1 ¢ 1 = 1 = 1 v2 1v2 v2 In a similar way, each Rn is a vector space with the usual operations of vector addition and scalar multiplication. (In R1, we usually do not write the members as column vectors, i.e., we usually do not write ‘(¼)’. Instead we just write ‘¼’.) 1.4 Example This subset of R3 that is a plane through the origin 0 1 x ¯ P = f@yA ¯ x + y + z = 0g z is a vector space if ‘+’ and ‘¢’ are interpreted in this way. 0 1 0 1 0 1 0 1 0 1 x1 x2 x1 + x2 x rx @y1A + @y2A = @y1 + y2 A r ¢ @yA = @ryA z1 z2 z1 + z2 z rz The addition and scalar multiplication operations here are just the ones of R3, reused on its subset P . We say that P inherits these operations from R3. This example of an addition in P 0 1 0 1 0 1 1 ¡1 0 @ 1 A + @ 0 A = @ 1 A ¡2 1 ¡1 illustrates that P is closed under addition. We’ve added two vectors from P — that is, with the property that the sum of their three entries is zero — and the result is a vector also in P . Of course, this example of closure is not a proof of closure. To prove that P is closed under addition, take two elements of P 0 1 0 1 x1 x2 @y1 A @y2 A z1 z2 Edited by Foxit PDF Editor Copyright (c) by Foxit Software Company, 2004 For Evaluation Only. Section I. Definition of Vector Space 83 (membership in P means that x1 + y1 + z1 = 0 and x2 + y2 + z2 = 0), and observe that their sum 0 1 x1 + x2 @y1 + y2 A z1 + z2 is also in P since its entries add (x1 + x2) + (y1 + y2) + (z1 + z2) = (x1 + y1 + z1) + (x2 + y2 + z2) to 0.
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