Linear Algebra

116 Chapter Two. Vector Spaces2.2 Remark One reason for sticking to finite-dimensional spaces is so that therepresentation of a vector with respect to a basis is a finitely-tall vector and wecan easily write it. Another reason is that the statement ‘any infinite-dimensionalvector space has a basis’ is equivalent to a statement called the Axiom of Choice[Blass 1984] and so covering this would move us far past this book’s scope. (Adiscussion of the Axiom of Choice is in the Frequently Asked Questions list forsci.math, and another accessible one is [Rucker].)To prove the main theorem we shall use a technical result, the ExchangeLemma. We first illustrate it with an example.2.3 Example Here is a basis for R 3 and a vector given as a linear combination ofmembers of that basis.⎛ ⎞ ⎛1⎜ ⎟ ⎜1⎞ ⎛⎟ ⎜0⎞ ⎛⎟ ⎜1⎞ ⎛⎟ ⎜1⎞ ⎛⎟ ⎜1⎞ ⎛⎟ ⎜0⎞⎟B = 〈 ⎝0⎠ , ⎝1⎠ , ⎝0⎠〉⎝2⎠ = (−1) · ⎝0⎠ + 2 ⎝1⎠ + 0 · ⎝0⎠0 0 2 00 0 2Two of the basis vectors have non-zero coefficients. Pick one, for instance thefirst. Replace it with the vector that we’ve expressed as the combination⎛ ⎞ ⎛1⎜ ⎟ ⎜1⎞ ⎛ ⎞0⎟ ⎜ ⎟ˆB = 〈 ⎝2⎠ , ⎝1⎠ , ⎝0⎠〉0 0 2and the result is another basis for R 3 .2.4 Lemma (Exchange Lemma) Assume that B = 〈⃗β 1 , . . . , ⃗β n 〉 is a basis for avector space, and that for the vector ⃗v the relationship ⃗v = c 1⃗β 1 + c 2⃗β 2 + · · · +c n⃗β n has c i ≠ 0. Then exchanging ⃗β i for ⃗v yields another basis for the space.Proof Call the outcome of the exchange ˆB = 〈⃗β 1 , . . . , ⃗β i−1 ,⃗v, ⃗β i+1 , . . . , ⃗β n 〉.We first show that ˆB is linearly independent. Any relationship d 1⃗β 1 + · · · +d i ⃗v + · · · + d n⃗β n = ⃗0 among the members of ˆB, after substitution for ⃗v,d 1⃗β 1 + · · · + d i · (c 1⃗β 1 + · · · + c i⃗β i + · · · + c n⃗β n ) + · · · + d n⃗β n = ⃗0(∗)gives a linear relationship among the members of B. The basis B is linearlyindependent, so the coefficient d i c i of ⃗β i is zero. Because we assumed that c i isnonzero, d i = 0. Using this in equation (∗) above gives that all of the other d’sare also zero. Therefore ˆB is linearly independent.We finish by showing that ˆB has the same span as B. Half of this argument,that [ˆB] ⊆ [B], is easy; we can write any member d 1⃗β 1 +· · ·+d i ⃗v+· · ·+d n⃗β n of [ˆB]as d 1⃗β 1 +· · ·+d i ·(c 1⃗β 1 +· · ·+c n⃗β n )+· · ·+d n⃗β n , which is a linear combinationof linear combinations of members of B, and hence is in [B]. For the [B] ⊆ [ˆB] halfof the argument, recall that when ⃗v = c 1⃗β 1 +· · ·+c n⃗β n with c i ≠ 0, then we canrearrange the equation to ⃗β i = (−c 1 /c i )⃗β 1 + · · · + (1/c i )⃗v + · · · + (−c n /c i )⃗β n .Now, consider any member d 1⃗β 1 + · · · + d i⃗β i + · · · + d n⃗β n of [B], substitute for