Linear Algebra

Section III. Basis and Dimension 117⃗β i its expression as a linear combination of the members of ˆB, and recognize,as in the first half of this argument, that the result is a linear combination oflinear combinations of members of ˆB, and hence is in [ˆB].QED2.5 Theorem In any finite-dimensional vector space, all bases have the samenumber of elements.Proof Fix a vector space with at least one finite basis. Choose, from amongall of this space’s bases, one B = 〈⃗β 1 , . . . , ⃗β n 〉 of minimal size. We will showthat any other basis D = 〈⃗δ 1 ,⃗δ 2 , . . .〉 also has the same number of members, n.Because B has minimal size, D has no fewer than n vectors. We will argue thatit cannot have more than n vectors.The basis B spans the space and ⃗δ 1 is in the space, so ⃗δ 1 is a nontrivial linearcombination of elements of B. By the Exchange Lemma, we can swap ⃗δ 1 for avector from B, resulting in a basis B 1 , where one element is ⃗δ 1 and all of then − 1 other elements are ⃗β’s.The prior paragraph forms the basis step for an induction argument. Theinductive step starts with a basis B k (for 1 k < n) containing k members of Dand n − k members of B. We know that D has at least n members so there is a⃗δ k+1 . Represent it as a linear combination of elements of B k . The key point: inthat representation, at least one of the nonzero scalars must be associated witha ⃗β i or else that representation would be a nontrivial linear relationship amongelements of the linearly independent set D. Exchange ⃗δ k+1 for ⃗β i to get a newbasis B k+1 with one ⃗δ more and one ⃗β fewer than the previous basis B k .Repeat the inductive step until no ⃗β’s remain, so that B n contains ⃗δ 1 , . . . ,⃗δ n .Now, D cannot have more than these n vectors because any ⃗δ n+1 that remainswould be in the span of B n (since it is a basis) and hence would be a linearcombination of the other ⃗δ’s, contradicting that D is linearly independent. QED2.6 Definition The dimension of a vector space is the number of vectors in anyof its bases.2.7 Example Any basis for R n has n vectors since the standard basis E n has nvectors. Thus, this definition generalizes the most familiar use of term, that R nis n-dimensional.2.8 Example The space P n of polynomials of degree at most n has dimensionn+1. We can show this by exhibiting any basis — 〈1, x, . . . , x n 〉 comes to mind —and counting its members.2.9 Example A trivial space is zero-dimensional since its basis is empty.Again, although we sometimes say ‘finite-dimensional’ as a reminder, inthe rest of this book we assume that all vector spaces are finite-dimensional.An instance of this is that in the next result the word ‘space’ means ‘finitedimensionalvector space’.