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Theorem 1.5.0 The following conditions for a finite subset X of a vector space are equivalent:

X is a maximal linearly independent set;

X is a minimal spanning set;

X is a linearly independent spanning set.

Proof. We show that each of (a) and (b) is equivalent to (c).

(c) ⇒ (a): If X is a basis for V, then every vector in V is a linear combination of X; so adding any vector to X gives a linearly dependent set, which means that X is maximal.

(a) ⇒ (c): If X is maximal independent, then any vector v added to X gives a linearly dependent set. In a dependence relation, the coefficient of v must be non-zero (else X would be linearly dependent); so we can use the relation to express v as a linear combination of X. So X spans V, and is a basis.

b) ⇒ (c): If X is a minimal spanning set, then no element of X is a linear combination of the others (or it could be dropped without losing the spanning property); so X is linearly independent. □

Corollary 1.5.11. Let V be a vector space and suppose that one basis has n elements, and another basis W has m elements. Then n = m.

Proof. Since a basis is both a maximal linearly independent set and minimal spanning set, we must have V = W. □

Corollary 1.5.1. Let V be a vector space and suppose that one basis has n elements, and another basis W has m elements. Then n = m.

Proof. Since a basis is both a maximal linearly independent set and minimal spanning set, we must have V = W. □

Definition 1.5.2 Let V be a vector space having a basis consisting of n elements. We shall say that n is the dimension of V. If V consists of 0 alone, then V does not have a basis, and we hall say that V has dimension 0.

Sat, 15 Apr 2023 09:57 GMT