Linear Algebra (VII)
Linear Algebra (VII) Yijia Chen 1. Review Basis and Dimension. We fix a vector space V. Lemma 1.1. Let A, B ⊆ V be two finite sets of vectors in V, possibly empty. If A is linearly indepen- dent and can be represented by B. Then jAj 6 jBj. Theorem 1.2. Let S ⊆ V and A, B ⊆ S be both maximally linearly independent in S. Then jAj = jBj. Definition 1.3. Let e1,..., en 2 V. Assume that – e1,..., en are linearly independent, – and every v 2 V can be represented by e1,..., en. Equivalently, fe1,..., eng is maximally linearly independent in V. Then fe1,..., eng is a basis of V. Note that n = 0 is allowed, and in that case, it is easy to see that V = f0g. By Theorem 1.2: 0 0 Lemma 1.4. If fe1,..., eng and fe1,..., emg be both bases of V with pairwise distinct ei’s and with 0 pairwise distinct ei, then n = m. Definition 1.5. Let fe1,..., eng be a basis of V with pairwise distinct ei’s. Then the dimension of V, denoted by dim(V), is n. Equivalently, if rank(V) is defined, then dim(V) := rank(V). 1 Theorem 1.6. Assume dim(V) = n and let u1,..., un 2 V. (1) If u1,..., un are linearly independent, then fu1,..., ung is a basis. (2) If every v 2 V can be represented by u1,..., un, then fu1,..., ung is a basis. Steinitz exchange lemma. Theorem 1.7. Assume that dim(V) = n and v1,..., vm 2 V with 1 6 m 6 n are linearly indepen- dent.
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