- 567 Suppose That V V2 Vk Is An Orthonormal Sub Set Of R Combine Theorem 4 4 On Page 245 With The Gram Schmid 1 (69.4 KiB) Viewed 29 times
567 Suppose that (V₁, V2,..., Vk) is an orthonormal sub- set of R". Combine Theorem 4.4 on page 245 with the Gram-Schmidt process to prove that this set can be extended to an orthonormal basis (V1, V2, Vk, ...,Vn} for R".
THEOREM 4.4 (Extension Theorem) Let S be a linearly independent subset of a nonzero sub- space V of R". Then S can be extended to a basis for V by inclusion of additional vectors. In particular, every nonzero subspace has a basis. PROOF Let S = {u₁, U₂,..., uz} be a linearly independent subset of V. If the span of S is V, then S is a basis for V that contains S, and we are done. Otherwise, there exists a vector v₁ in V that is not in the span of S. Property 3 on page 81 implies that S' = {u₁, u₂,..., uk, V₁} is linearly independent. If the span of S' is V, then S' is a basis for V that contains S, and again we are done. Otherwise, there exists a vector v₂ in V that is not in the span of S'. As before, S" = {u₁, U₂,..., uk, V₁, V2]) is linearly independent. We continue this process of selecting larger linearly independent subsets of V containing S until one of them is a generating set for V (and hence a basis for V that contains S). Note that this process must stop in at most n steps, because every subset of R" containing more than n vectors is linearly dependent by property 4 on page 81. To prove that V actually has a basis, let u be a nonzero vector in V. Applying the Extension Theorem to S = {(u), which is a linearly independent subset of V (property 1 on page 81), we see that V has a basis. We have seen that a nonzero subspace of R" has infinitely many bases. Although the vectors in two bases for a nonzero subspace may be different, the next theorem shows that the number of vectors in each basis for a particular subspace must be the same.