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ʼn each part, apply the Gram-Schmidt process to the given subset S of the inner product space o obtain an orthogonal basis for span(S). Then normalize the vectors in this basis to obtain an onormal basis ß for span(S), and compute the Fourier coefficients of the given vector tive to ß. Finally, use Theorem 6.5 to verify your result. V = R¹, S = {(2, -1, −2, 4), (−2, 1, −5, 5), (−1, 3, 7, 11)}, and x = (–11, 8, –4, 18)

ʼn each part, apply the Gram-Schmidt process to the given subset S of the inner product space o obtain an orthogonal basis for span(S). Then normalize the vectors in this basis to obtain an onormal basis ß for span(S), and compute the Fourier coefficients of the given vector tive to ß. Finally, use Theorem 6.5 to verify your result. V = R¹, S = {(2, -1, −2, 4), (−2, 1, −5, 5), (−1, 3, 7, 11)}, and x = (–11, 8, –4, 18)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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2. In each part, apply the Gram-Schmidt process to the given subset S of the inner product space V to obtain an orthogonal basis for span(S). Then normalize the vectors in this basis to obtain an orthonormal basis for span(S), and compute the Fourier coefficients of the given vector relative to ß. Finally, use Theorem 6.5 to verify your result. (e) V = R¹, S = {(2,−1, −2, 4), (−2, 1, −5, 5), (-1, 3, 7, 11)}, and x = (-11, 8, -4, 18) (j) V = C¹, S = {(1, i, 2 — i, −1), (2 + 3i, 3i, 1 — i, 2i), (−1 + 7i, 6 + 10i, 11 – 4i, 3 + 4i)}, and x = = (-2+7i, 6+9i, 9 - 3i, 4 + 4i)
Transcribed Image Text:2. In each part, apply the Gram-Schmidt process to the given subset S of the inner product space V to obtain an orthogonal basis for span(S). Then normalize the vectors in this basis to obtain an orthonormal basis for span(S), and compute the Fourier coefficients of the given vector relative to ß. Finally, use Theorem 6.5 to verify your result. (e) V = R¹, S = {(2,−1, −2, 4), (−2, 1, −5, 5), (-1, 3, 7, 11)}, and x = (-11, 8, -4, 18) (j) V = C¹, S = {(1, i, 2 — i, −1), (2 + 3i, 3i, 1 — i, 2i), (−1 + 7i, 6 + 10i, 11 – 4i, 3 + 4i)}, and x = = (-2+7i, 6+9i, 9 - 3i, 4 + 4i)
Theorem 6.5. Let V be a nonzero finite-dimensional inner product space. Then V has an orthonormal basis 3. Furthermore, if ß = {V₁, V2, ..., Un} and x € V, then x = Proof. Let Bo be an ordered basis for V. Apply Theorem 6.4! to obtain an orthogonal set 3' of nonzero vectors with span (3′) = span (Bo) = V. By normalizing each vector in ß', we obtain an orthonormal set 3 that generates V. By Corollary 2 to Theorem 6.3 , ß is linearly independent; therefore 3 is an orthonormal basis for V. The remainder of the theorem follows from Corollary 1 to Theorem 6.3 = {w₁, W2,..., wn} be a linearly independent Theorem 6.4. Let V be an inner product space and S subset of V. Define S' = {V1, V2,..., Un}, where v₁ = w₁ and k-1 Vk = Wk - Σ k-1 (Vk, Vi) = (Wk, Vi) - Σ j=1 n Σ (x, vi) vi. i=1 (Wk, Vj) Vj ||v₂||² j=1 Then S' is an orthogonal set of nonzero vectors such that span (S') = span(S). Proof. The proof is by mathematical induction on n, the number of vectors in S. For k = 1,2,..., n, let Sk = {w₁, W2, ..., Wk}. If n 1, then the theorem is proved by taking S₁ S₁; i.e., v₁ w₁0. Assume then that the set Sk-1 = {V₁, V₂, ..., Uk-1} with the desired properties has been constructed by the repeated use of (1). We show that the set S% = {V₁, V₂, ..., Uk-1, Uk} also has the desired properties, where U is obtained from S-1 by (1). If U = 0, then (1) implies that we € span (S-1) = span (Sk-1), which contradicts the assumption that S is linearly independent. For 1 ≤ i ≤k- 1, it follows from (1) that (Wk, Vj) || 03 || 2 y = (Vj, Vi) = k Σ i=1 for 2 ≤ k ≤n. (y, vi), ||v₂||² (wk, Vi). since (v₁, vi) = 0 if i ‡ j by the induction assumption that S-1 is orthogonal. Hence is an orthogonal set of nonzero vectors. Now, by (1), we have that span (S%) span (Sk). But by Corollary 2 to Theorem 6.3 2 S is linearly independent; so dim (span (S%)) = dim (span (Sk)) =k. Therefore span (S) = span (Sk). The construction of {v₁, V2,..., Un} by the use of Theorem 6.4 is called the Gram-Schmidt process. Theorem 6.3. Let V be an inner product space and S = {V₁, V₂, ..., Uk} be an orthogonal subset of V consisting of nonzero vectors. If y E span(S), then (Wk, Vi) ||v₂||² - Vi ||v₂||² = 0,
Transcribed Image Text:Theorem 6.5. Let V be a nonzero finite-dimensional inner product space. Then V has an orthonormal basis 3. Furthermore, if ß = {V₁, V2, ..., Un} and x € V, then x = Proof. Let Bo be an ordered basis for V. Apply Theorem 6.4! to obtain an orthogonal set 3' of nonzero vectors with span (3′) = span (Bo) = V. By normalizing each vector in ß', we obtain an orthonormal set 3 that generates V. By Corollary 2 to Theorem 6.3 , ß is linearly independent; therefore 3 is an orthonormal basis for V. The remainder of the theorem follows from Corollary 1 to Theorem 6.3 = {w₁, W2,..., wn} be a linearly independent Theorem 6.4. Let V be an inner product space and S subset of V. Define S' = {V1, V2,..., Un}, where v₁ = w₁ and k-1 Vk = Wk - Σ k-1 (Vk, Vi) = (Wk, Vi) - Σ j=1 n Σ (x, vi) vi. i=1 (Wk, Vj) Vj ||v₂||² j=1 Then S' is an orthogonal set of nonzero vectors such that span (S') = span(S). Proof. The proof is by mathematical induction on n, the number of vectors in S. For k = 1,2,..., n, let Sk = {w₁, W2, ..., Wk}. If n 1, then the theorem is proved by taking S₁ S₁; i.e., v₁ w₁0. Assume then that the set Sk-1 = {V₁, V₂, ..., Uk-1} with the desired properties has been constructed by the repeated use of (1). We show that the set S% = {V₁, V₂, ..., Uk-1, Uk} also has the desired properties, where U is obtained from S-1 by (1). If U = 0, then (1) implies that we € span (S-1) = span (Sk-1), which contradicts the assumption that S is linearly independent. For 1 ≤ i ≤k- 1, it follows from (1) that (Wk, Vj) || 03 || 2 y = (Vj, Vi) = k Σ i=1 for 2 ≤ k ≤n. (y, vi), ||v₂||² (wk, Vi). since (v₁, vi) = 0 if i ‡ j by the induction assumption that S-1 is orthogonal. Hence is an orthogonal set of nonzero vectors. Now, by (1), we have that span (S%) span (Sk). But by Corollary 2 to Theorem 6.3 2 S is linearly independent; so dim (span (S%)) = dim (span (Sk)) =k. Therefore span (S) = span (Sk). The construction of {v₁, V2,..., Un} by the use of Theorem 6.4 is called the Gram-Schmidt process. Theorem 6.3. Let V be an inner product space and S = {V₁, V₂, ..., Uk} be an orthogonal subset of V consisting of nonzero vectors. If y E span(S), then (Wk, Vi) ||v₂||² - Vi ||v₂||² = 0,
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