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115_solns_6

# 115_solns_6 - 115A PROBLEMS FROM CLASS NOV 29 Sec 6.2 15(a...

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Unformatted text preview: 115A PROBLEMS FROM CLASS, NOV 29 Sec 6.2. 15. (a) Parseval’s identity. Let β = {v1 , . . . , vn } be an orthonormal basis for an inner product space V . For any x, y ∈ V prove that n x, y = i=1 x, vi y , vi . By Theorem 6.5, since β is an orthonormal basis, we can write the vectors x and y as n n x= i=1 x, vi vi , y= i=1 y , vi vi . Thus, n n x, y = i=1 n n x, vi vi , j =1 y , vj vj = i=1 j =1 n n x, vi vi , y , vj vj x, vi y , vj vi , vj i=1 j =1 n = = i=1 x, vi y , vi 16. (a) Bessel’s inequality. Let V be an inner product space, and let S = {v1 , . . . , vn } be an orthonormal subset of V . Prove that for any x ∈ V we have n x 2 ≥ i=1 | x, vi | . 2 Following the hint, let W = span(S ). Then S is an orthnormal basis for W (why?). By Theorem 6.6, there exist unique u ∈ W and z ∈ W ⊥ such that x = u+z . Moreover, we have that n u= i=1 1 x, vi vi . 2 115A PROBLEMS FROM CLASS, NOV 29 2 By Pythagoras’ theorem, since u ⊥ z , we have that x It follows that 2 2 2 x =u +z ≥u = i=1 n n 2 n n = u+z 2 =u 2 +z . 2 x, vi vi , j =1 x, vj vj = i=1 j =1 n n x, vi vi , x, vj vj x, vi x, vj vi , vj i=1 j =1 n = = i=1 | x, vi | . 2 (b) Observe that the inequality is an equality precisely when z = 0. Thus, in this case, x = u ∈ W = span(S ). References Friedberg, et. al. Linear Algebra. Department of Mathematics, University of California, Los Angeles, 90095 E-mail address : [email protected] ...
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115_solns_6 - 115A PROBLEMS FROM CLASS NOV 29 Sec 6.2 15(a...

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