This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Email address : [email protected] ...
View
Full
Document
This note was uploaded on 10/20/2010 for the course MATH MAT443 taught by Professor Jones during the Fall '09 term at ASU.
 Fall '09
 Jones
 Linear Algebra, Algebra

Click to edit the document details