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Inner Product Spaces §6.3 Orthonormal Bases

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Inner Product Spaces §6.3 Orthonormal Bases Preview Orthonormal Bases Example Inner Product Spaces x6.3 Orthonormal Bases Satya Mandal, KU Summer 2017 Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Goals I Define Orthonormal Basis of an Inner Product Spaces I Discuss Gram-Schmidt Method of finding an Orthonormal Basis Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Definition: Orthonormal Basis Definition Suppose (V ; h−; −i) is an Inner product space. I A subset S ⊆ V is said to be an Orthogonal subset, if hu; vi = 0, for all u; v 2 S, with u 6= v. That means, if elements in S are pairwise orthogonal. I An Orthogonal subset S ⊆ V is said to be an Orthonormal subset if, in addition, kuk = 1, for all u 2 S. I If an Orthonormal set S is also a basis of V , then it is called an Orthonormal Basis. That means, if 8 < hu; vi = 0 8 u; v 2 S; u 6= v kuk = 1 u 2 S : S is a basis of V Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Continued I We remark that, we did not require that S is a finite set. However, we would mainly be considering finite such subsets S = fu1; u2;:::; ung ⊆ V . I So, a finite subset S = fu1; u2;:::; ung ⊆ V is an orthonormal basis, if 8 < hui ; uj i = 0 8 i; j = 1; 2;:::; n i 6= j kui k = 1 8 i; = 1; 2;:::; n : S is a basis of V Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Example 6.3.1 Most obvious example of an orthonormal basis is standard n basis S = fe1; e2;:::; eng ⊆ R , where 8 > e1 = (1; 0; 0;:::; 0) > <> e2 = (0; 1; 0;:::; 0) e3 = (0; 0; 1;:::; 0) (1) > ··· > : en = (0; 0; 0;:::; 1) Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Lemma 6.3.1 Suppose (V ; h−; −i) is an inner product space. Suppose S = fu1; u2;:::; ung ⊆ V is an orthogonal set, consisting of nonzero vectors. Then, S is linearly independent. Proof. Suppose c1u1 + c2u2 + ··· + cnun = 0 where c1;:::; cn 2 R: =) hu1; c1u1 + c2u2 + ··· + cnuni = hu1; 0i = 0 =) c1hu1; u1i + c2hu1; u2i + ··· + cnhu1; uni = 0 =) c1hu1; u1i + c20 + ··· + cn0 = 0: =) c1hu1; u1i = 0: =) c1 = 0: Likewise; c2 = ··· = cn = 0 This completes the proof. Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Recall: Orthogonal Projections Suppose (V ; h−; −i) is an inner product space and u; v 2 V . Then the orthogonal projective of u along v is hu; vi Projv(u) = v kvk2 In the next frame, we would discuss Gram-Schmidt Orthogonalization Process. Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Theorem 6.3.2: Gram-Schmidt Orthogonalization Process Let (V ; h−; −i) be an inner product space and S = fu1; u2;:::; ung be a basis of V . Construct the following sequence of vectors: 8 > v1 = u1 > v = u − Proj (u ) <> 2 2 v1 2 v3 = u3 − Projv1 (u3) − Projv2 (u3) > > ······ :> vn = un − Projv1 (un) − Projv2 (un) − · · · − Projvn−1 (un) (2) Then, fv1;:::; vng is an Orthogonal Basis of V . Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Proof. Proof.It is easy to see, inductively, spanfv1;:::; vng = spanfu1; u2;:::; ung = V : Since dim V = n, the theorem follows from Theorem 4.5.5. Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Continued In this frame, we write Equation 2, more explicitly. 8 > v1 = u1 > hu2;v1i > v2 = u2 − 2 v1 > kv1k < hu3;v1i hu3;v2i v3 = u3 − 2 v1 − 2 v2 (3) kv1k kv2k > ······ > > hun;v1i hun;v2i hun;v2i : vn = un − 2 v1 − 2 v2 − · · · − 2 vn−1 kv1k kv2k kvn−1k Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Continued In this frame, we show first two terms geometrically: • ? O u2 u2−projv1 (u2)=:v2 • / • / • v1=u1 projv1 u2 Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Corollary 6.3.3 Let (V ; h−; −i) be an inner product space and S = fu1; u2;:::; ung be a basis of V . Let fv1; v2;:::; vng be as in Equation 2, which is an orthogonal basis of V . Then, v v v 1 ; 2 ; ··· ; n kv1k kv2k kvnk is an Orthonormal Basis of V . Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Example 6.3.2 Let u1 = (1; 1; 1), u2 = (1; −1; 1), u3 = (1; 1; −1). Use Gram-Schmidt Orthogonalization process, to compute an orthonormal basis of R3, in two steps, using Theorem 6.3.2 and Corollary 6.3.3. Solution: We have 2 2 2 ku1k = ku2k = ku3k = 3; hu1; u2i = hu1; u3i = 1; hu2; u3i = −1 Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Continued 8 > v1 = u1 = (1; 1; 1) > hu2;v1i 1 2 4 2 > v2 = u2 − 2 v1 = (1; −1; 1) − 3 (1; 1; 1) = 3 ; − 3 ; 3 > kv1k > kv k2 = 24 ; hu ; v i = − 4 <> 2 9 3 2 3 hu3;v1i hu3;v2i v3 = u3 − 2 v1 − 2 v2 kv1k kv2k > −4=3 > = (1; 1; −1) − 1 (1; 1; 1) − 2 ; − 4 ; 2 > 3 24=9 3 3 3 > = (1; 1; −1) − 1 (1; 1; 1) + 1 ; − 2 ; 1 = (1; 1; −1) + (0; −1; 0) > 3 3 3 3 :> = (1; 0; −1) Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Continued So, an Orthogonal Basis of R3 is 2 4 2 (1; 1; 1); ; − ; ; (1; 0; −1) 3 3 3 p p 2 2 p Also; kv1k = 3; kv2k = p kv3k = 2 3 Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases Preview Orthonormal Bases Example Continued So, an Orthonormal basis of R3 is ( p ) 1 1 1 3 2 4 2 1 1 p ; p ; p ; p ; − ; ; p ; 0; −p 3 3 3 2 2 3 3 3 2 2 1 1 1 1 2 1 1 1 = p ; p ; p ; p ; −p ; p ; p ; 0; −p 3 3 3 6 6 6 2 2 Satya Mandal, KU Inner Product Spaces x6.3 Orthonormal Bases.
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