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Inner Product Spaces §6.2 Inner Product Spaces Preview Inner Product Spaces Examples Inner Product Spaces x6.2 Inner product spaces Satya Mandal, KU Summer 2017 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Goals 2 n I Concept of length, distance, and angle in R or R is extended to abstract vector spaces V . Such a vector space will be called an Inner Product Space. I An Inner Product Space V comes with an inner product that is like dot product in Rn. n I The Euclidean space R is only one example of such Inner Product Spaces. Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Inner Product Definition Suppose V is a vector space. I An inner product on V is a function h∗; ∗i : V × V ! R that associates to each ordered pair (u; v) of vectors a real number hu; vi, such that for all u; v; w in V and scalar c; we have 1. hu; vi = hv; ui: 2. hu; v + wi = hu; vi + hu; wi: 3. chu; vi = hcu; vi: 4. hv; vi ≥ 0 and v = 0 () hv; vi = 0: I The vector space V with such an inner product is called an inner product space. Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Theorem 6.2.1: Properties Let V be an inner product space. Let u; v 2 V be two vectors and c be a scalar, Then, 1. h0; vi = 0 2. hu + v; wi = hu; wi + hv; wi 3. hu; cvi = chu; vi Proof. We would have to use the properties in the definition. 1. Use (3): h0; vi = h00; vi = 0h0; vi = 0. 2. Use commutativity (1) and (2): hu+v; wi = hw; u+vi = hw; ui+hw; vi = hu; wi+hv; wi 3. Use (1) and (3): hu; cvi = hcv; ui = chv; ui = chu; vi The proofs are complete. Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Definitions Definitions Let V be an inner product space and u; v 2 V . 1. The length or norm of v is defined as kvk = phv; vi: 2. The distance between u; v 2 V is defined as d(u; v) = ku − vk 3. The angle θ vectors u; v 2 V is defined by the formula: hu; vi cos θ = 0 ≤ θ ≤ π: kuk kvk A version of Cauchy-Swartz inequality, to be given later, would assert that right side is between -1 and 1. Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Theorem(s) 6.2.2 Let V be an inner product space and u; v 2 V . Then, 1. Cauchy-Schwartz Inequality: jhu; vij ≤ kuk kvk : 2. Triangle Inequality: ku + vk ≤ kuk + kvk : 3. (Definition) We say that u; v are (mutually) orthogonal or perpendicular, if hu; vi = 0: We write u ? v: 4. Pythagorean Theorem. If u; v are orthogonal, then ku + vk2 = kuk2 + kvk2 : Proof. Exactly similar to the corresponding theorems in x6.1 for Rn: Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Orthogonal Projection Definition. Let V be an inner product space. Suppose v 2 V is a non-zero vector. Then, for u 2 V define Orthogonal Projection of u on to v: proj (u) = hv;ui v v kvk2 • ? u projv(u)−u • / ••/ projvu v Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Theorem 6.2.3 Let V be an inner product space. Suppose v 2 V is a non-zero vector. Then, (u − projv(u)) ? projv(u): Proof. hv; ui hv; ui hu − projv(u); projv(u)i = u − v ; v kvk2 kvk2 hv; ui hv; ui hv; ui = u; v − v ; v kvk2 kvk2 kvk2 hv; ui2 hv; ui2 = − hv; vi = 0 kvk2 kvk4 The proof is complete. Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Examples 6.2.1 I Remark. If v = (1; 0) (or on x−axis) and u = (x; y), then projvu = (x; 0): I (1) The Obvious Example: With dot product as the inner product, the Euclidean n−space Rn is an inner product space. Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Examples 6.2.2: Integration Integration is a great way to define inner product. Let V = C[a; b] be the vector space of all continuous functions f :[a; b] ! R: For f ; g 2 C[a; b]; define inner product Z b hf ; gi = f (x)g(x)dx: a It is easy to check that hf ; gi satisfies the properties of inner product spaces. Namely, 1. hf ; gi = hg; f i; for all f ; g 2 C[a; b]: 2. hf ; g + hi = hf ; gi + hf ; hi; for all f ; g; h 2 C[a; b]: 3. chf ; gi = hcf ; gi; for all f ; g 2 C[a; b] and c 2 R: 4. hf ; f i ≥ 0 for all f 2 C[a; b] and f = 0 , hf ; f i = 0: Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Continued Accordingly, for f 2 C[a; b]; we can define length (or norm) s Z b kf k = phf ; f i = f (x)2dx: a This 'length' of continuous functions would have all the properties that you expect "length" or "magnitude" to have. Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Examples 6.2.2A: Double Integration Let D ⊆ R2 be any connected region. Let V = C(D) be the vector space of all bounded continuous functions f (x; y): D ! R: For f ; g 2 V define inner product ZZ hf ; gi = f (x; y)g(x; y)dxdy: D As in Example 6.2.2, it is easy to check that hf ; gi satisfies the properties of inner product spaces. In this case, length or norm of f 2 V is given by sZZ kf k = phf ; f i = f (x; y)2dxdy: D Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview Inner Product Spaces Examples Continued In particular: I Example a: If D = [a; b] × [c; d], then Z d Z b hf ; gi = f (x; y)g(x; y)dxdy: c a I Example b: If D is the unit disc: D = f(x; y): x 2 + y 2 ≤ 1g, then for f ; g 2 C(D) is: ZZ hf ; gi = f (x; y)g(x; y)dxdy: D p Z 1 Z 1−y 2 = p f (x; y)g(x; y)dxdy −1 − 1−y 2 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Example 6.2.3 In R2; define an inner product (as above): for u = (u1; u2); v = (v1; v2) define hu; vi = 2(u1v1 + u2v2). It is easy to check that this is an Inner Product on R2 (we skip the proof.) Let u = (1; 3); v = (2; −2). I (1) Compute hu; vi: Solution: hu; vi = 2(u1v1 + u2v2) = 2(2 − 6) = −8 I (2) Compute kuk : Solution: p p p p kuk = hu; ui = 2(u1u1 + u2u2) = 2(1 + 9) = 20: Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued I (3) Compute kvk : Solution: kvk = phv; vi = p2(4 + 4) = 4 I (4) Compute d(u; v): Solution: d(u; v) = ku − vk = k(−1; 5)k p = p2(1 + 25) = 52: Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Example 6.2.4 Let V = C[0; 1] with inner product Z 1 hf ; gi = f (x)g(x)dx for f ; g; 2 V : 0 Let f (x) = 2x and g(x) = x 2 + x + 1: I (1) Compute hf ; gi: Solution: We have Z 1 Z 1 hf ; gi = f (x)g(x)dx = 2 x 3 + x 2 + x dx 0 0 x 4 x 3 x 2 1 1 1 1 13 = 2 + + = 2 + + − 0 = 4 3 2 x=0 4 3 2 6 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued I (2) Compute norm kf k : Solution: We have s s Z 1 Z 1 kf k = phf ; f i = f (x)2dx = 4x 2dx 0 0 s x 3 1 r4 r1 = 4 = − 0 = 2 3 x=0 3 3 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued I (3) Compute norm kgk : Solution: We have s s Z 1 Z 1 kgk = phg; gi = g(x)2dx = (x 2 + x + 1)2dx 0 −1 s Z 1 = (x 4 + 2x 3 + 3x 2 + 2x + 1) dx 0 s x 5 x 4 x 3 x 2 1 = + 2 + 3 + 2 + x 5 4 3 2 0 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued s 1 1 1 1 r37 = + 2 + 3 + 2 + 1 − 0 = 5 4 3 2 10 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued I (4) Compute d(f ; g): Solution: We have d(f ; g) = kf − gk = s Z 1 phf − g; f − gi = (−x 2 + x − 1)2dx 0 s Z 1 = (x 4 − 2x 3 + 3x 2 − 2x + 1) dx 0 s x 5 x 4 x 3 x 2 1 = − 2 + 3 − 2 + x 5 4 3 2 0 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued s 1 1 1 1 r 7 = − 2 + 3 − 2 + 1 − 0 = 5 4 3 2 10 Satya Mandal, KU Inner Product Spaces x6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Example 6.2.4 Let V = C[−π; π] with inner product hf ; gi as in Example 6.2.2 (by definite integral).
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