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Inner Product Spaces §6.2 Inner product spaces

Satya Mandal, KU

Summer 2017

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview Inner Product Spaces Examples Goals

2 n I Concept of length, , and in R or R is extended to abstract vector spaces V . Such a vector will be called an .

I An Inner Product Space V comes with an inner product that is like in Rn. n I The R is only one example of such Inner Product Spaces.

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview Inner Product Spaces Examples Inner Product

Definition Suppose V is a .

I An inner product on V is a

h∗, ∗i : V × V → R that associates to each (u, v) of vectors a real hu, vi, such that for all u, v, w in V and 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 §6.2 Inner product spaces Preview Inner Product Spaces Examples Theorem 6.2.1: Properties

Let V be an inner product space. Let u, v ∈ 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 §6.2 Inner product spaces Preview Inner Product Spaces Examples Definitions

Definitions Let V be an inner product space and u, v ∈ V . 1. The length or of v is defined as kvk = phv, vi. 2. The distance between u, v ∈ V is defined as d(u, v) = ku − vk 3. The angle θ vectors u, v ∈ V is defined by the formula: hu, vi cos θ = 0 ≤ θ ≤ π. kuk kvk A version of Cauchy-Swartz , to be given later, would assert that right side is between -1 and 1.

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview Inner Product Spaces Examples Theorem(s) 6.2.2

Let V be an inner product space and u, v ∈ V . Then, 1. Cauchy-Schwartz Inequality: |hu, vi| ≤ kuk kvk . 2. : ku + vk ≤ kuk + kvk . 3. (Definition) We say that u, v are (mutually) orthogonal or , if

hu, vi = 0. We write u ⊥ v.

4. . If u, v are orthogonal, then

ku + vk2 = kuk2 + kvk2 .

Proof. Exactly similar to the corresponding theorems in §6.1 for Rn. Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview Inner Product Spaces Examples Orthogonal Projection

Definition. Let V be an inner product space. Suppose v ∈ V is a non-zero vector. Then, for u ∈ 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 §6.2 Inner product spaces Preview Inner Product Spaces Examples Theorem 6.2.3

Let V be an inner product space. Suppose v ∈ 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 §6.2 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 §6.2 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 ∈ 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 ∈ C[a, b]. 2. hf , g + hi = hf , gi + hf , hi, for all f , g, h ∈ C[a, b]. 3. chf , gi = hcf , gi, for all f , g ∈ C[a, b] and c ∈ R. 4. hf , f i ≥ 0 for all f ∈ C[a, b] and f = 0 ⇔ hf , f i = 0.

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview Inner Product Spaces Examples Continued

Accordingly, for f ∈ 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 ”” to have.

Satya Mandal, KU Inner Product Spaces §6.2 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 ∈ 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 ∈ V is given by

sZZ kf k = phf , f i = f (x, y)2dxdy. D

Satya Mandal, KU Inner Product Spaces §6.2 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 = {(x, y): x 2 + y 2 ≤ 1}, then for f , g ∈ C(D) is: ZZ hf , gi = f (x, y)g(x, y)dxdy. D √ Z 1 Z 1−y 2 = √ f (x, y)g(x, y)dxdy −1 − 1−y 2

Satya Mandal, KU Inner Product Spaces §6.2 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 √ kuk = hu, ui = 2(u1u1 + u2u2) = 2(1 + 9) = 20.

Satya Mandal, KU Inner Product Spaces §6.2 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 √ = p2(1 + 25) = 52.

Satya Mandal, KU Inner Product Spaces §6.2 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, ∈ 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 §6.2 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 §6.2 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 §6.2 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 §6.2 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 §6.2 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 §6.2 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 ). Let f (x) = x 3 and g(x) = x 2 − 3. Show that f and g are orthogonal. Solution: We have to show that hf , gi = 0. We have hf , gi =

Z π Z π Z π f (x)g(x)dx = x 3(x 2 − 3)dx = (x 5 − 3x 3dx −π −π −π

x 6 x 4 π = − = 0. 6 4 −π So, f ⊥ g.

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Example 6.2.5 √ √ Exercise Let u = ( 2, 2) and v = (3, −4). I Compute projv(u) and proj√u(v) √ √ I Solution. First hu, vi = 2 ∗ 3 − 2 ∗ 4 = − 2,

q√ 2 √ 2 kuk = 2 + 2 = 4, kvk = p32 + (−4)2 = 5

I hv, ui √  √ √  projv(u) = v = − 2(3, −4) = −3 2, 4 2 kvk2

I hu, vi √ √ √ proju(v) = u = − 2( 2, 2) = (−2, −2) kuk2

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Example 6.2.6

Let V = C[0, 1] with inner product

Z 1 hf , gi = f (x)g(x)dx for f , g, ∈ V . 0 Let f (x) = 2x and g(x) = x 2 + x + 1. Compute the orthogonal projection of f onto g, and the orthogonal projection of g onto f . Solution From Example 6.2.4, where we worked these two functions f , g, we have

13 r1 r37 hf , gi = , kf k = 2 , kgk = 6 2 10

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued

13 hg, f i 6 13 projf (g) = f = (2x) = x kf k2 2 6 Also,

13 hg, f i 6 2 130 2 projg (f ) = 2 g = 37 (x + x + 1) = (x + x + 1) kgk 10 222

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Example 6.2.7

Let V = C[0, 1] with inner product hf , gi as in Example 6.2.2 (by definite integral). Let f (x) = x 3 + x and g(x) = 2x + 1. Compute the orthogonal projection of f onto g. Solution Recall the definition: proj (u) = hv,ui v So, v kvk2

hg, f i projg (f ) = g kgk2

I First compute hg, f i =

Z 1 Z 1 (x 3 + x)(2x + 1)dx (2x 4 + x 3 + 2x 2 + x)dx 0 0

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued

I  x 5 x 4 x 3 x 2 1 109 = 2 + + 2 + = 5 4 3 2 0 60 109 I So hg, f i = 60

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued

2 I Now compute kgk =

Z 1 Z 1  x 3 x 2 1 (2x+1)2dx = (4x 2+4x+1)dx = 4 + 4 + x 0 −1 3 2 0  1 1  13 = 4 + 4 + 1 − 0 = 3 2 3

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces Preview By Integration Inner Product Spaces Orthogonal Projections Examples Continued

I So,

109 hg, f i 60 109 projg (f ) = 2 g = 13 (2x + 1) = (2x + 1) kgk 3 260

Satya Mandal, KU Inner Product Spaces §6.2 Inner product spaces