Inner Product Spaces 5.2 Inner product spaces

Inner Product Spaces 5.2 Inner product spaces November 15

Goals Concept of length, distance, and angle in R 2 or R n is extended to abstract vector spaces V. Sucn a vector space will be called an Inner Product Space. An Inner Product Space V comes with an inner product that is like dot product in R n. The Euclidean space R n is only one example of such Inner Product Spaces.

Inner Product Definition Suppose V is a vector space. An inner product on V is a function, : V V R that associates to each odered pair (u,v) of vectors a real number u,v is associated, such that for all u,v,w in V and scalar c, we have 1. u,v v,u. 2. u,v+w u,v + u,w. 3. c u,v cu,v. 4. v,v 0 and v 0 v,v 0. The vector space V with such an inner product is called and inner product space.

Properties: Theorem 5.7 Theorem 5.7 Let V be an inner product space. Let u,v V be two vectors and c be a scalar, Then, 1. 0,v 0 2. u+v,w u,w + v,w 3. u,cv c u,v Proof. We would have to use the properties in the definition. 1. Use (3): 0,v 00,v 0 0,v 0. 2. Use commutativity (1) and (2): u+v,w w,u+v w,u + w,v u,w + v,w 3. Use (1) and (3): u,cv cv,u c v,u c u,v The proofs are complete.

Definitions Definitions Let V be an inner product space and u,v V. 1. The length or norm of v is defined as v v,v. 2. The distance between u,v V is defined as d(u,v) u v 3. The angle θ vectors u,v V is defined by the formula: cosθ u,v u v 0 θ π. A version of Cauchy-Swartz inequality, to be given later, would assert that right side is between -1 and 1.

Theorem(s) 5.8 Theorem(s) 5.8 Let V be an inner product space and u,v V. Then, 1. Cauchy-Schwartz Inequality: u, v u v. 2. Triangle Inequality: u+v u + v. 3. (Definition) We say that u, v are (mutually) orthogonal or perpendicular, if u,v 0. We write u v. 4. Pythagorean Theorem. If u, v are orthogonal, then u+v 2 u 2 + v 2. Proof. Exactly similar to the corresponding theorem in 5.1 for R n.

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 unto v: proj v (u) v,u v 2 v 888888888888888 u 8 proj vu proj v(u) u v

Continued Theorem. In fact (u proj v (u)) proj v (u). Proof. ( ) v,u u proj v (u),proj v (u) u v 2 v, v,u v 2 v u, v,u ( v,u v 2 v v 2 v ) v,u 2 v 2 v,u 2 v 4 v,v 0, v,u v 2 v The proof is complete. Remark. If v (1,0) (or on x axis) and u (x,y), then proj v u (x,0).

Examples (1) The Obvious Example: With dot product as the inner product, the Euclidean n space R n is an inner product space. (2) Example 2 (Textbook): On R 2 for u (u 1,u 2 ),v (v 1,v 2 ) define u,v u 1 v 1 +2u 2 v 2. This defines an inner product on R 2. Proof. One directly checks: (1) u,v v,u, (2) u+v,w v,w + u,w, (3) cu,v c v,u, (4) v,v 0 and v,v 0 v 0.

Examples (3) 5.2 Example 5 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 f,g b a f(x)g(x)dx. It is easy to check that f,g satisfies the properties ofinner product space. Namely, 1. f,g g,f, for all f,g C[a,b]. 2. f,g +h f,g + f,h, for all f,g,h C[a,b]. 3. c f,g cf,g, for all f,g C[a,b] and c R. 4. f,f 0 for all f C[a,b] and f 0 f,f 0.

Continued Accordingly, for f C[a,b], we can define length (or norm) f b f,f f(x) 2 dx. This length of continuous functions would have all the properties that you expect length to have. a

Exercise 20 Exercise 20 In R 2, define an inner product (as above): for u (u 1,u 2 ), v (v 1,v 2 ) define u,v u 1 v 1 +2u 2 v 2. Let u (0, 6), v (,1). (1) Compute u, v. Solution: u,v u 1 v 1 +2u 2 v 2 0 ()+2( 6) 1 2. (2) Compute u. Solution: u u,u u 1 u 1 +2u 2 u 2 0 0+2( 6) ( 6) 72.

Continued (3) Compute v. Solution: v v,v ()()+2(1) (1) 3. (4) Compute d(u, v). Solution: d(u,v) u v (1, 7) 1 1+2( 7)( 7) 99.

Exercise 40 Exercise 40 Let V C[,1] with inner product f,g 1 f(x)g(x)dx for f,g, V. Let f(x) x and g(x) x 2 x +2. (1) Compute f,g. Solution: We have 1 1 ( f,g f(x)g(x)dx x 3 +x 2 2x ) dx [ ] x 4 1 4 + x3 3 2x2 2 x [ 4 + 1 ] [ 3 21 2 4 + ] 3 21 2 2 3.

Continued (2) Compute norm f. Solution: We have f 1 1 f,f f(x) 2 dx x 2 dx [x 3 3 ] 1 x ( 1 3 1 ) 2 3 3

Continued (3) Compute norm g. Solution: We have g 1 1 g,g g(x) 2 dx (x 2 x +2) 2 dx 1 (x 4 2x 3 +5x 2 4x +4)dx [x 5 5 2x4 4 +5x3 3 4x2 2 +4x ] 1 2 5 + 10 3 +8.

Continued (4) Compute d(f,g). Solution: We have 1 d(f,g) f g f,f ( x 2 2) 2 dx 1 (x 4 +4x 2 +4)dx [x 5 2 5 + 8 3 +8. 5 +4x3 3 +4x ] 1

Exercise 66 Exercise 66 Let V C[,1] with inner product f,g as in Exercise 40 (by definite integral). Let f(x) x and g(x) 3x2. Show that f and g are orthogonal. 2 Solution: We have to show that f,g 0. We have f,g 1 f(x)g(x)dx 1 ( ) 3x 2 1 1 x dx 2 2 (3x3 x)dx [ )] 1 1 (3 x4 2 4 x2 1 ( 3 1 2 2 4 1 ) 1 ( 3 1 2 2 4 1 ) 0. 2 So, f g.

Exercise Exercise Let u (2, 2) and v (3,1). Compute proj v (u) and proj u (v) Solution. First u,v 2 3 2 1 4, u 2 2 +( 2) 2 8, v 3 2 +1 2 10 proj v (u) v,u v 2 v 4 ( 6 10 (3,1) 5, 2 ) 5 proj u (v) u,v u 2 u 4 (2, 2) (1,) 8

Exercise 78 Exercise 78 Let V C[,1] with inner product f,g as in Exercise 40 (by definite integral). Let f(x) x 3 x and g(x) 2x. Compute the orthogonal projection of f onto g. Solution Recall the definition: proj v (u) v,u v 2 v So, First compute g,f 1 (x 3 x)(2x )dx proj g (f) g,f g 2 g 1 (2x 4 x 3 2x 2 +x)dx

Continued ( 2 5 1 4 21 3 + 1 2 [2 x5 5 x4 4 2x3 ) ] 1 3 + x2 2 ( 2 5 1 4 2 3 + 1 2 ) 8 15 So g,f 8 15

Continued Now compute g 2 1 1 ] 1 (2x) 2 dx (4x 2 4x+1)dx [4 x3 3 4x2 2 +x (4 13 ) [ 412 +1 4 ] 3 41 2 +() 14 3

Continued So, proj g (f) g,f g 2 g 8 15 14 3 (2x ) 4 (2x ) 35

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