Chapter 5: Inner Product Spaces

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1 Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples of ier product spaces prove properties of ier product spaces defie the orm of a vector prove properties of orms I Chapter we itroduced the idea of ier products for a Euclidea space. I this chapter we exted the cocept of ier product to geeral vector spaces. his sectio is more difficult tha the equivalet sectio (B) i chapter because we caot visualise the vectors. A Defiitio of Ier Product How did we defie the ier product i Chapter? It was defied as follows: u v u Let u ad v v be vectors i the the ier product of u ad v u v deoted by u v is (.9) u v uv + uv + uv + + uv Remember the aswer was a scalar ot a vector. he ier product was amed the dot product (also called the scalar product) i. his is the usual (or stadard) ier product i but there are may other ier products i. For the geeral vector space the ier product u v is deoted by u, v rather tha u v. For the geeral vector space the defiitio of ier product is based o Propositio (.) of Chapter ad is give by: Defiitio (5.). A ier product o a real vector space V is a operatio which assigs to each pair of vectors, u ad v, a uique real umber u, v which satisfies the followig axioms for all vectors u, v ad w i V ad all scalars k. (i) u, v v, u [Commutative Law] (ii) u+ v, w u, w + v, w [Distributive Law] (iii) kuv, k u, v (iv) u, u ad we have u, u if ad oly if u O A real vector space which satisfies these axioms is called a real ier product space. Note that evaluatig, gives a real umber (scalar) ot a vector. Next we give some examples of ier product spaces.

2 Chapter 5: Ier Product Spaces A Examples of Ier Product Example Show that the Euclidea space,, with the dot product as defied i (.9) above is ideed a ier product space. Solutio See Propositio (.) of Chapter. Example Let V be the Euclidea space,, ad A. 5 Show that u, w u Aw is a ier product for. Solutio. What are u ad v equal to? hey are vectors i which meas they ca be writte as u w u ad w u w What is u equal to? u u is the vector u trasposed, that is u ( u u). How do we show u u, w u Aw is a ier product? By checkig all 4 axioms of Defiitio (5.) give above. Check (i): w u, w u Aw ( u u) 5 w w+ w u u Matrix Multiplicatio w+ 5w u( w+ w) + u( w+ 5w) uw + uw + uw + 5uw [ Expadig] Goig the other way we have u w, u w Au ( w w) 5 u u+ u ( w w) [ Matrix Multiplicatio] u+ 5u w( u+ u) + w( u+ 5u) uw + uw + wu + 5wu [ Expadig] Comparig these two, u, w u Aw uw + uw + uw + 5 uw ad w, u w Au uw + uw + wu + 5 wu, we have [ ]

3 Chapter 5: Ier Product Spaces u, w w, u Hece part (i) of defiitio (5.) is satisfied. Check (ii): v Let v the we have v u+ v, w u+ v Aw u+ v w u + v 5 w w ( u+ v u + v) 5 w w+ w ( u+ v u + v) w+ 5w ( u+ v)( w+ w) + ( u + v)( w+ 5w) uw + uw + vw + vw + uw + 5uw + vw + 5vw Similarly by usig the results of part (i) above we have u, w + v, w uw + uw + uw + 5uw + vw + vw + vw + 5vw u, w v, w [ ] u+ v, w By Above Part (ii) of defiitio (5.) is satisfied. Check (iii): Need to check that ku, w k u, w ku, w ku Aw k u Aw Because k is a scalar therefore k k kuaw k u, w We are give u Aw u, w Hece part (iii) is satisfied. Check (iv): Need to show u, u ad we have u, u if ad oly if u O: u u, u u Au ( u u) 5 u u+ u ( u u) u+ 5u u u + u + u u + 5u We ca rewrite the last lie as ( ) ( ) ( u) uu uu 5( u) ( u ) 4uu 5( u )

4 Chapter 5: Ier Product Spaces 4 Also ( u ) uu ( u ) u, u ( u u ) + u + uu + u + u ( u u ) ( u ) [ Because we have square umbers] + + u, u u + u + u u u his meas that u O. Hece part (iv) is fulfilled therefore all 4 axioms are satisfied, so we coclude that u, w u Aw is a ier product for. Example Let be the quadratic polyomials. Let P be polyomials i P p c + c x+ c x ad q d + d x+ d x. Show that p, q cd + cd + cd defies a ier product o P. [For example if p + x + 5 x ad q 4 x+ 7x the p, q ] Solutio. How do we show p, q cd + cd + cd is a ier product o P? Checkig all 4 axioms of Defiitio (5.). (i) Need to check p, q q, p is true for p, q cd + cd + cd : p, q cd + cd + cd dc dc dc Remember order of multiplicatio does ot matter + + q, p (ii) Need to check p+ q, r p, r + q, r : Let r e + e x+ e x We have ad ( c d ) ( c d ) x ( c d ) x p+ q p+ q, r c + d + c + d x+ c + d x, e + e x+ e x c + d e + c + d e + c + d e ce + de + ce + de + ce + de Evaluatig the other ier product p, r + q, r ce + ce + ce + de + de + de p, r q, r ce + de + ce + de + ce + de p+ q, r F Hece part (ii) is satisfied. [ Rearragig] [ rom Above]

5 Chapter 5: Ier Product Spaces 5 (iii) Need to check kp, q k p, q : kp, q k c + c x+ c x, d + d x+ d x, [ Opeig Brackets] [ Usig the Defiitio of I.P. ] [ Factorizig] kc + kc x+ kc x d + d x+ d x kc d + kc d + kc d k c d + cd + c d k p, q Hece part (iii) is satisfied. (iv) Need to show p, p ad we have p, p if ad oly if p O: p, p c + cx+ c x, c + cx+ cx cc + cc + cc c + c + c Also p, p ( c) + ( c) + ( c) c c c If c c c the p O. All 4 axioms are satisfied, therefore p, q cd + cd + cd is a ier product o P. I the above example if w e defie p, q ( c + d ) + ( c+ d) + ( c + d) the p, q is ot a ier product. Why ot? Because axiom (iv) of Defiitio (5.) fails, tha t is p, p is false. For example if p x x the p, p x x, x x + + < Hece p, p [Not greater tha or equal to zero]. Next we state a ier product o the vector space of matrices, M. A example of a ier product o the vector space of matrices is the followig: he trace of a matrix is the sum of its leadig diagoal elemets, that is a b trace a d c d + Let M be the vector space of by matrices ad ier product o M be defied by A, B tr B A where tr is the trace. his is a ier product o M. here is a questio o this ier product i Exercise 5a. If we defie A, B AB o M the this is ot a ier product. Why ot? Because we do ot have the commutative law, that is A, B AB BA B, A [Not Equal]

6 Chapter 5: Ier Product Spaces 6 here are may other examples of ier product spaces which you eed to show i Exercise 5a. Next we move oto properties of ier products. A Properties of Ier Products Propositio (5.). Let u, v ad w be vectors i a real ier product space V ad k be ay scalar. We have the followig: (i) u, O O, v (ii) u, kv k u, v (iii) u, v+ w u, v + u, w Proof of (i). We ca write the zero vector as O because O. Usig the axioms of defiitio (5.) we have u O u O,, ( O) O, u By Part (i) of (5.) which is u, v v, u O, u By Part (iii) of (5.) which is ku, v k u, v Similarly O, v. Proof of (ii). Sice ier product is commutative, u, v v, u, by axiom (i) of defiitio (5.) therefore proof of part (ii) is straightforward. u, kv kv, u By Part (i) of ( 5.) which is u, v v, u k v, u By Part (iii) of (5.) which is ku, v k u, v k u, v By Part (i) of (5.) which is u, v v, u Proof of (iii). We have u, v+ w v+ w, u By Part (i) of (5.) which is u, v v, u By Part (ii) of (5.) which is v, u + w, u v+ w, u v, u + w, u u, v + u, w By Part (i) of (5.) which is u, v v, u A4 Norm or Legth of a Vector Do you remember how the Euclidea orm was defied? he orm of a vector u i was defied by uu. he orm is defied i a similar maer for the geeral vector space V. Let u be a vector i V the the orm deoted by u is defied as (5.) u u, u [Positive Root] Note that for the geeral vector space we caot use the dot product because that is

7 Chapter 5: Ier Product Spaces 7 oly defied for Euclidea space,, ad i this chapter we are examiig ier products o geeral vector spaces. he orm of a vector u is a real umber which gives the size of the vector u. Geerally to fid the orm u it is easier to determie u u, u ad the take the square root of your result. As before a vector with orm is called a uit vector. I the example below we apply the orm of a vector to vector spaces of cotiuous fuctios. Example,,. Let C [ ] be the vector space of cotiuous fuctios o the closed iterval [ ] Let f ( x ) x ad g( x) x be fuctios i C [, ]. Let the ier product o C [, ] be give by f, g f x g x dx Determie (i) f, g (ii) f (iii) g (iv) f g Solutio. (i) Usig f, g f ( x) g( x) dx with f ( x) x g( x) x (ii) f, g f x g x dx ad we have x( x ) dx f ( x) x g( x) x 4 Substitutig ad x x ( x x) dx As stated above to fid f it is easier to determie the sq uare root of your result: f f, f What is f equal to? f x f x dx f f, f ad the take xxdx Substitutig f x x x dx Because xx x x

8 Chapter 5: Ier Product Spaces 8 Square root of f which is (iii) Similarly we have g g, g f. x x dx g x x 4 ( x x ) dx Expadig ( x )( x ) g x g x dx Substitutig + 5 x x + x [ Itegratig] What is g equal to? We eed to take the square root of 8 5 to fid g. Hece 8 g. 5 (iv) Similarly to fid f g we first determie f g ad the take the square root of our aswer: f g f g, f g Hece 7 f g. f x g x f x g x dx [ ] x x x x dx Substitutig ( x x x x 4 x x x x ) dx [ Expadig] 4 ( x x x x ) dx [ Simplifyig] x x 5 4 x x x x + + x Itegratig x x + x he distace betwee two vectors u ad v is deoted as d ( u, ) (5.4) d ( u, v) u v [ ] v ad is defied as I Example 4 above the distace betwee f ad g was give by f g 7.

9 Chapter 5: Ier Product Spaces 9 Note that this does ot mea there is a distace of 7 betwee the two graphs of f ad g. It is the result of applyig the defiitio of a ier product give above. Example 5 Determie orm p of the vector i Solutio. P where the ier product is give by p, q cd + cd + cd By defiitio (5.) above we have p p, p but first we fid the take the square root: p, p cc + cc+ cc c + c + c herefore p p, p ( c ) + ( c ) + ( c ). p p, p ad Some orms are ot defied i terms of the ier product as the followig example shows: x Cosider the Euclidea space ad let x be a vector i the the x followig are importat orms i this space: x x + x + x + + x [his is called the oe orm] x x + x + x + + x [his is called the two orm] ( x j ) Note that ( x j ) x max where j,,, [his is called the ifiity orm] x max where j,,, meas we select the maximum absolute value out of x, x, x, ad x. 4 For example if we have the vector 4 x i the 9 7 x max (, 4, 9, 7 ) max (, 4, 9, 7) 9 x dp x A5 Properties of the Norm of a Vector Next we state certai properties of the orm of a vector. Propositio (5.5). Let V be a ier product space ad u ad v be vectors i V. If k is ay scalar the we have followig properties: (i) u [ No-egative] (ii) u u O (iii) ku k u

10 Chapter 5: Ier Product Spaces [Note that for a scalar k we have Proof of part (i). k k where k is the modulus of k]. By the above defiitio (5.) we have u u, u where the square root is the positive root, that is u u, u Proof of part (ii). Agai from defiitio (5.) we have ad Proof of part (iii). Applyig the defiitio (5.) we first fid u u, u uo u O u O, O ku ku, ku k u, ku ku ad the we take the square root: kk u, u k u, u akig the square root gives ku k u, u k u, u k u his is our required result. SUMMARY (5.). A ier product is a fuctio o a vector space which satisfies the followig: (i) u, v v, u [Commutative Law] (ii) u+ v, w u, w + v, w [Distributive Law] (iii) kuv, k u, v (iv) u, u ad we have u, u if ad oly if u O (5.)he ier product has the followig properties: (i) u, O O, v (ii) u, kv k u, v (iii) u, v+ w u, v + u, w he orm of a vector deoted u ad is defied as (5.) u u, u Propositio (5.5). Let V be a ier product space the we have the followig: (i) u [ No-egative] (ii) u u O (iii) ku k u

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