Lecture 5. Inner Product

Transcription

1 Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right ngle. Then, the point of intersection is the point on the line closest to P. Let us now tke plne L R 3 nd point outside the plne. How cn we find the point u L closest to P? The nswer is the sme s before, go from P so tht you meet the plne in right ngle. Observtion In ech of the bove exmples we needed two things: A1 We hve to be ble to sy wht the length of vector is. B1 Sy wht right ngle is. Both of these things cn be done by using the dot-product (or inner product) in R n. Definition. Let (x 1, x,..., x n ), (y 1, x,..., y n ) R n. Then, the dot-product of these vectors is given by the number: ((x 1, x,..., x n ), (y 1, x,..., y n )) x 1 y 1 + x y x n y n. 7

2 8 LECTURE 5. INNER PRODUCT The norm (or length) of the vector u (x 1, x,..., x n ) R n is the nonnegtive number: u (u, u) x 1 + x x n. Exmples Exmple. () ((1,, 3), (1, 1, 1)) (b) ((1,, 1), (, 1, 3)) Perpendiculr Becuse, x 1 y 1 + x y x n y n x 1 + x x n y1 + y y n or u v we hve tht (for u, v ) 1 u v 1. Hence we cn define: cos( ) u v. In prticulr, u v (u is perpendiculr to v) if nd only if. Questions Exmple. Let L be the line in R given by y x. Thus, L {r(1, ) : r R}. Let P (, 1). Consider the following questions.

3 9 Question 1: Wht is the point on L closest to P? Answer: Becuse u L, we cn write u (r, r). Furthermore, v u ( r, 1 r) is perpendiculr to L. Hence, ((1, ), ( r, 1 r)) r + 4r 4 5r. Hence, r 4 5 nd v ( 4 5, 8 5 ). Question : Wht is the distnce of P from the line? Answer: The length of the vector v u, i.e. v u. First we hve to find out wht v u is. We hve done lmost ll the work: v u (, 1) ( 4 5, 8 5 ) (6 5, 3 5 ). The distnce therefore is: Properties of the Inner Product 1. (positivity)to be ble to define the norm, we used tht (u, u).. (zero length)all non-zero vectors should hve non-zero length. Thus, (u, u) only if u. 3. (linerity)if the vector v R n is fixed, then mp u from R n to R is liner. Tht is, (ru + sw, v) r + s(w, v). 4. (symmetry) For ll u, v R n we hve: (v, u). We will use the properties bove to define n inner product on rbitrry vector spces. Definition Let V be vector spce. An inner product on V is mp (.,.) : V V R stisfying the following properties: 1. (positivity) (u, u), for ll v V.. (zero length) (u, u) only if u.

4 3 LECTURE 5. INNER PRODUCT 3. (linerity)if v V is fixed, then mp u from V to R is liner. 4. (symmetry) (v, u), for ll u, v V. Definition. We sy tht u nd v re perpendiculr if. Definition. If (.,.) is n inner product on the vector spce V, then the norm of vector v V is given by: u (u, u). Properties of the Norm Lemm. The norm stisfies the following properties: 1. u, nd u only if u.. ru r u. Proof. We hve tht ru (ru, rv) (r r r u. Exmples Exmple. Let < b, I [, b], nd V P C([, b]). Define: (f, g) Then, (.,.) is n inner product on V. Proof. Let r, s R, f, g, h V. Then: b f(t)g(t) dt 1. (f, f) b f(t) dt. As f(t), it follows tht b f(t) dt.. If (f, f), then f(t) for ll t, i.e f.

5 31 3. b (rfsg)(t)h(t) dt b rf(t)h(t) + sg(t)h(t) dt r b f(t)h(t) dt + s r(f, h) + s(g, h). b g(t)h(t) dt Hence, liner in the first fctor. 4. As f(t)g(t) g(t)f(t), it follows tht (f, g) (g, f). Notice tht the norm is: b f f(t) dt. Exmple. Let, b 1 in the previous exmple. g(t) t 3t. Then: Tht is, f(t) t nd (f, g) t (t 3t ) dt t 3 3t 4 ) dt Also, the norms re: f t 4 dt 1. 5 g (t 3t ) dt t 6t 3 + 9t 4 dt

6 3 LECTURE 5. INNER PRODUCT Exmple. Let f(t) cos πt nd g(t) sin πt. Then: (f, g) 1 4π cos πt sin πt dt [ (sin πt) ] 1. So, cos πt is perpendiculr to sin πt on the intervl [, 1]. Exmple. Let f(t) χ [,1/) χ [1/,1) nd g(t) χ [,1). Then: (f, g) / (χ [,1/) (t) χ [1/,1) (t))(χ [,1) ) dt χ [,1/) (t) dt One cn lso esily show tht f g 1. Problem χ [1/,1) (t) dt dt dt 1 1/ 1. Problem: Find polynomil f(t) + bt tht is perpendiculr to the polynomil g(t) 1 t. Answer: We re looking for numbers nd b such tht: (f, g) ( + bt)(1 t) dt + bt t bt dt + b b 3 + b 6. Thus, 3 + b. So, we cn tke f(t) 1 3t. Importnt Fcts We stte now two importnt fcts bout the inner product on vector spce V. Recll tht in R we hve: cos(θ) u v.

7 33 where u, v re two non-zero vectors in R nd θ is the ngle between u nd v. In prticulr, becuse 1 cos θ 1, we must hve: u v. We will show now tht this comes from the positivity nd linerity of the inner product. Theorem. Let V be vector spce with inner product (.,.). Then: for ll u, v V. u v Proof. We cn ssume tht u, v becuse otherwise both the LHS nd the RHS will be zero. By the positivity of the inner product we get: Thus, (v, u) (v, u) (v u, v u u u) (positivity) (v, u) (v, u) (v, u) (v, v) (v, u) + 4 (u, u) u u u (linerity) v (v, u) + u u v u. (symmetry) u v or u v. Note Notice tht: (v u, v u u u) only if v u u i.e. v u u. Thus, v nd u hve to be on the sme line through.

8 34 LECTURE 5. INNER PRODUCT A Lemm We cn therefore conclude: Lemm. u v if nd only if u nd v re on the sme line through. Theorem The following sttement is generliztion of Pythgors Theorem. Theorem. Let V be vector spce with inner product (.,.). Then: u + v u v for ll u, v V. Furthermore, u + v u + v if nd only if. Proof. If, then (*) reds: u + v (u + v, u + v) (u, u) + + (v, v) ( ) u + u v + v ( u + v ). u + v u + v On the other hnd, if u + v u + v, we see from (*) tht. Exmple. Let u (1,, 1), v (,, 4). Then: 4 4 nd u , v Also, u + v (1, 4, 3) nd finlly: u + v u + v.