Lecture 5. Inner Product


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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 dotproduct (or inner product) in R n. Definition. Let (x 1, x,..., x n ), (y 1, x,..., y n ) R n. Then, the dotproduct 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 nonzero vectors should hve nonzero 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 nonzero 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.
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