4.11 Inner Product Spaces


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1 314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible Inner Product Spces We now extend the fmilir ide of dot product for geometric vectors to n rbitrry vector spce V. This enbles us to ssocite mgnitude with ech vector in V nd lso to define the ngle between two vectors in V. The mjor reson tht we wnt to do this is tht, s we will see in the next section, it enbles us to construct orthogonl bses in vector spce, nd the use of such bsis often simplifies the representtion of vectors. We begin with brief review of the dot product. Let x = (x 1,x 2,x 3 ) nd y = (y 1,y 2,y 3 ) be two rbitrry vectors in R 3, nd consider the corresponding geometric vectors x = x 1 i + x 2 j + x 3 k, y = y 1 i + y 2 j + y 3 k. The dot product of x nd y cn be defined in terms of the components of these vectors s An equivlent geometric definition of the dot product is x y = x 1 y 1 + x 2 y 2 + x 3 y 3. (4.11.1) (x 1, x 2, x 3 ) x x z (y 1, y 2, y 3 ) Figure : Defining the dot product in R 3. y y x y = x y cos θ, (4.11.2) where x, y denote the lengths of x nd y respectively, nd 0 θ π is the ngle between them. (See Figure ) Tking y = x in Equtions (4.11.1) nd (4.11.2) yields x 2 = x x = x x2 2 + x2 3, so tht the length of geometric vector is given in terms of the dot product by x = x x = x1 2 + x2 2 + x2 3. Furthermore, from Eqution (4.11.2), the ngle between ny two nonzero vectors x nd y is cos θ = x y x y, (4.11.3) which implies tht x nd y re orthogonl (perpendiculr) if nd only if x y = 0. In generl vector spce, we do not hve geometricl picture to guide us in defining the dot product, hence our definitions must be purely lgebric. We begin by considering the vector spce R n, since there is nturl wy to extend Eqution (4.11.1) in this cse. Before proceeding, we note tht from now on we will use the stndrd terms inner product nd norm in plce of dot product nd length, respectively.
2 4.11 Inner Product Spces 315 DEFINITION Let x = (x 1,x 2,...,x n ) nd y = (y 1,y 2,...,y n ) be vectors in R n. We define the stndrd inner product in R n, denoted x, y, by x, y =x 1 y 1 + x 2 y 2 + +x n y n. The norm of x is x = x, x = x x x2 n. Exmple If x = (1, 1, 0, 2, 4) nd y = (2, 1, 1, 3, 0) in R 5, then x, y =(1)(2) + ( 1)(1) + (0)(1) + (2)(3) + (4)(0) = 7, x = ( 1) = 22, y = = 15. Bsic Properties of the Stndrd Inner Product in R n In the cse of R n, the definition of the stndrd inner product ws nturl extension of the fmilir dot product in R 3. To generlize this definition further to n rbitrry vector spce, we isolte the most importnt properties of the stndrd inner product in R n nd use them s the defining criteri for generl notion of n inner product. Let us exmine the inner product in R n more closely. We view it s mpping tht ssocites with ny two vectors x = (x 1,x 2,...,x n ) nd y = (y 1,y 2,...,y n ) in R n the rel number x, y =x 1 y 1 + x 2 y 2 + +x n y n. This mpping hs the following properties: For ll x, y, nd z in R n nd ll rel numbers k, 1. x, x 0. Furthermore, x, x =0 if nd only if x = y, x = x, y. 3. kx, y =k x, y. 4. x + y, z = x, z + y, z. These properties re esily estblished using Definition For exmple, to prove property 1, we proceed s follows. From Definition , x, x =x x x2 n. Since this is sum of squres of rel numbers, it is necessrily nonnegtive. Further, x, x =0 if nd only if x 1 = x 2 = = x n = 0 tht is, if nd only if x = 0. Similrly, for property 2, we hve y, x =y 1 x 1 + y 2 x 2 + +y n x n = x 1 y 1 + x 2 y 2 + +x n y n = x, y. We leve the verifiction of properties 3 nd 4 for the reder.
3 316 CHAPTER 4 Vector Spces Definition of Rel Inner Product Spce We now use properties 1 4 s the bsic defining properties of n inner product in rel vector spce. DEFINITION Let V be rel vector spce. A mpping tht ssocites with ech pir of vectors u nd v in V rel number, denoted u, v, isclledninner product in V, provided it stisfies the following properties. For ll u, v, nd w in V, nd ll rel numbers k, 1. u, u 0. Furthermore, u, u =0 if nd only if u = v, u = u, v. 3. ku, v =k u, v. 4. u + v, w = u, w + v, w. The norm of u is defined in terms of n inner product by u = u, u. A rel vector spce together with n inner product defined in it is clled rel inner product spce. y y [f(x)] 2 Figure : f, f gives the re between the grph of y =[f(x)] 2 nd the xxis, lying over the intervl [,b]. b x Remrks 1. Observe tht u = u, u tkes welldefined nonnegtive rel vlue, since property 1 of n inner product gurntees tht the norm evlutes the squre root of nonnegtive rel number. 2. It follows from the discussion bove tht R n together with the inner product defined in Definition is n exmple of rel inner product spce. One of the fundmentl inner products rises in the vector spce C 0 [,b] of ll relvlued functions tht re continuous on the intervl [,b]. In this vector spce, we define the mpping f, g by f, g = f(x)g(x) dx, (4.11.4) for ll f nd g in C 0 [,b]. We estblish tht this mpping defines n inner product in C 0 [,b] by verifying properties 1 4 of Definition If f is in C 0 [,b], then f, f = [f(x)] 2 dx. Since the integrnd, [f(x)] 2, is nonnegtive continuous function, it follows tht f, f mesures the re between the grph y =[f(x)] 2 nd the xxis on the intervl [,b]. (See Figure ) Consequently, f, f 0. Furthermore, f, f =0 if nd only if there is zero re between the grph y =[f(x)] 2 nd the xxis tht is, if nd only if [f(x)] 2 = 0 for ll x in [,b].
4 4.11 Inner Product Spces 317 y f(x) 0 for ll x in [,b] Figure : f, f =0if nd only if f is the zero function. b x Hence, f, f =0 if nd only if f(x) = 0, for ll x in [,b], sof must be the zero function. (See Figure ) Consequently, property 1 of Definition is stisfied. Now let f, g, nd h be in C 0 [,b], nd let k be n rbitrry rel number. Then g, f = g(x)f(x) dx = Hence, property 2 of Definition is stisfied. For property 3, we hve kf, g = s needed. Finlly, (kf )(x)g(x) dx = kf (x)g(x) dx = k f(x)g(x) dx = f, g. f(x)g(x) dx = k f, g, f + g, h = = b (f + g)(x)h(x) dx = f (x)h(x) dx + [f(x)+ g(x)]h(x) dx g(x)h(x) dx = f,h + g,h, so tht property (4) of Definition is stisfied. We cn now conclude tht Eqution (4.11.4) does define n inner product in the vector spce C 0 [,b]. Exmple Use Eqution (4.11.4) to determine the inner product of the following functions in C 0 [0, 1]: f(x)= 8x, g(x) = x 2 1. Also find f nd g. Solution: From Eqution (4.11.4), Moreover, we hve f, g = 1 0 8x(x 2 1)dx = [ 2x 4 4x 2 ] 1 0 = 2. nd f = x 2 dx = g = (x 2 1) 2 dx = (x 4 2x 2 + 1)dx = We hve lredy seen tht the norm concept generlizes the length of geometric vector. Our next gol is to show how n inner product enbles us to define the ngle between two vectors in n bstrct vector spce. The key result is the CuchySchwrz inequlity estblished in the next theorem. Theorem (CuchySchwrz Inequlity) Let u nd v be rbitrry vectors in rel inner product spce V. Then u, v u v. (4.11.5)
5 318 CHAPTER 4 Vector Spces Proof Let k be n rbitrry rel number. For the vector u + kv, wehve But, using the properties of rel inner product, 0 u + kv 2 = u + kv, u + kv. (4.11.6) u + kv, u + kv = u, u + kv + kv, u + kv = u + kv, u + u + kv,kv = u, u + kv, u + u, kv + kv, kv = u, u +2 kv, u +k v, kv = u, u +2 kv, u +k kv, v = u, u +2 kv, u +k 2 v, v = u 2 + 2k v, u +k 2 v 2. Consequently, (4.11.6) implies tht v 2 k u, v k + u 2 0. (4.11.7) The lefthnd side of this inequlity defines the qudrtic expression The discriminnt of this qudrtic is P(k) = v 2 k u, v k + u 2. = 4( u, v ) 2 4 u 2 v 2. If >0, then P(k)hs two rel nd distinct roots. This would imply tht the grph of P crosses the kxis nd, therefore, P would ssume negtive vlues, contrry to (4.11.7). Consequently, we must hve 0. Tht is, 4( u, v ) 2 4 u 2 v 2 0, or equivlently, Hence, ( u, v ) 2 u 2 v 2. u, v u v. If u nd v re rbitrry vectors in rel inner product spce V, then u, v is rel number, nd so (4.11.5) cn be written in the equivlent form u v u, v u v. Consequently, provided tht u nd v re nonzero vectors, we hve 1 u, v u v 1. Thus, ech pir of nonzero vectors in rel inner product spce V determines unique ngle θ by cos θ = u, v, u v 0 θ π. (4.11.8)
6 4.11 Inner Product Spces 319 We cll θ the ngle between u nd v. In the cse when u nd v re geometric vectors, the formul (4.11.8) coincides with Eqution (4.11.3). Exmple Determine the ngle between the vectors u = (1, 1, 2, 3) nd v = ( 2, 1, 2, 2) in R 4. Solution: Using the stndrd inner product in R 4 yields u, v = 5, u = 15, v = 13, so tht the ngle between u nd v is given by cos θ = = , 0 θ π. Hence, ( ) 195 θ = rccos rdins Exmple Use the inner product (4.11.4) to determine the ngle between the functions f 1 (x) = sin 2x nd f 2 (x) = cos 2x on the intervl [ π, π]. Solution: f 1,f 2 = Using the inner product (4.11.4), we hve π π sin 2x cos 2x dx= 1 2 π Consequently, the ngle between the two functions stisfies π cos θ = 0, 0 θ π, sin 4x dx= 1 8 ( cos 4x) π π = 0. which implies tht θ = π/2. We sy tht the functions re orthogonl on the intervl [ π, π], reltive to the inner product (4.11.4). In the next section we will hve much more to sy bout orthogonlity of vectors. Complex Inner Products 9 The preceding discussion hs been concerned with rel vector spces. In order to generlize the definition of n inner product to complex vector spce, we first consider the cse of C n. By nlogy with Definition , one might think tht the nturl inner product in C n would be obtined by summing the products of corresponding components of vectors in C n in exctly the sme mnner s in the stndrd inner product for R n. However, one reson for introducing n inner product is to obtin concept of length of vector. In order for quntity to be considered resonble mesure of length, we would wnt it to be nonnegtive rel number tht vnishes if nd only if the vector itself is the zero vector (property 1 of rel inner product). But, if we pply the inner product in R n given in Definition to vectors in C n, then, since the components of vectors in C n re complex numbers, it follows tht the resulting norm of vector in 9 In the reminder of the text, the only complex inner product tht we will require is the stndrd inner product in C n, nd this is needed only in Section 5.10.
7 320 CHAPTER 4 Vector Spces C n would be complex number lso. Furthermore, pplying the R 2 inner product to, for exmple, the vector u = (1 i, 1 + i), we obtin u 2 = (1 i) 2 + (1 + i) 2 = 0, which mens tht nonzero vector would hve zero length. To rectify this sitution, we must define n inner product in C n more crefully. We tke dvntge of complex conjugtion to do this, s the definition shows. DEFINITION If u = (u 1,u 2,...,u n ) nd v = (v 1,v 2,...,v n ) re vectors in C n, we define the stndrd inner product in C n by 10 u, v =u 1 v 1 + u 2 v 2 + +u n v n. The norm of u is defined to be the rel number u = u, u = u u u n 2. The preceding inner product is mpping tht ssocites with the two vectors u = (u 1,u 2,...,u n ) nd v = (v 1,v 2,...,v n ) in C n the sclr u, v =u 1 v 1 + u 2 v 2 + +u n v n. In generl, u, v will be nonrel (i.e., it will hve nonzero imginry prt). The key point to notice is tht the norm of u is lwys rel number, even though the seprte components of u re complex numbers. Exmple If u = (1 + 2i, 2 3i) nd v = (2 i, 3 + 4i), find u, v nd u. Solution: Using Definition , u, v =(1 + 2i)(2 + i) + (2 3i)(3 4i) = 5i 6 17i = 6 12i, u = u, u = (1 + 2i)(1 2i) + (2 3i)(2 + 3i) = = 3 2. The stndrd inner product in C n stisfies properties (1), (3), nd (4), but not property (2). We now derive the pproprite generliztion of property (2) when using the stndrd inner product in C n. Let u = (u 1,u 2,...,u n ) nd v = (v 1,v 2,...,v n ) be vectors in C n. Then, from Definition , v, u =v 1 u 1 + v 2 u 2 + +v n u n = u 1 v 1 + u 2 v 2 + +u n v n = u, v. Thus, v, u = u, v. We now use the properties stisfied by the stndrd inner product in C n to define n inner product in n rbitrry (tht is, rel or complex) vector spce. 10 Recll tht if z = + ib, then z = ib nd z 2 = zz = ( + ib)( ib) = 2 + b 2.
8 4.11 Inner Product Spces 321 DEFINITION Let V be (rel or complex) vector spce. A mpping tht ssocites with ech pir of vectors u, v in V sclr, denoted u, v, is clled n inner product in V, provided it stisfies the following properties. For ll u, v nd w in V nd ll (rel or complex) sclrs k, 1. u, u 0. Furthermore, u, u =0 if nd only if u = v, u = u, v. 3. ku, v =k u, v. 4. u + v, w = u, w + v, w. The norm of u is defined in terms of the inner product by u = u, u. Remrk Notice tht the properties in the preceding definition reduce to those in Definition in the cse tht V is rel vector spce, since in such cse the complex conjugtes re unnecessry. Thus, this definition is consistent extension of Definition Exmple Use properties 2 nd 3 of Definition to prove tht in n inner product spce u, kv =k u, v for ll vectors u, v nd ll sclrs k. Solution: From properties 2 nd 3, we hve u,kv = kv, u =k v, u =k v, u =k u, v. Notice tht in the prticulr cse of rel vector spce, the foregoing result reduces to u,kv =k u, v, since in such cse the sclrs re rel numbers. Exercises for 4.11 Key Terms Inner product, Axioms of n inner product, Rel (complex) inner product spce, Norm, Angle, CuchySchwrz inequlity. Skills Know the four inner product spce xioms. Be ble to check whether or not proposed inner product on vector spce V stisfies the inner product spce xioms. Be ble to compute the inner product of two vectors in n inner product spce. Be ble to find the norm of vector in n inner product spce. Be ble to find the ngle between two vectors in n inner product spce. TrueFlse Review For Questions 1 7, decide if the given sttement is true or flse, nd give brief justifiction for your nswer. If true, you cn quote relevnt definition or theorem from the text. If flse, provide n exmple, illustrtion, or brief explntion of why the sttement is flse.
9 322 CHAPTER 4 Vector Spces 1. If v nd w re linerly independent vectors in n inner product spce V, then v, w =0. 2. In ny inner product spce V,wehve kv, kw =k v, w. 3. If v 1, w = v 2, w =0 in n inner product spce V, then c 1 v 1 + c 2 v 2, w =0. 4. In ny inner product spce V, x + y, x y < 0if nd only if x < y. 5. In ny vector spce V, there is t most one vlid inner product, tht cn be defined on V. 6. The ngle between the vectors v nd w in n inner product spce V is the sme s the ngle between the vectors 2v nd 2w. 7. If p(x) = x+ 2 x 2 nd q(x) = b 0 +b 1 x+b 2 x 2, then we cn define n inner product on P 2 vi p, q = 0 b 0. Problems 1. Use the stndrd inner product in R 4 to determine the ngle between the vectors v = (1, 3, 1, 4) nd w = ( 1, 1, 2, 1). 2. If f(x)= sin x nd g(x) = x on [0,π], use the function inner product defined in the text to determine the ngle between f nd g. 3. If v = (2+i, 3 2i, 4+i)nd w = ( 1+i, 1 3i, 3 i), use the stndrd inner product in C 3 to determine, v, w, v, nd w. 4. Let A = [ ] [ ] b11 b, B = b 21 b 22 be vectors in M 2 (R). Show tht the mpping A, B = 11 b b b b 22 (4.11.9) defines n inner product in M 2 (R). 5. Referring to A nd B in the previous problem, show tht the mpping A, B = 11 b b b b 11 does not define vlid inner product on M 2 (R). For Problems 6 7, use the inner product (4.11.9) to determine A, B, A, nd B. [ ] [ ] A =,B = A = [ ] 32,B = 2 4 [ ] Let p 1 (x) = + bx nd p 2 (x) = c + dx be vectors in P 1. Determine mpping p 1,p 2 tht defines n inner product on P 1. Consider the vector spce R 2. Define the mpping, by v, w =2v 1 w 1 + v 1 w 2 + v 2 w 1 + 2v 2 w 2 ( ) for ll vectors v = (v 1,v 2 ) nd w = (w 1,w 2 ) in R 2. This mpping is required for Problems Verify tht Eqution ( ) defines n inner product on R 2. For Problems 10 12, determine the inner product of the given vectors using () the inner product ( ), (b) the stndrd inner product in R v = (1, 0), w = ( 1, 2). 11. v = (2, 1), w = (3, 6). 12. v = (1, 2), w = (2, 1). 13. Consider the vector spce R 2. Define the mpping, by v, w =v 1 w 1 v 2 w 2, ( ) for ll vectors v = (v 1,v 2 ) nd w = (w 1,w 2 ). Verify tht ll of the properties in Definition except (1) re stisfied by ( ). The mpping ( ) is clled pseudoinner product in R 2 nd, when generlized to R 4, is of fundmentl importnce in Einstein s specil reltivity theory. 14. Using Eqution ( ), determine ll nonzero vectors stisfying v, v =0. Such vectors re clled null vectors. 15. Using Eqution ( ), determine ll vectors stisfying v, v < 0. Such vectors re clled timelike vectors.
10 4.12 Orthogonl Sets of Vectors nd the GrmSchmidt Process Using Eqution ( ), determine ll vectors stisfying v, v > 0. Such vectors re clled spcelike vectors. 17. MkesketchofR 2 nd indicte the position of the null, timelike, nd spcelike vectors. 18. Consider the vector spce R n, nd let v = (v 1,v 2,...,v n ) nd w = (w 1,w 2,...,w n ) be vectors in R n. Show tht the mpping, defined by v, w =k 1 v 1 w 1 + k 2 v 2 w 2 + +k n v n w n is vlid inner product on R n if nd only if the constnts k 1,k 2,...,k n re ll positive. 19. Prove from the inner product xioms tht, in ny inner product spce V, v, 0 =0 for ll v in V. 20. Let V be rel inner product spce. () Prove tht for ll v, w V, v + w 2 = v v, w + w 2. [Hint: v + w 2 = v + w, v + w.] (b) Two vectors v nd w in n inner product spce V re clled orthogonl if v, w =0. Use () to prove the generl Pythgoren theorem: If v nd w re orthogonl in n inner product spce V, then v + w 2 = v 2 + w 2. (c) Prove tht for ll v, w in V, (i) v + w 2 v w 2 = 4 v, w. (ii) v + w 2 + v w 2 = 2( v 2 + w 2 ). 21. Let V be complex inner product spce. Prove tht for ll v, w in V, v + w 2 = v 2 + 2Re( v, w ) + v 2, where Re denotes the rel prt of complex number Orthogonl Sets of Vectors nd the GrmSchmidt Process The discussion in the previous section hs shown how n inner product cn be used to define the ngle between two nonzero vectors. In prticulr, if the inner product of two nonzero vectors is zero, then the ngle between those two vectors is π/2 rdins, nd therefore it is nturl to cll such vectors orthogonl (perpendiculr). The following definition extends the ide of orthogonlity into n rbitrry inner product spce. DEFINITION Let V be n inner product spce. 1. Two vectors u nd v in V re sid to be orthogonl if u, v =0. 2. A set of nonzero vectors {v 1, v 2,...,v k } in V is clled n orthogonl set of vectors if v i, v j =0, whenever i j. (Tht is, every vector is orthogonl to every other vector in the set.) 3. A vector v in V is clled unit vector if v = An orthogonl set of unit vectors is clled n orthonorml set of vectors. Thus, {v 1, v 2,...,v k } in V is n orthonorml set if nd only if () v i, v j =0 whenever i j. (b) v i, v i =1 for ll i = 1, 2,...,k.
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