601-jra-p393-402 - INNER-PRODUCT SPACES, ORTHOGONAL BAsnS,...

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Unformatted text preview: INNER-PRODUCT SPACES, ORTHOGONAL BAsnS, AND PROJECTIONS (OPTIONAL) Up to now we have considered a vector space solely as an entity with an algebraic structure. We know, however, that R" possesses more than just an algebraic structure; in particular, we know that we can measure the size or length of a vector x in R" by the quantity ||x|| = Similarly, we can define the distance from x to y as ||x — y H. The ability to measure distances means that R” has a geometric structure, which supplements the algebraic Structure. The geometric structure can be employed to study problems of convergence, continuity, and the like. In this section we briefly describe how a suitable measure of distance might be imposed on a general vector space. Our‘development will be brief, and we will leave most of the details to the reader; but the ideas parallel those in Sections 3.6 and 3.8—3.9. Inner-Product Spaces To begin, we observe that the geometric structure for R" is based on the scalar product xTy. Essentially the scalar product is a real-valued function of two vector variables: Given x and y in R”, the scalar product produces a number xTy. Thus to derive a geometric structure for a vector space V, we should look for a generalization of the scalar-product function. A consideration of the properties of the scalar-product function leads to the definition of an inner-product function for a vector space. (With reference to Definition 7, which follows, we note that the expression uTv does not make sense in a general vector space V. Thus not only does the nomenclature change—scalar product becomes inner product—but also the notation changes as well, with (u, v) denoting the inner product of u and v.) , , DEFllzuuguZ ‘ l I EXAMPI I; 1 Solution An inner product on a real vector space V is a functiOn that assigns a real number, s ' ' (u, v), to eaCh‘ pair of vectors u and v in V, and that‘satisfies-ftheseproperties: 7 f 1’.» “(u, ’u‘iigoand (u, u) = 0 if and only ifu = 0;; ‘ ‘ ' 7 2. (u, v)‘1-.—-;l(v, u). ' ' ‘ " " i (auiv): a(u, v).‘_ 4. (u, ‘v”+ 'w) = (u,“v) + (u, w). The usual scalar product in R" is an inner product in the sense of Definition 7, where (x, y) = xTy. To illustrate the flexibility 'of Definition 7, we also note that there are other sorts of inner products for R". The following example gives another inner product for R2. Let V be the vector space R2, and let A be the (2 x 2) matrix 3 2 A = . 2 4 Verify that the function (u, v) = uTAv is an inner product for R2. Let u be a vector in R2: Then (11 u)—uTAu—[ ] 3 2 u] a — — ul9u2 2 4 “2 s so (u, u) = 3a? + 4u1u2 + 414% = 214% + (u1 + 2u2)2. Thus (u, u) z 0 and (u, u) = 0 if and only if u; = uz = 0. This shows that property 1 of Definition 7 is satisfied. To see that property 2 of Definition 7 holds, note that A is symmetric; thatis, AT = A. Also observe that if u and v are in R2, then uTAv is a (1 x 1) matrix, so (uTAv)T = uTAv. It now follows that (u, v) = uTAv = (uTAv)T = vTAT(uT)T = vTATu = (v, 11). Properties 3 and 4 of Definition 7 follow easily from the properties of matrix mul- tiplication, so (u, v) is an inner product for R2. fl In Example 1, an inner product 'for R2 was defined in terms of a matrix A: (u, v) = uTAv. In general, we might ask the following question: “For what (11 x n) matrices, A, does the operation uTAv define an inner product on R" ? ” IEXAMDLL- 2 Solution Figure 5.5 The value (p, p) is equal to the area under the graph of y = p(t)2. IEXAMPIL 3 Solution The answer to this question is suggested by the solution to Example 1. In particular (see Exercises 3 and 32), the operation (u, v) = uTAv is an inner product for R" if and only if A is a symmetric positive-definite matrix. There are a number of ways in which inner products can be defined on spaces of functions. For example, Exercise 6 will show that (p, q) = p(0)q(0) + p(l)q(l) + p(2)q(2) defines one inner product for P2. The following example gives yet another inner product for 732. For p(t) and q(t) in P2, verify that I (p, q) =fo p(t)q(t)dt is an inner product. To check property 1 of Definition 7, note that l (p.12) =/0 p(t)2dt. and p(t)2 2 O for 0 s t 5 1. Thus (p, p) is the area under the curve p(t)2, 0 5 t 5 1. Hence (p, p) z 0, and equality holds if and only if p(t) = 0,0 _<_ t _<_ 1 (see Fig. 5.5). Properties 2, 3, and 4 of Definition Tare straightforward to verify, and we include here only the verification of property 4. If p(t), q(t), and r(t) are in 132, then I l (p,q+r) = /0 p(t)[q(t)+r(t)]dt= f0 [p(t)q(t)+p(t)r(t)]dt l 1 =1) P(t)q(t)dt +f0 P(I)r(t)dt = (P, 4) + (P, r), as required by property 4. Ii After the key step of defining a vector-space analog of the scalar product, the rest is routine. For purposes of reference we call a vector space with an inner product an inner-product space. As in R", we can use the inner product as a measure of size: If V is an inner-product space, then for each v in V we define ||vll (the norm of v) as "VII = Note that (v, v) z 0 for all v in V, so the norm function is always defined. (v, v). Use the inner product for P2 defined in Example 2 to determine l|t2 II. By definition, ":2" = ./(t2,:2). 'But(12,t2) = fo'tztzdt = [0' m1: = 1/5. so Ilt2||=1/~/§- Ii Before continuing, we pause to illustrate one way in which the inner-product space framework is used in practice. One of the many inner products for the vector space C[0, l] is l (f, g) = [0 f(x)g(x)dx. l I ’ liteonw 'lO ‘ l I ‘lTHLIORLM 'l'l EXAMPLE il Solution If f is a relatively complicated function in C [0, l], we might wish to approximate f by a simpler function, say a polynomial. For definiteness suppose we want to find a polynomial p in P2 that is a good approximation to f. The phrase “good approximation” is too vague to be used in any calculation, but the inner-product space framework allows us to measure size and thus to pose some meaningful problems. In particular, we can ask for a polynomial p* in P2 such that llf -P*Il _<. Ilf - PII for all p in P2. Finding such a polynomial p* in this setting is equivalent to minimizing l ftfo)—pon%m 0 among all p in P2. We will present a procedure for doing this shortly. Orthogonal Bases If u and v are vectors in an inner-product space V, we say that u and v are orthogonal if (u, v) = 0. Similarly, B = {v1, v2, . . . , vp} is an orthogonal set in V if (v;, vj) = 0 when i yé j. In addition, if an orthogonal set of vectors B is a basis for V, we call B an orthogonal basis. The next two theorems correspond to their analogs in R", and we leave the proofs to the exercises. [See Eq. (5a), Eq. (5b), and Theorem 14 of Section 3.6.] Let B = {v., v2, . . . , v,,} be an orthogonal basis for an inner-product space V. If u is any vector in V, then v ,u v ,u v,,, u u=<. )W+(2 >h+nn+< >". 1 (VI a VI) (v2, v2) (vns Vn) Gram-Schmidt Orthogonalization Let V be an inner-product space, and let {u1,u2, . . . , u") be a basis for V. Letv, = u), and for2 5 k 5 n define vk by k-l (“In Vk = 11k — V'. 1;: (WAG) 1 Then {v. , V2, .. . , v,,} is an orthogonal basis for V. I! Let the inner product on P2 be the one given in Example 2. Starting with the natural basis {1, x, x2}, use Gram—Schmidt orthogonalization to obtain an orthogonal basis for P2. If we let {p0, pl, 192} denote the orthogonal basis, we have po(x) = l and find p.(x) from (P0. 36) (p03 p0) P10) =X - P006)- We calculate l l (posx)=f xdx=ll2andlpmpol=f dx=1; 0 0 so 191 (x) = x - 1/2. The next step of the Gram—Schmidt orthogonalization process is to form (p..x2> p (x) _ (pawl) (Pi, Pi) I (P0, P0) 19200 = x2 - po(x). The required constants are I (p.,x2) =f (x3 —x2/2)dx =1/12 0 I (p1,p|)=/ (xZ—x+l/4)dx=l/12 0 1 (120,23) = / xzdx =1/3 0 1 (170.170) =/0 dx :1. Therefore, p2(x) = x2 — p.(x) — po(x)/3 = x2 — x + 1/6, and {po, 1);, p2} is an orthogonal basis for 'P2 with respect to the inner product. d i ‘ I EXAMPLE 5 Let B = {p0, [11, p2} be the orthogonal basis for 1’; obtained in Example 4. Find the coordinates of x2 relative to B. ' Solution By Theorem 10, x2 = agp0(x) + alp. (x) + agp2(x), where do = (P0, x2)/(Po. P0) 01=(P1.x2)/(P1,P1) 02 = (P2, x2)/(P2, P2)- The necessary calculations are l (po,x2)=/ xzdx= 1/3 0 1 (121,262) =/ [x3 — (1/2)x2]dx =1/12 0 l (p2,x2) =/ [x4 —x3 +(l/6)x2]dx =1/180 0 l (P09P0) =f dx :1 0 . 1 (pl, p1) = [x2 — x + 1/4]dx =1/12 I (P2.p2)=/ [xZ—x+l/6]2dx=1/180. 0 Thus a0 = l/3,a. = I, and a2 = 1. We can easily check that x2 = (l/3)po(x)+ 121(x)+ mo). 1 | ‘ THEOREM 12 l I THEOREM 13 Orthogonal Projections We return now to the previously discussed problem of finding a polynomial p* in P2 that is the best approximation ofa function f in C [0, 1]. Note that the problem amounts to determining a vector p* in a subspace of an inner-product space, where p* is closer to f than any other vector in the subspace. The essential aspects of this problem can be stated formally as the following general problem: Let V be an inner—product space and let W be a subspace of V. Given a vector v in V, find a vector w* in W such that llv —— w*|| 5 llv — w|| for all win W. (1) A vector w* in W satisfying inequality (1) is called the projection of v onto W, or (frequently) the best least-squares approximation to v. Intuitively w* is the nearest vector in W to v. . The solution process for this problem is almost exactly the same as that for the least- squares problem in R”. One distinction in our general setting is that the subspace W might not be finite dimensional. If W is an infinite-dimensional subspace of V, then there may or may not be a projection of v onto W. If W is finite dimensional, then a projection always exists, is unique, and can be found explicitly. The next two theorems outline this concept, and again we leave the proofs to the reader since they parallel the proof of Theorem 18 of Section 3.9. Let V be an inner-product space, and let W be a subspace of V. Let v be a vector in V, and suppose w* is a vector in W such that (v — w“, w) = 0 for all w in W. Then ||v — w*|| 5 llv — wll for all w in W with equality holding only for w = w“. d Let V be an inner-product space, and let v be a vector in V. Let W be an n-dimensional subspace of V, and let {ul , uz, . . . , u,,} be an orthogonal basis for W. Then ||v — w*|[ 5 ||v — w|| for all w in W if and only if (vi "2) (I12, uz) w, _ (V. m) (v, u") _ u". (2) ("1, ui) (nu, un) d ur+ u2+---+ In view of Theorem 13, it follows that when W is a finite-dimensional subspace of an inner-product space V, we can always find projections by first finding an orthogonal basis for W (by using Theorem 11) and then calculating the projection w* from Eq. (2). To illustrate the process of finding a projection, we return to the inner-product space C [0, l] with the subspace 732. As a specific but rather unrealistic function, f, we choose f (x) = cosx, x in radians. The inner product is l (f, g) = [0 f(x)g(x)dx. 8 Chapter 5 , EXAMPLE 6 Solution | EXAMPLE 7 Vector Spaces and Linear 'Ii'ansformations In the vector space C [0, 1], let f (x) space 732. = cos x. Find the projection of f onto the sub- Let { p0, pl, p2} be the orthogonal basis for 732 found in Example 4. (Note that the inner product used in Example 4 coincides with the present inner product on C [0, 1]. By Theorem 13, the projection of f onto P2 is the polynomial [2* defined by (f! p0) (19o, p0) P*(x) = where P0(X) + (f! p2) ([72, p2) (flPi) (PhPl) P206). P10) + l (f, m) =f cos(x)dx 2 .841471 0 l (f. P1) =] (x — 1/2) cos(x)dx 2 .038962 0 l (f, P2) = f (x2 —x + 1/6) cos(x) dx 2 —.002394_ 0 From Example 5, we have (p0, p0) Therefore, p*(x) is given by = 1.1mm) = 1/12. and (192.192) = 1/180. P*(x) = (f1P0)P0(x)+ 12(faP11Pi(x) + 180(f1P2)P2(x) 2 .841471p0(x) — .467544pl(x) — .430920p2(x). In order to assess how well p*(x) approximates cosx in the interval [0, l], we can tabulate p*(x) and cosx at various values of x (see Table 5.1). H The function Si(x) (important in applications such as optics) is defined as follows: Si(x) =f smudu, forx #0. 0 (3) ll The integral in (3) is not an elementary one and so, for a given value of x, Si(x) must be evaluated using a numerical integration procedure. In this example, we approximate Table 5.1 x p*(x) cos x 0.0 1.0034 1.000 0.2 .9789 .9801 0.4 .9198 .921 1 0.6 .8263 .8253 0.8 .6983 .6967 1.0 .5359 .5403 p*(x) — cosx .0034 —.0012 —.9013 .0010 .0016 —.0044 5.6 Inner-Product Spaces, Orthogonal Bases, and Projections (Optional) 399 Si(x) by a cubic polynomial for 0 5 x 5 1. In particular, it can be shown that if we define Si(0) = 0, then Si(x) is continuous for all x. Thus we can ask: "What is the projection of Si(x) onto the subspace 733 of C [0, 1]?” This projection will serve as an approximation to Si(x) for 0 5 x 5 1. Solution We used the computer algebra system Derive to carry out the calculations. Some of the steps are shown in Fig. 5.6. To begin, let {p0, pl, [)2, p3} be the orthogonal basis for 733 found by the Gram—Schmidt process. From Example 4, we already know that 9 3 10 P1 (x) 3? P2 (X) (1:) P3 (x) dx 2800 1 J'x I SIN (u) 49. o 180 P2 (x) o ———:I——— du dx 50: -0.0804033 1 x I 51: J 2800 93 (x) J- im du dx 0 O u 52: -0.0510442 Figure 5.6 Some of the steps used by Derive to generate the projection of Si(x) onto 733 in Example 7 U Lll'dplel' D chtUl‘ opacca ullu ulllcal llanauu umuuua po(x) = 1, p1(x) = x — 1/2, and p2(x) = x2 — x + 1/6. To find p3, we first calculate the inner products (p09 x3): (p19 x3), (p2! x3) (see steps 6—9 in Fig. 5.6 for (p1, x3) and (p2, x3)). Using Theorem 11, we find p3 and, for later use, (p3, p3): p3(x) = x3 — (3/2)):2 + (3/5)): — 1/20 (£73,173) = 1/2800 (see steps 15—18 in Fig. 5.6). Finally, by Theorem 13, the projection of Si(x) onto 733 is the polynomial p* defined by (Si. Po) (Si. PI) (Si. 172) (Si. p3) (120.190) (pl. pt) (172.172) (1)3. 123) = (Si. polpo(x) + 12(Si. P1)Pl(x) + 180(Si, p2)p2(x) + 2800(Si. p3)p3(x). p*(x) = po(x) + pt(x) + p2(x) + p3(x) In the expression above for p*, the inner products (Si, pk) for k = 0, 1, 2, and 3 are given by l I x . (Si, pk): f0 pk(x)Si(x)dx= f0 pk(x){fo fldu} dx u (see steps 49—52 in Fig. 5.6 for 180(Si, p2) and 2800(Si, p3)). Now, since Si(x) must be estimated numerically, it follows that the inner products (Si, pk) must be estimated as well. Using Derive to approximate the inner products, we obtain the projection (or best least-squares approximation) p*(x) = .486385po(x) + .951172p|(x) — .0804033p2(x) —- .0510442p3(x). To assess how well p* (x) approximates Si(x) in [0, l], we tabulate each function at a few selected points (see Table 5.2). As can be seen from Table 5.2, it appears that p* (x) is a very good approximation to Si(x). Ii Table 5.2 x p*(x) Si(x) p*(x) — Si(x) 0.0 .000049 .000000 .000049 0.2 . 199578 . 199556 .000028 0.4 .396449 .396461 —.0000 12 0.6 .588113 .588128 —.000015 0.8 .7721 19 .772095 .000024 1.0 .946018 .946083 —.000065 1. .Prove that (p,q) = tie EXERCISES Prove that (x, y) = 4x, y. +xzy2 is an inner product on R2, where x = and y = . x2 Y2 . Prove that (x, y) = alxlyl +a2x2y2 +- - « +a,.x,,y,, is an inner product on R", where a1, a2, . . positive real numbers and where .,a,, are .,x,.]T and 9 Ynlr- x=[x|1x2! -' Y=[)’i,y2,.-- . A real (n x n) symmetric matrix A is called positive definite if xTAx > 0 for all x in R", x ¢ 0. Let A be a symmetric positive-definite matrix, and verify that (x, y) = x’Ay defines an inner product on R"; that is, verify that the four properties of Definition 7 are satisfied. . Prove that the following symmetric matrix A is pos- itive definite. Prove this by choosing an arbitrary vector x in R2, x 95 0, and calculating xTAx. A): ;] o1“ P21911006) = 00 +aix +a2x2 and q(x) =.bo + blx +b2x2. Prove that (p, q) = aobo +alb1+a2b2 is an inner product on ’P2. p(0)q(0) + p(l)q(1) + p(2)q (2) is an inner product on 792. . Let A = (00') and B = (by) he (2 x 2) matrices. ShOWIhat (A. B) = 0111711+012b12+021b21+022b22 is an inner product for the vector space of all (2 x 2) matrices. . Forx = [1, —-2]T andy = [0, HT in R2, find (x, y), ||x|l, Ily II, and llx — y|| using the inner product given in Exercise 1. . Repeat Exercise 8 with the inner product defined in Exercise 3 and the matrix A given in Exercise 4. . In 73’; let p(x) = -~l + 2x + x2 and q(x) = l — x + 2x2. Using the inner product given in Ex- ercise 5. find (p, q). llpll, llqll. and HP - qll. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. Repeat Exercise 10 using the inner product defined in Exercise 6. Show that {1, x, x2} is an orthogonal basis for ’P2 with the inner product defined in Exercise 5 but not with the inner product in Exercise 6. In R2 let S =: {x: llxll = 1}. Sketch a graph ofS if (x, y) = xTy. Now graph S using the inner product given in Exercise 1. Let A be the matrix given in Exercise 4, and for x, y in R2 define (x, y) = xTAy (see Exercise 3). Start- ing with the natural basis {e1, e2), use Theorem 11 to obtain an orthogonal basis {uh uzl for R2. Let {u1,u2) be the orthogonal basis for R2 obtained in Exercise 14 and let v = [3, 417. Use Theorem 10 to find scalars a1, a; such that v = arul + azuz. Use Theorem 11 to calculate an orthogonal basis {p0, pi, p;} for P2 with respect to the inner product in Exercise 6. Start with the natural basis {1, x, x2) for P2. Use Theorem 10 to write q(x) = 2 + 3x — 4.:2 in terms of the orthogonal basis {p0, pl, [12} obtained in Exercise 16. Show that the function defined in Exercise 6 is not an inner product for 733. [Hint Find p(x) in ’P3 such that (p, p) = 0, but p 7E 0.] Starting with the natural basis {1, x, x , x3, x4}, generate an orthogonal basis for ’P4 with respect to the inner product (p.q) = Z pom). 2 If V is an inner-product space, show that (v, 0) = 0 for each vector v in V. Let V be an inner-product space, and let u be a vec- tor in V such that (u, v) = O for every vector v in V. Show that u =' 0. Let a be a scalar and v a vector in an inner-product space V. Prove that ||av|| = |aH|v||. Prove that if {vl , v2, . . . , vk} is an orthogonal set of nonzero vectors in an inner-product space, then this set is linearly independent. . Prove Theorem 10. 402 25. 26. 27. 28. 29. Chapter 5 Vector Spaces and Linear Transformations Approximate x3 with a polynomial in 733. [Hints Use the inner product 1 (1319') = / ptt)q(r)a‘r, .o and let {p0, pl, 1);} be the orthogonal basis for P3 obtained in Example 4. Now apply Theorem 13.] In Examples 4 and 7 we found p0(,v), . . . ._ {)3(.\']. which are orthogonal with respect to l o: g) = f from) at. 0 Continue the process, and find p.1(x) so that {pmph ....p4] is an orthogonal basis for 7%. (Clearly there is an infinite sequence of polynomials p”. p.. ...,p,,, . .. that satisfy 1 f p;(.r)py(.r)d.r : O, isl: j. 0 These are called the Legendre polynomials.) With the orthogonal basis for P3 obtained in Ex— ample 7, use Theorem 13 to find the projection of f(,r) = cos x in P3. Construct a table similar to Table 5.! and note the improvement. An inner product on Cl—l, l] is 2 ' f(.r)s(x) _ (1.x. (1‘33) = — — 3T "1 ‘r 1 _ I2 Starting with the set {l,.r,.r2..r3....), use the Gram—Schmidt process to find polynomials Tn(.r). T. (.r), EU), and '1'3(.r) such that {1'}, Ty) = 0 when i # j. These polynomials are called the Chebyshev polynomials of (he first kind. [Hatr- Make a change of variables .r : cos 6.! A sequence of orthogonal polynomials usually sat- islics a three—term recurrence relation. For example, the Chebyshev polynomials are related by Ylt+l(x) : ZX'THL‘.) _ Til—l (-11)! n 2 1| 2; - - - . ttx’t where T0(.r) = l and EU) = .r. Verify that the polynomials defined by the relation (R) above are indeed orthogonal in Ciel, l] with respect to the inner product in Exercise 28. Verify this as follows: a) Make the change of variables x = cos (-9. and use induction to show that it}; (cos 0) : cos M}. k = 0. l. . . ., where Ill—(x) is defined by (R). Is) to) Using part a). show that U}. Ty) = 0 when t' y’: j. c) Use induction to show that 710:) is a polyno~ mial of degree k. k :0,1..... (1) Use (R) to calculate T3, T3. T4. and T5. . Let C[— l, l] have the inner product of Exercise 28. and let f be in C[—l, 1]. Use Theorem l3 to prove that llf — p*l| 5 ||f — p“ feral] pin “P” if W. I an n , A I...) (,1) = E —l— gay-Fifi). where of = (f. Ty). j : 0.1.....n. . The iterated trapezoid rule provides a good esti- mate of fab f(.r) (I); when f(.r) is periodic in [(1, b]. In particular, let N be a positive integer, and let it = (b—a)/N. Next, define .r; by .r; = a -l-t't't.r' : U. l. . . . . N. and suppose f(.r) is in C[a. b]. If we define rtt'f) by N—l 2" l are = gm.) + t: Z In.) + grit-M. i=1 then A(f) is the iterated trapezoid rule applied to f(.r). Using the result in Exercise 30, write a com- puter program that generates a good approximation to f(.r) in C[~—1, 1]. That is, for an input function f(.r} and a specified value of”. calculate estimates of (to. a]. . . . . on, where fit-:l_f.7itl1 Allin). To do this calculation, make the usual change of variables x = cos 6’ so that ') i"! at. = :/ ftcostfleos(kt9)d0. k : 0,1.....n. Ti 0 Use the iterated trapezoid rule to estimate each at. Test your program on ft'x') = (23" and note that (R) can be used to evaluate p"‘(_r) at any point x in [—l.l]. . Show that if xi is a real (it x a) matrix and if the expression (11. v) = ule defines an inner product on R”. then A must be symmetric and positive del- inite [sec Exercise 3 for the definition of positive definite). [Hints Consider (Cf, ey).] ...
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This note was uploaded on 11/08/2011 for the course MATH 601 taught by Professor Osu during the Fall '11 term at Cornell University (Engineering School).

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601-jra-p393-402 - INNER-PRODUCT SPACES, ORTHOGONAL BAsnS,...

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