MATH 304Linear AlgebraLecture 28:Inner product spaces.NormThe notion of norm generalizes the notion of lengthof a vector in Rn.Definition. Let V be a vector space. A functionα : V → R is called a norm on V if it has thefollowing properties:(i) α(x) ≥ 0, α(x) = 0 only for x = 0 (positivi ty)(ii) α(rx) = |r|α(x) for all r ∈ R (homogeneity)(iii) α(x + y) ≤ α(x) + α(y) (triangle inequality)Notation. The norm of a vector x ∈ V is usuallydenoted kxk. Different norms on V aredistinguished by subscripts, e.g., kxk1and kxk2.Examples. V = Rn, x = (x1, x2, . . . , xn) ∈ Rn.• kxk∞= max(|x1|, |x2|, . . . , |xn|).• kxkp=|x1|p+ |x2|p+ ··· + |xn|p1/p, p ≥ 1.In particular, kxk2= |x|.Examples. V = C [a, b], f : [a, b] → R.• kf k∞= maxa≤x≤b|f (x)|.• kf kp=Zba|f (x)|pdx1/p, p ≥ 1 .Inner productThe notion of inner product generalizes the noti onof dot product of vectors in Rn.Definition. Let V be a vector space. A functionβ : V ×V → R, usually denoted β(x, y) = hx, yi, iscalled an inner product on V if it is positive,symmetric, and bilinear. That is, if(i) hx, xi ≥ 0, hx, xi = 0 only for x = 0 (positivity)(ii) hx, yi = hy, xi (symmetry)(iii) hrx, yi = r hx, yi (homogeneity)(iv) hx + y, zi = hx, zi + hy, zi (distributive law)An inner product space is a vector space endowedwith an inner product.Examples. V = Rn.• hx, yi = x · y = x1y1+ x2y2+ ··· + xnyn.• hx, yi = d1x1y1+ d2x2y2+ ··· + dnxnyn,where d1, d2, . . . , dn> 0.Example. V = Pn, polynom ials of degree < n.• hp, qi = p(x1)q(x1) + p(x2)q(x2) + ···+ p(xn)q(xn),where x1, x2, . . . , xnare di stinct points on R.We have hp, pi = 0 =⇒ p = 0 since a nonzeropolynomi al of degree less than n cannot have nroots.Examples. V = C [a, b].• hf , gi =Zbaf (x)g(x) dx.• hf , gi =Zbaf (x)g (x)w (x) dx,where w is bounded, piecewis e continuous, andw > 0 everywhere on [a, b].w is call ed the weight function.Theorem Suppose hx, yi is an inner product on avector space V . Thenhx, yi2≤ hx, xihy, yi for all x, y ∈ V .Proof: For any t ∈ R let vt= x + ty. Thenhvt, vti = hx, xi + 2thx, yi+ t2hy, yi.The right-hand side is a quadratic polynomial in t(provided that y 6= 0). Since hvt, vti ≥ 0 for all t,the discrim inant D is no npositive. ButD = 4hx, yi2− 4hx, xihy, yi.Cauchy-Schwarz Inequality:|hx, yi| ≤phx, xiphy, yi.Cauchy-Schwarz Inequality:|hx, yi| ≤phx, xiphy, yi.Corollary 1 |x · y| ≤ |x||y| for all x, y ∈ Rn.Equivalently, for all xi, yi∈ R,(x1y1+ ··· + xnyn)2≤ (x21+ ··· + x2n)(y21+ ··· + y2n).Corollary 2 For any f , g ∈ C [ a, b],Zbaf (x)g (x) dx2≤Zba|f (x)|2dx ·Zba|g( x)|2dx.Norms induced by inner productsTheorem Suppose hx, yi is an inner product on avector space V . Then kxk =phx, xi is a norm.Proof: Positi vity is obvious. Homogeneity:krxk =phrx, rxi =pr2hx, xi = |r|phx, xi.Triangle inequality (follows fro m Cauchy-Schwarz’s):kx + yk2= hx + y, x + yi= hx, xi + hx, yi + hy, xi + hy, yi≤ hx, xi+ |hx, yi| + |hy, xi| + hy, yi≤ kxk2+ 2kxkkyk+ kyk2= (kxk + kyk)2.Examples. • The length of a vector in Rn,|x| =px21+ x22+ ··· + x2n,is the norm induced by the dot productx · y = x1y1+ x2y2+ ··· + xnyn.• The norm kf k2=Zba|f (x)|2dx1/2on thevector space C[a, b] is induced by the inner producthf , gi =Zbaf (x)g (x) dx.AngleSince |hx, yi| ≤ k xkkyk, we can defi ne the anglebetween no nz ero vectors in any vector space withan inner product (and induced norm):∠(x, y) = arccoshx, yikxkkyk.Then hx, yi = kxkkykcos ∠(x, y).In particular, vectors x and y are orthogonal(denoted x ⊥ y) if hx, yi = 0.Problem. Find the angle between functionsf1(x) = x and f2(x) = x2in the inner space C[0, 1]with the inner pr oducthf , gi =Z10f (x)g (x) dx.hf1, f2i =Z10x · x2dx =14,hf1, f1i =Z10x2dx =13, hf2, f2i =Z10(x2)2dx =15.cos ∠(f1, f2) =hf1, f2ikf1kkf2k=1/4p1/3p1/5=√154sin ∠(f1, f2) =p1 − cos2∠(f1, f2) = 1/4∠(f1, f2) = arcsin14xx + yyPythagorean Theorem:x ⊥ y =⇒ kx + yk2= kxk2+ kyk2Proof: kx + yk2= hx + y, x + yi= hx, xi + hx, yi + hy, xi + hy, yi= hx, xi + hy, yi = kxk2+ kyk2.xxx + yx − yyyParallelogram Identity:kx + yk2+ kx − yk2= 2kxk2+ 2kyk2Proof: kx+yk2= hx+y, x+yi = hx, xi + hx, yi+ hy, xi+ h y, yi.Similarly, kx−yk2= hx, xi − hx, yi−hy, xi + hy, yi.Then kx+yk2+ kx−yk2= 2hx, xi+ 2hy, yi = 2kxk2+ 2kyk2.Example. Norms on Rn, n ≥ 2:• kxk∞= max(|x1|, |x2|, . . . , |xn|),• kxkp=|x1|p+ |x2|p+ ··· + |xn|p1/p, p ≥ 1.Theorem The norms k xk∞and kxkp, p 6= 2 donot satisfy the Parallelogram Identity. Hence theyare not induced by any inner product on Rn.Proof: A counterexample to the Parallelogram Identity isprovided by vectors e1= (1, 0, 0, . . . , 0) and e2= (0, 1, 0, . . . , 0).ke1k∞= k e2k∞= 1, ke1kp= k e2kp= 1 for any p ≥ 1.ke1+ e2k∞= ke1− e2k∞= 1,ke1+ e2kp= k e1− e2kp= 21/pfor any p ≥ 1.Thus 2ke1k2∞+ 2ke2k2∞= 2 ke1k2p+ 2ke2k2p= 4.On the other hand, ke1+ e2k2∞+ ke1− e2k2∞= 2 6= 4 andke1+ e2k2p+ ke1− e2k2p= 2(21/p)2= 21+2/p6= 4 unless p =
View Full Document
Unlocking...