MATH 304Linear AlgebraLecture 26:Norms and inner products.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 (positivity)(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|).Positivity and homogeneity are obvious.The triangle inequality:|xi+ yi| ≤ |xi| + |yi| ≤ maxj|xj| + maxj|yj|=⇒ maxj|xj+ yj| ≤ maxj|xj| + maxj|yj|• kxk1= |x1| + |x2| + · · · + |xn|.Positivity and homogeneity are obvious.The triangle inequality: |xi+ yi| ≤ |xi| + |yi|=⇒Pj|xj+ yj| ≤Pj|xj| +Pj|yj|Examples. V = Rn, x = (x1, x2, . . . , xn) ∈ Rn.• kxkp=|x1|p+ |x2|p+ · · · + |xn|p1/p, p > 0.Remark. kxk2= Euclidean length of x.Theorem kxkpis a norm on Rnfor any p ≥ 1.Positivity and homogeneity are still obvious (andhold for any p > 0). The triangle inequality forp ≥ 1 is known as the Minkowski inequality:|x1+ y1|p+ |x2+ y2|p+ · · · + |xn+ yn|p1/p≤≤|x1|p+ · · · + |xn|p1/p+|y1|p+ · · · + |yn|p1/p.Normed vector spaceDefinition. A normed vector space is a vectorspace endowed with a norm.The norm defin es a distance function on the normedvector space: dist(x, y) = kx − yk.Then we say that a sequence x1, x2, . . . convergesto a vector x if dist(x, xn) → 0 as n → ∞.Also, we say that a vector x is a goodapproximation of a vector x0if dist(x, x0) is small.Unit circle: kxk = 1kxk = (x21+ x22)1/2blackkxk =12x21+ x221/2greenkxk = |x1| + |x2| bluekxk = max(|x1|, |x2|) redExamples. V = C [a, b], f : [a, b] → R.• kf k∞= maxa≤x≤b|f (x)|.• kf k1=Zba|f (x)| dx.• kf kp=Zba|f (x)|pdx1/p, p > 0.Theorem kf kpis a norm on C [a, b] for any p ≥ 1.Inner productThe notion of inner product generalizes the notionof 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,is called 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.• hx, yi = (Dx) · (Dy),where D is an invertible n×n matrix.Remarks. (a) Invertibility of D is necessary to showthat hx, xi = 0 =⇒ x = 0.(b) The second example is a particular case of thethird one when D = diag(d1/21, d1/22, . . . , d1/2n).Counterexamples. V = R2.• hx, yi = x1y1− x2y2.Let v = (1, 2), then hv, vi = 12− 22= −3.hx, yi is symmetric and bilinear, but not positive.• hx, yi = 2x1y1+ x1x2+ 2x2y2+ y1y2.v = (1, 1), w = (1, 0) =⇒ hv, wi = 3, h2v, wi = 8.hx, yi is positive and symmetric, but not bilinear.• hx, yi = x1y1+ x1y2− x2y1+ x2y2.v = (1, 1), w = (1, 0) =⇒ hv, wi = 0, hw, vi = 2.hx, yi is positive and bilinear, but not symmetric.Problem. Find an inner product on R2such thathe1, e1i = 2, he2, e2i = 3, and he1, e2i = −1,where e1= (1, 0), e2= (0, 1).Let x = (x1, x2), y = (y1, y2) ∈ R2.Then x = x1e1+ x2e2, y = y1e1+ y2e2.Using bilinearity, we obtainhx, yi = hx1e1+ x2e2, y1e1+ y2e2i= x1he1, y1e1+ y2e2i + x2he2, y1e1+ y2e2i= x1y1he1, e1i + x1y2he1, e2i + x2y1he2, e1i + x2y2he2, e2i= 2x1y1− x1y2− x2y1+ 3x2y2.It remains to check that hx, xi > 0 for x 6= 0.hx, xi = 2x21− 2x1x2+ 3x22= (x1− x2)2+ x21+ 2x22.Example. V = Mm,n(R), space of m×n matrices.• hA, Bi = trace (ABT).If A = (aij) and B = (bij), then hA, Bi =mPi=1nPj=1aijbij.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, piecewise continuous, andw > 0 everywhere on [a, b].w is called the weight func
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