MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets. Norm The notion of norm generalizes the notion of length of a vector in Rn. Definition. Let V be a vector space. A function α : V → R is called a norm on V if it has the following 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 usually denoted kxk. Different norms on V are distinguished by subscripts, e.g., kxk1 and kxk2. n n Examples. V = R , x = (x1, x2,..., xn) ∈ R . • kxk∞ = max(|x1|, |x2|,..., |xn|). p p p 1/p • kxkp = |x1| + |x2| + ··· + |xn| , p ≥ 1. Examples. V = C[a, b], f : [a, b] → R. • kf k∞ = max |f (x)|. a≤x≤b b 1/p p • kf kp = |f (x)| dx , p ≥ 1. Za Normed vector space Definition. A normed vector space is a vector space endowed with a norm. The norm defines a distance function on the normed vector space: dist(x, y) = kx − yk. Then we say that a sequence x1, x2,... converges to a vector x if dist(x, xn) → 0 as n → ∞. Also, we say that a vector x is a good approximation of a vector x0 if dist(x, x0) is small. Inner product The notion of inner product generalizes the notion of dot product of vectors in Rn. Definition. Let V be a vector space. A function β : 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 = rhx, yi (homogeneity) (iv) hx + y, zi = hx, zi + hy, zi (distributive law) An inner product space is a vector space endowed with 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. Problem. Find an inner product on R2 such that he1, e1i = 2, he2, e2i = 3, and he1, e2i = −1, where e1 = (1, 0), e2 = (0, 1). 2 Let x = (x1, x2), y = (y1, y2) ∈ R . Then x = x1e1 + x2e2, y = y1e1 + y2e2. Using bilinearity, we obtain hx, 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. 2 2 2 2 2 hx, xi = 2x1 − 2x1x2 + 3x2 =(x1 − x2) + x1 + 2x2 . Example. V = Mm,n(R), space of m×n matrices. • hA, Bi = trace (ABT ). m n If A = (aij ) and B = (bij ), then hA, Bi = aij bij . i=1 j=1 P P Examples. V = C[a, b]. b • hf , gi = f (x)g(x) dx. Za b • hf , gi = f (x)g(x)w(x) dx, Za where w is bounded, piecewise continuous, and w > 0 everywhere on [a, b]. w is called the weight function. Theorem Suppose hx, yi is an inner product on a vector space V . Then hx, yi2 ≤hx, xihy, yi for all x, y ∈ V . Proof: For any t ∈ R let vt = x + ty. Then 2 hvt, vti = hx, xi + 2thx, yi + t hy, yi. The right-hand side is a quadratic polynomial in t (provided that y =6 0). Since hvt, vti≥ 0 for all t, the discriminant D is nonpositive. But D = 4hx, yi2 − 4hx, xihy, yi. Cauchy-Schwarz Inequality: |hx, yi| ≤ hx, xi hy, yi. p p Cauchy-Schwarz Inequality: |hx, yi| ≤ hx, xi hy, yi. p p Corollary 1 |x · y|≤|x||y| for all x, y ∈ Rn. Equivalently, for all xi , yi ∈ R, 2 2 2 2 2 (x1y1 + ··· + xnyn) ≤ (x1 + ··· + xn )(y1 + ··· + yn ). Corollary 2 For any f , g ∈ C[a, b], b 2 b b f (x)g(x) dx ≤ |f (x)|2 dx · |g(x)|2 dx. Za Za Za Norms induced by inner products Theorem Suppose hx, yi is an inner product on a vector space V . Then kxk = hx, xi is a norm. Proof: Positivity is obvious. Homogeneity:p krxk = hrx, rxi = r 2hx, xi = |r| hx, xi. Triangle inequalityp (followsp from Cauchy-Schwarz’s):p kx + yk2 = hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi ≤hx, xi + |hx, yi| + |hy, xi| + hy, yi ≤ kxk2 + 2kxk kyk + kyk2 = (kxk + kyk)2. Examples. • The length of a vector in Rn, 2 2 2 |x| = x1 + x2 + ··· + xn , is the norm inducedp by the dot product x · y = x1y1 + x2y2 + ··· + xnyn. b 1/2 2 • The norm kf k2 = |f (x)| dx on the Za vector space C[a, b] is induced by the inner product b hf , gi = f (x)g(x) dx. Za Angle Since |hx, yi| ≤ kxk kyk, we can define the angle between nonzero vectors in any vector space with an inner product (and induced norm): hx, yi ∠(x, y) = arccos . kxk kyk Then hx, yi = kxk kyk cos ∠(x, y). In particular, vectors x and y are orthogonal (denoted x ⊥ y) if hx, yi = 0. x x + y y Pythagorean Law: x ⊥ y =⇒ kx + yk2 = kxk2 + kyk2 Proof: kx + yk2 = hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = hx, xi + hy, yi = kxk2 + kyk2. y x − y x x + y x y Parallelogram Identity: kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2 Proof: kx+yk2 = hx+y, x+yi = hx, xi + hx, yi + hy, xi + hy, yi. Similarly, kx−yk2 = hx, xi−hx, yi−hy, xi + hy, yi. Then kx+yk2 + kx−yk2 = 2hx, xi + 2hy, yi = 2kxk2 + 2kyk2. Orthogonal sets Let V be an inner product space with an inner product h·, ·i and the induced norm k · k. Definition. A nonempty set S ⊂ V of nonzero vectors is called an orthogonal set if all vectors in S are mutually orthogonal. That is, 0 ∈/ S and hx, yi = 0 for any x, y ∈ S, x =6 y. An orthogonal set S ⊂ V is called orthonormal if kxk = 1 for any x ∈ S. Remark. Vectors v1, v2,..., vk ∈ V form an orthonormal set if and only if 1 if i = j hv , v i = i j 0 if i =6 j Examples. • V = Rn, hx, yi = x · y. The standard basis e1 = (1, 0, 0,..., 0), e2 = (0, 1, 0,..., 0), . , en = (0, 0, 0,..., 1). It is an orthonormal set. • V = R3, hx, yi = x · y. v1 = (3, 5, 4), v2 = (3, −5, 4), v3 = (4, 0, −3). v1 · v2 = 0, v1 · v3 = 0, v2 · v3 = 0, v1 · v1 = 50, v2 · v2 = 50, v3 · v3 = 25. Thus the set {v1, v2, v3} is orthogonal but not orthonormal. An orthonormal set is formed by v1 v2 1 2 normalized vectors w = kv1k, w = kv2k, v3 3 w = kv3k. π • V = C[−π, π], hf , gi = f (x)g(x) dx. Z−π f1(x) = sin x, f2(x) = sin 2x, ..., fn(x) = sin nx, ... π π if m = n hfm, fni = sin(mx)sin(nx) dx = −π 0 if m =6 n Z Thus the set {f1, f2, f3,... } is orthogonal but not orthonormal. It is orthonormal with respect to a scaled inner product 1 π hhf , gii = f (x)g(x) dx. π Z−π Orthogonality =⇒ linear independence Theorem Suppose v1, v2,..., vk are nonzero vectors that form an orthogonal set. Then v1, v2,..., vk are linearly independent. Proof: Suppose t1v1 + t2v2 + ··· + tk vk = 0 for some t1, t2,..., tk ∈ R. Then for any index 1 ≤ i ≤ k we have ht1v1 + t2v2 + ··· + tk vk , vi i = h0, vi i = 0. =⇒ t1hv1, vi i + t2hv2, vi i + ··· + tk hvk , vi i = 0 By orthogonality, ti hvi , vi i = 0 =⇒ ti = 0. Orthonormal bases Let v1, v2,..., vn be an orthonormal basis for an inner product space V . Theorem Let x = x1v1 + x2v2 + ··· + xnvn and y = y1v1 + y2v2 + ··· + ynvn, where xi , yj ∈ R. Then (i) hx, yi = x1y1 + x2y2 + ··· + xnyn, 2 2 2 (ii) kxk = x1 + x2 + ··· + xn . Proof: (ii) followsp from (i) when y = x. n n n n hx, yi = xi vi , yj vj = xi vi , yj vj * i=1 j=1 + i=1 * j=1 + X n nX Xn X = xi yj hvi , vj i = xi yi . i=1 j=1 i=1 X X X Orthogonal projection Theorem Let V be an inner product space and V0 be a finite-dimensional subspace of V . Then any vector x ∈ V is uniquely represented as x = p + o, where p ∈ V0 and o ⊥ V0. The component p is the orthogonal projection of the vector x onto the subspace V0. We have kok = kx − pk = min kx − vk. v∈V0 That is, the distance from x to the subspace V0 is kok. x o p V0 Let V be an inner product space. Let p be the orthogonal projection of a vector x ∈ V onto a finite-dimensional subspace V0. If V0 is a one-dimensional subspace spanned by a hx, vi vector v then p = v. hv, vi If v1, v2,..., vn is an orthogonal basis for V0 then hx, v1i hx, v2i hx, vni p = v1 + v2 + ··· + vn. hv1, v1i hv2, v2i hvn, vni n hx, vj i hx, vi i Indeed, hp, vi i = hvj , vi i = hvi , vi i = hx, vi i hvj , vj i hvi , vi i j=1 X =⇒ hx−p, vi i =0 =⇒ x−p ⊥ vi =⇒ x−p ⊥ V0..