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 R n. 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 x. Different norms on V are distinguished by subscripts, e.g., x 1 and x 2.
Examples. V = R n, x = (x 1, x 2,...,x n ) R n. x = max( x 1, x 2,..., x n ). x p = ( x 1 p + x 2 p + + x n p) 1/p, p 1. Examples. V = C[a, b], f : [a, b] R. f = max f (x). a x b ( b 1/p f p = f (x) dx) p, p 1. a
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) = x y. Then we say that a sequence x 1,x 2,... converges to a vector x if dist(x,x n ) 0 as n. Also, we say that a vector x is a good approximation of a vector x 0 if dist(x,x 0 ) is small.
Inner product The notion of inner product generalizes the notion of dot product of vectors in R n. Definition. Let V be a vector space. A function β : V V R, usually denoted β(x,y) = x,y, is called an inner product on V if it is positive, symmetric, and bilinear. That is, if (i) x,x 0, x,x = 0 only for x = 0 (positivity) (ii) x, y = y, x (symmetry) (iii) r x, y = r x, y (homogeneity) (iv) x + y,z = x,z + y,z (distributive law) An inner product space is a vector space endowed with an inner product.
Examples. V = R n. x,y = x y = x 1 y 1 + x 2 y 2 + + x n y n. x,y = d 1 x 1 y 1 + d 2 x 2 y 2 + + d n x n y n, where d 1, d 2,...,d n > 0. x,y = (Dx) (Dy), where D is an invertible n n matrix.
Problem. Find an inner product on R 2 such that e 1,e 1 = 2, e 2,e 2 = 3, and e 1,e 2 = 1, where e 1 = (1, 0), e 2 = (0, 1). Let x = (x 1, x 2 ), y = (y 1, y 2 ) R 2. Then x = x 1 e 1 + x 2 e 2, y = y 1 e 1 + y 2 e 2. Using bilinearity, we obtain x,y = x 1 e 1 + x 2 e 2, y 1 e 1 + y 2 e 2 = x 1 e 1, y 1 e 1 + y 2 e 2 + x 2 e 2, y 1 e 1 + y 2 e 2 = x 1 y 1 e 1,e 1 + x 1 y 2 e 1,e 2 + x 2 y 1 e 2,e 1 + x 2 y 2 e 2,e 2 = 2x 1 y 1 x 1 y 2 x 2 y 1 + 3x 2 y 2. It remains to check that x,x > 0 for x 0. x,x = 2x 2 1 2x 1 x 2 + 3x 2 2 = (x 1 x 2 ) 2 + x 2 1 + 2x 2 2.
Example. V = M m,n (R), space of m n matrices. A, B = trace (AB T ). If A = (a ij ) and B = (b ij ), then A, B = m Examples. V = C[a, b]. f, g = f, g = b a b a f (x)g(x) dx. f (x)g(x)w(x) dx, i=1 j=1 where w is bounded, piecewise continuous, and w > 0 everywhere on [a, b]. w is called the weight function. n a ij b ij.
Theorem Suppose x,y is an inner product on a vector space V. Then x,y 2 x,x y,y for all x,y V. Proof: For any t R let v t = x + ty. Then v t,v t = x,x + 2t x,y + t 2 y,y. The right-hand side is a quadratic polynomial in t (provided that y 0). Since v t,v t 0 for all t, the discriminant D is nonpositive. But D = 4 x,y 2 4 x,x y,y. Cauchy-Schwarz Inequality: x,y x,x y,y.
Cauchy-Schwarz Inequality: x,y x,x y,y. Corollary 1 x y x y for all x,y R n. Equivalently, for all x i, y i R, (x 1 y 1 + + x n y n ) 2 (x1 2 + + x2 n)(y1 2 + + y2 n). Corollary 2 For any f, g C[a, b], ( b 2 b f (x)g(x) dx) f (x) 2 dx a a b a g(x) 2 dx.
Norms induced by inner products Theorem Suppose x,y is an inner product on a vector space V. Then x = x,x is a norm. Proof: Positivity is obvious. Homogeneity: rx = rx, rx = r 2 x,x = r x,x. Triangle inequality (follows from Cauchy-Schwarz s): x + y 2 = x + y,x + y = x,x + x,y + y,x + y,y x,x + x,y + y,x + y,y x 2 + 2 x y + y 2 = ( x + y ) 2.
Examples. The length of a vector in R n, x = x1 2 + x2 2 + + x2 n, is the norm induced by the dot product x y = x 1 y 1 + x 2 y 2 + + x n y n. ( b 1/2 The norm f 2 = f (x) dx) 2 on the vector space C[a, b] is induced by the inner product f, g = b a a f (x)g(x) dx.
Angle Since x,y x y, we can define the angle between nonzero vectors in any vector space with an inner product (and induced norm): (x,y) = arccos x,y x y. Then x,y = x y cos (x,y). In particular, vectors x and y are orthogonal (denoted x y) if x,y = 0.
x x + y y Pythagorean Law: x y = x + y 2 = x 2 + y 2 Proof: x + y 2 = x + y,x + y = x,x + x,y + y,x + y,y = x,x + y,y = x 2 + y 2.
y x y x x + y x y Parallelogram Identity: x + y 2 + x y 2 = 2 x 2 + 2 y 2 Proof: x+y 2 = x+y,x+y = x,x + x,y + y,x + y,y. Similarly, x y 2 = x,x x,y y,x + y,y. Then x+y 2 + x y 2 = 2 x,x + 2 y,y = 2 x 2 + 2 y 2.
Orthogonal sets Let V be an inner product space with an inner product, and the induced norm. 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 x,y = 0 for any x,y S, x y. An orthogonal set S V is called orthonormal if x = 1 for any x S. Remark. Vectors v 1,v 2,...,v k V form an orthonormal set if and only if { 1 if i = j v i,v j = 0 if i j
Examples. V = R n, x,y = x y. The standard basis e 1 = (1, 0, 0,...,0), e 2 = (0, 1, 0,...,0),..., e n = (0, 0, 0,...,1). It is an orthonormal set. V = R 3, x,y = x y. v 1 = (3, 5, 4), v 2 = (3, 5, 4), v 3 = (4, 0, 3). v 1 v 2 = 0, v 1 v 3 = 0, v 2 v 3 = 0, v 1 v 1 = 50, v 2 v 2 = 50, v 3 v 3 = 25. Thus the set {v 1,v 2,v 3 } is orthogonal but not orthonormal. An orthonormal set is formed by normalized vectors w 1 = v 1 v 1, w 2 = v 2 w 3 = v 3 v 3. v 2,
π V = C[ π, π], f, g = f (x)g(x) dx. π f 1 (x) = sin x, f 2 (x) = sin 2x,..., f n (x) = sin nx,... f m, f n = π π sin(mx) sin(nx) dx = { π if m = n 0 if m n Thus the set {f 1, f 2, f 3,... } is orthogonal but not orthonormal. It is orthonormal with respect to a scaled inner product f, g = 1 π π π f (x)g(x) dx.
Orthogonality = linear independence Theorem Suppose v 1,v 2,...,v k are nonzero vectors that form an orthogonal set. Then v 1,v 2,...,v k are linearly independent. Proof: Suppose t 1 v 1 + t 2 v 2 + + t k v k = 0 for some t 1, t 2,...,t k R. Then for any index 1 i k we have t 1 v 1 + t 2 v 2 + + t k v k,v i = 0,v i = 0. = t 1 v 1,v i + t 2 v 2,v i + + t k v k,v i = 0 By orthogonality, t i v i,v i = 0 = t i = 0.
Orthonormal bases Let v 1,v 2,...,v n be an orthonormal basis for an inner product space V. Theorem Let x = x 1 v 1 + x 2 v 2 + + x n v n and y = y 1 v 1 + y 2 v 2 + + y n v n, where x i, y j R. Then (i) x,y = x 1 y 1 + x 2 y 2 + + x n y n, (ii) x = x 2 1 + x2 2 + + x2 n. Proof: (ii) follows from (i) when y = x. n n n x,y = x i v i, y j v j = x i v i, i=1 = n j=1 n x i y j v i,v j = i=1 i=1 j=1 i=1 n x i y i. n y j v j j=1
Orthogonal projection Theorem Let V be an inner product space and V 0 be a finite-dimensional subspace of V. Then any vector x V is uniquely represented as x = p + o, where p V 0 and o V 0. The component p is the orthogonal projection of the vector x onto the subspace V 0. We have o = x p = min v V 0 x v. That is, the distance from x to the subspace V 0 is o.
o x p V 0
Let V be an inner product space. Let p be the orthogonal projection of a vector x V onto a finite-dimensional subspace V 0. If V 0 is a one-dimensional subspace spanned by a vector v then p = x,v v,v v. If v 1,v 2,...,v n is an orthogonal basis for V 0 then p = x,v 1 v 1,v 1 v 1 + x,v 2 v 2,v 2 v 2 + + x,v n v n,v n v n. Indeed, p,v i = n j=1 x,v j v j,v j v j,v i = x,v i v i,v i v i,v i = x,v i = x p,v i = 0 = x p v i = x p V 0.