MATH 304 Linear Algebra Lecture 28: Orthogonal bases. The Gram-Schmidt orthogonalization process.

Transcription

1 MATH 304 Linear Algebra Lecture 28: Orthogonal bases. The Gram-Schmidt orthogonalization process.

2 Orthogonal sets Let V be an inner product space with an inner product, and the induced norm v = v,v. Definition. Nonzero vectors v 1,v 2,...,v k V form an orthogonal set if they are orthogonal to each other: v i,v j = 0 for i j. If, in addition, all vectors are of unit norm, v i = 1, then v 1,v 2,...,v k is called an orthonormal set. Theorem Any orthogonal set is linearly independent.

3 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 x n v n and y = y 1 v 1 + y 2 v y n v n, where x i, y j R. Then (i) x,y = x 1 y 1 + x 2 y x n y n, (ii) x = x x 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

4 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.

5 o x p V 0

6 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 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.

7 The Gram-Schmidt orthogonalization process Let V be a vector space with an inner product. Suppose x 1,x 2,...,x n is a basis for V. Let v 1 = x 1, v 2 = x 2 x 2,v 1 v 1,v 1 v 1, v 3 = x 3 x 3,v 1 v 1,v 1 v 1 x 3,v 2 v 2,v 2 v 2,... v n = x n x n,v 1 v 1,v 1 v 1 x n,v n 1 v n 1,v n 1 v n 1. Then v 1,v 2,...,v n is an orthogonal basis for V.

8 v 3 x 3 p 3 Span(v 1,v 2 ) = Span(x 1,x 2 )

9 Any basis x 1,x 2,...,x n Orthogonal basis v 1,v 2,...,v n Properties of the Gram-Schmidt process: v k = x k (α 1 x α k 1 x k 1 ), 1 k n; the span of v 1,...,v k is the same as the span of x 1,...,x k ; v k is orthogonal to x 1,...,x k 1 ; v k = x k p k, where p k is the orthogonal projection of the vector x k on the subspace spanned by x 1,...,x k 1 ; v k is the distance from x k to the subspace spanned by x 1,...,x k 1.

10 Normalization Let V be a vector space with an inner product. Suppose v 1,v 2,...,v n is an orthogonal basis for V. Let w 1 = v 1 v 1, w 2 = v 2 v 2,..., w n = v n v n. Then w 1,w 2,...,w n is an orthonormal basis for V. Theorem Any finite-dimensional vector space with an inner product has an orthonormal basis. Remark. An infinite-dimensional vector space with an inner product may or may not have an orthonormal basis.

11 Orthogonalization / Normalization An alternative form of the Gram-Schmidt process combines orthogonalization with normalization. Suppose x 1,x 2,...,x n is a basis for an inner product space V. Let v 1 = x 1, w 1 = v 1 v 1, v 2 = x 2 x 2,w 1 w 1, w 2 = v 2 v 2, v 3 = x 3 x 3,w 1 w 1 x 3,w 2 w 2, w 3 = v 3 v 3,... v n = x n x n,w 1 w 1 x n,w n 1 w n 1, w n = v n v n. Then w 1,w 2,...,w n is an orthonormal basis for V.

12 Problem. Let V 0 be a subspace of dimension k in R n. Let x 1,x 2,...,x k be a basis for V 0. (i) Find an orthogonal basis for V 0. (ii) Extend it to an orthogonal basis for R n. Approach 1. Extend x 1,...,x k to a basis x 1,x 2,...,x n for R n. Then apply the Gram-Schmidt process to the extended basis. We shall obtain an orthogonal basis v 1,...,v n for R n. By construction, Span(v 1,...,v k ) = Span(x 1,...,x k ) = V 0. It follows that v 1,...,v k is a basis for V 0. Clearly, it is orthogonal. Approach 2. First apply the Gram-Schmidt process to x 1,...,x k and obtain an orthogonal basis v 1,...,v k for V 0. Secondly, find a basis y 1,...,y m for the orthogonal complement V0 and apply the Gram-Schmidt process to it obtaining an orthogonal basis u 1,...,u m for V0. Then v 1,...,v k,u 1,...,u m is an orthogonal basis for R n.

13 Problem. Let Π be the plane in R 3 spanned by vectors x 1 = (1, 2, 2) and x 2 = ( 1, 0, 2). (i) Find an orthonormal basis for Π. (ii) Extend it to an orthonormal basis for R 3. x 1,x 2 is a basis for the plane Π. We can extend it to a basis for R 3 by adding one vector from the standard basis. For instance, vectors x 1, x 2, and x 3 = (0, 0, 1) form a basis for R 3 because = = 2 0.

14 Using the Gram-Schmidt process, we orthogonalize the basis x 1 = (1, 2, 2), x 2 = ( 1, 0, 2), x 3 = (0, 0, 1): v 1 = x 1 = (1, 2, 2), v 2 = x 2 x 2,v 1 v 1,v 1 v 1 = ( 1, 0, 2) 3 (1, 2, 2) 9 = ( 4/3, 2/3, 4/3), v 3 = x 3 x 3,v 1 v 1,v 1 v 1 x 3,v 2 v 2,v 2 v 2 = (0, 0, 1) 2 4/3 (1, 2, 2) ( 4/3, 2/3, 4/3) 9 4 = (2/9, 2/9, 1/9).

15 Now v 1 = (1, 2, 2), v 2 = ( 4/3, 2/3, 4/3), v 3 = (2/9, 2/9, 1/9) is an orthogonal basis for R 3 while v 1,v 2 is an orthogonal basis for Π. It remains to normalize these vectors. v 1,v 1 = 9 = v 1 = 3 v 2,v 2 = 4 = v 2 = 2 v 3,v 3 = 1/9 = v 3 = 1/3 w 1 = v 1 / v 1 = (1/3, 2/3, 2/3) = 1 3 (1, 2, 2), w 2 = v 2 / v 2 = ( 2/3, 1/3, 2/3) = 1 3 ( 2, 1, 2), w 3 = v 3 / v 3 = (2/3, 2/3, 1/3) = 1 3 (2, 2, 1). w 1,w 2 is an orthonormal basis for Π. w 1,w 2,w 3 is an orthonormal basis for R 3.