# 6. Orthogonality and Least-Squares

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1 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 1 6. Orthogonality and Least-Squares 6.1 Inner product, length, and orthogonality Orthogonal sets Orthogonal projections The Gram-Schmidt process Least-squares problems Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU Inner product, length, and orthogonality Definition The inner product of two vector u and v in R n is written u.v. If u =, v =, then u.v = u 1 v 1 + u 2 v u n v n. Theorem 1 (a) u.v = v.u (b) (u + v).w = u.w + v.w (c) (cu).v = c(u.v) = u.(cv) (d) u.u 0, and u.u = 0 u = 0.

2 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 3 Definition The length (or norm) of v is the non-negative scalar v defined by Note: cv = c v. Definition The distance between u and v, written as dist(u, v), is the length of the vector u - v. That is, dist(u, v) = u - v. Orthogonal vectors Definition Two vectors u and v in R n are orthogonal if u.v = 0. Theorem 2 (The Pythagorean Theorem, 畢氏定理 ) Two vectors u and v are orthogonal if and only if u + v 2 = u 2 + v 2. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 4 (Lcos ) 2 + (Lsin ) 2 = L 2 L

3 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 5 Orthogonal complements Definition The set of all vectors u that are orthogonal to every vector w in W, then we say that the set is the orthogonal complement of W, and denote by W. (next page) Note 1. A vector x is in W if and only if x is orthogonal to every vector in a set that spans subspace W. 2. W is a subspace of R n for any subset W of R n. 3. Zero vector is orthogonal to any vector. Theorem 3 Let A be an m n matrix. Then (i) (Row A) = Nul A. (ii) (Col A) = Nul A T. ({x A T x = 0} {x x T A = 0} left null space of A) Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 6 R n W w 1 w 2 w i w 3 W = { x x R n, x w i s}

4 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 7 W is a subspace of R n for any subset W of R n. If w is an arbitrary vector in W, then 1. 0 W 2. If u W, then cu.w = c (u.w) = 0; thus cu W 3. If u and v W, then (u+v).w = u.w + v.w = 0; thus (u+v) W If W is a subset of R n, then (W ) = W? If W is a subspace of R n, then (W ) = W? Exercises of Section 6.1. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU Orthogonal sets A set of vectors {u 1,, u n } in R n is said to be an orthogonal set if u i u j = 0 for all i j. Theorem 4 If S = {u 1,, u p } is an orthogonal set of nonzero vectors in R n, then S is linearly independent and hence is a basis for the subspace spanned by S. An orthogonal basis for a subspace W of R n is a basis for W that is also an orthogonal set. u i u j

5 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 9 Theorem 5 Let {u 1,, u p } be an orthogonal basis for a subspace W of R n. Then each y in W has a unique representation as a linear combination of u 1,, u p. In fact, if y = c 1 u c p u p then c j =, j = 1, 2,, p. Note 投影分量 (1) y j = c j u j = u j is the orthogonal projection of y onto u j. (2) y - y j = y - u j is component of y orthogonal to u j. u 2 y u 1 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 10 A set {u 1,, u p } is an orthonormal set if it is an orthogonal set of unit vectors. Normalize the length of vector u = [u 1 u n ] T to u Theorem 6 An m n matrix U has orthonormal columns if and only if U T U =. Proof. 內積

6 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 11 Theorem 7 Let U be an m n matrix with orthonormal columns, and let x and y be in R n. Then (a) Ux = x (preserving length) (b) (Ux).(Uy) = x.y (preserving orthogonality) (c) (Ux).(Uy) = 0 if and only if x.y = 0 Proof. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 12 (b) (Ux).(Uy) = (x 1 u x n u n ).(y 1 u y n u n ) = x 1 y 1 + x 2 y x n y n Exercises of Section 6.2.

7 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU Orthogonal projections Purpose To find the projection of a vector on a subspace. Theorem 8 (The Orthogonal Decomposition Theorem) Let W be a subspace of R n. Then each y in R n can be written uniquely in the form y = + z, (1) where is in W and z is in W. In fact, if { u 1,, u p } is any orthogonal basis of W, then u 3 W = Span {u 1, u 2 } y (2) z and z = y -. The vector in Eq.(1) is called the orthogonal projection of y onto W and is often written as proj w y. u 1 ŷ u 2 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 14 For example, The orthogonal projection of y onto W. y W 0 Ex.2. Let u 1 =, u 2 =, and y =. Observe that {u 1, u 2 } is an orthogonal basis for W = Span {u 1, u 2 }. Write y as the sum of vector in W and vector orthogonal to W.

8 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 15 Solution. The orthogonal projection of y onto W is Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 16 Properties of orthogonal projections If {u 1, u 2, u p } is an orthogonal basis for W and if y is in W = Span{u 1, u 2,, u p }, then proj W y = y. Theorem 9 (The Best approximation theorem) Let W be a subspace of R n, y be any vector in R n, and be the orthogonal projection of y onto W. Then is the closest point in W to y, in the sense that for all v in W distinct from. y (3) W 0

9 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 17 Ex.4. The distance from a point y in R n to a subspace W is defined as the distance from y to the nearest point in W. Find the distance from y to W = Span {u 1, u 2 }, where y =, u 1 =, u 2 =. Solution. By the best approximation theorem, the distance from y to W is, where = proj W y. Since {u 1, u 2 } is orthogonal basis for W, Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 18 Theorem 10 If { u 1,.., u p } is an orthonormal basis for a subspace W of R n, then proj W y = (y u 1 ) u 1 + (y u 2 ) u (y u p ) u p (4) If U = [u 1 u 2 u p ], then proj W y = U U T y for all y in R n. (5) Proof. proj W y = (y u 1 ) u 1 + (y u 2 ) u (y u p ) u p Since = U proj W y = U = U (U T y) = U U T y Exercises of Section 6.3.

10 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU The Gram - Schmidt process Purpose The Gram - Schmidt process is a simple algorithm for producing an orthogonal or orthonormal basis for nonzero subspace of R n. Theorem 11 (The Gram - Schmidt process) Let a basis { x 1,, x p } for a subspace W of R n, define Non-zero linear independent set Orthogonal set Then { v 1,, v p } is an orthogonal basis for W. In addition Span { v 1,, v k } = Span { x 1,, x k } for 1 k p. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 20 The geometric meaning of the Gram - Schmidt process (A three-dimensional case) x 2 v 2 v 1 = x 1? x3 v3 x2 Note that the angle between x i and v i is less than 90 o. v1 = x1 v2

11 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 21 v 2 x 2 x 1 = v 1 Orthonormal bases An orthonormal basis is constructed easily from an orthogonal basis { v 1,, v p }; simply normalize all v k by v k / v k v k. QR factorization of matrices This factorization is widely used in computer algorithm for various computations, such as solving equations and finding eigenvalues. (Exercise 23 of Section 5.2.) Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 22 Theorem 12 (The QR factorization) If A is an m n matrix with linearly independent columns, then A can be factored as A = QR, where Q is an m n matrix whose columns form an orthonormal basis for Col A and R is an n n upper triangular invertible matrix with positive entries on its diagonal. Ex.4.

12 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 23 (b) To find R. Since A = QR and Q T Q = (Theorem 6 of page 10) Q T A= Q T QR Q T A = R. Exercises of Section 6.4. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 24 Problem No! A = QR and Q T Q = Q T A= Q T QR Q T A = R QQ T A = QR = A QQ T = I?

13 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU Least-squares problems For a linear system, Ax = b, a solution is demanded but none exists. The best one can do is to find an x that makes Ax as close as possible to b. The general least-squares problem is just to find an x that makes b - Ax as small as possible. The term least-squares arises from the fact that b - Ax is the square root of a sum of squares. Definition If A is an m n matrix and b is in R m, a least-squares solution of Ax = b is an in R n such that b - A b - Ax for all x in R n. Theorem 9 on page 16 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 26 Note No matter that x is selected, the vector Ax will necessarily be in the column space Col A. So we seek an x that makes Ax the closest point in Col A to b. b Col A If is the orthogonal projection of b on Col A, then A =. It means that Ax = is consistent and there is a solution in R n. By the orthogonal decomposition principle, the projection has the property that b - is orthogonal to Col A, so b - A is orthogonal to each column of A. If a j is any column of A, then a j is orthogonal to (b A ) and a jt (b A ) = 0. Since each a jt is a row of A T, A T (b A ) = 0 A T b A T A = 0 A T A = A T b.

14 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 27 Theorem 13 The set of least-squares solutions of Ax = b coincides with the nonempty set of solutions of the normal equations A T A = A T b. If A T A is invertible, then = (A T A) -1 A T b. Ex.1. Find a least-squares solution of the inconsistent system In many calculations, A T A is invertible, but this is not always the case. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 28 Ex.2. Find a least-squares solution of Ax = b for where A T A is not invertible.

15 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 29 Theorem 14 The matrix A T A is invertible if and only if the columns of A are linearly independent. In this case, the equation Ax = b has only one least-squares solution, and it is given by = (A T A) -1 A T b. Note The distance from b to A is called the least-squares error. Alternative calculations of least-square solutions Meaning In some cases, the normal equations for a least-squares problem can be ill-conditioned; that is, small errors in the calculations of the entries of A T A can sometimes cause relatively large errors in the solution. If the columns of A are linearly independent, the least-squares solution can often be computed more reliably through a QR factorization of A. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 30 ill-conditioned The condition number associated with the linear equation Ax = b gives a bound on how inaccurate the solution x will be after approximate solution. Conditioning is a property of the matrix, not the algorithm or floating point accuracy of the computer used to solve the corresponding system. The rate at which the solution x will change with respect to a change in b. Thus, if the condition number is large, even a small error in b may cause a large error in x.

16 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 31 The condition number is the maximum ratio of the relative error in x divided by the relative error in b,, where x is 2 norm. For example, where condition number is The larger condition number is, the more ill conditioned the coefficient matrix is. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 32 Theorem 15 Given an m n matrix A with linearly independent columns, let A = QR be a QR factorization of A. Then for each b in R m, the equation Ax = b has a unique least-squares solution, = R -1 Q T b. Proof. = (A T A) -1 A T b since A = QR = (R T Q T QR) -1 R T Q T b = (R T R) -1 R T Q T b = (R T R) -1 R T Q T b = R -1 (R T ) -1 R T Q T b = R -1 Q T b.

17 Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 33 Note Since R is an upper triangular matrix, should be calculated from the equation R = Q T b. It is much faster to solve this equation by backsubstitution or row operations than to compute R -1 and use the preceding equation. Linear Algebra 6. Orthogonality and Least-Squares CSIE NCU 34 Ex.5. Find the least-squares solution of Ax = b for Exercises for Section 6.5.

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