Announcements Monday, November 20

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1 Announcements Monday, November 20 You already have your midterms! Course grades will be curved at the end of the semester. The percentage of A s, B s, and C s to be awarded depends on many factors, and will not be determined until all grades are in. Individual exam grades are not curved. WeBWorK 6.1, 6.2, 6.3 due the Wednesday after Thanksgiving. Reading day: Math is 1 3pm on December 6 in Clough 144 and 152. I ll be there for part of it. My office is Skiles 244. Rabinoffice hours are Monday, 1 3pm and Tuesday, 9 11am.

2 Section 6.2/6.3 Orthogonal Projections

3 Best Approximation Suppose you measure a data point x which you know for theoretical reasons must lie on a subspace W. x x y y W Due to measurement error, though, the measured x is not actually in W. Best approximation: y is the closest point to x on W. How do you know that y is the closest point? The vector from y to x is orthogonal to W : it is in the orthogonal complement W.

4 Orthogonal Decomposition Recall: If W is a subspace of R n, its orthogonal complement is W = { v in R n v is perpendicular to every vector in W } Theorem Every vector x in R n can be written as x = x W + x W for unique vectors x W in W and x W in W. The equation x = x W + x W (with respect to W ). is called the orthogonal decomposition of x The vector x W is the closest vector to x on W. x [interactive 1] [interactive 2] x W x W W

5 Orthogonal Decomposition Justification Theorem Every vector x in R n can be written as x = x W + x W for unique vectors x W in W and x W in W. Why? Uniqueness: suppose x = x W + x W = x W + x W for x W, x W in W and x W, x W in W. Rewrite: x W x W = x W x W. The left side is in W, and the right side is in W, so they are both in W W. But the only vector that is perpendicular to itself is the zero vector! Hence 0 = x W x W = x W = x W 0 = x W x W = x W = x W Existence: We will compute the orthogonal decomposition later using orthogonal projections.

6 Orthogonal Decomposition Example Let W be the xy-plane in R 3. Then W is the z-axis x = 2 = x W = 2 x W = a a 0 x = b = x W = b x W = 0. c 0 c This is just decomposing a vector into a horizontal component (in the xy-plane) and a vertical component (on the z-axis). x x W x W W [interactive]

7 Orthogonal Decomposition Computation? Problem: Given x and W, how do you compute the decomposition x = x W + x W? Observation: It is enough to compute x W, because x W = x x W. First we need to discuss orthogonal sets. Definition A set of nonzero vectors is orthogonal if each pair of vectors is orthogonal. It is orthonormal if, in addition, each vector is a unit vector. Lemma An orthogonal set of vectors is linearly independent. Hence it is a basis for its span. Suppose {u 1, u 2,..., u m} is orthogonal. We need to show that the equation c 1u 1 + c 2u c mu m = 0 has only the trivial solution c 1 = c 2 = = c m = 0. 0 = u 1 (c 1u 1 + c 2u c mu m) = c 1(u 1 u 1) Hence c 1 = 0. Similarly for the other c i.

8 Orthogonal Sets Examples Example: B = 1, 2, 0 is an orthogonal set. Check: = = = Example: B = {e 1, e 2, e 3} is an orthogonal set. Check: = = = Example: Let x = ( ) ( a b be a nonzero vector, and let y = b a orthogonal set: ( ) a b ( ) b = ab + ab = 0. a ). Then {x, y} is an

9 Orthogonal Projections Definition Let W be a subspace of R n, and let {u 1, u 2,..., u m} be an orthogonal basis for W. The orthogonal projection of a vector x onto W is proj W (x) def = m i=1 x u i u i u i u i = x u1 u 1 u 1 u 1 + x u2 u 2 u 2 u x un u n u n u n. This is a vector in W because it is in Span{u 1, u 2,..., u m}. Theorem Let W be a subspace of R n, and let x be a vector in R n. Then x W = proj W (x) and x W = x proj W (x). In particular, proj W (x) is the closest point to x in W. Why? Let y = proj W (x). We need to show that x y is in W. In other words, u i (x y) = 0 for each i. Let s do u 1: ( ) m x u i u 1 (x y) = u 1 x u i = u 1 x x u1 (u 1 u 1) 0 = 0. u i u i u 1 u 1 i=1

10 Orthogonal Projection onto a Line The formula for orthogonal projections is simple when W is a line. Let L = Span{u} be a line in R n, and let x be in R n. The orthogonal projection of x onto L is the point proj L (x) = x u u u u. x y x u [interactive] L y = proj L (x) Example: Compute the orthogonal projection of x = ( ) 6 4 onto the line L spanned by u = ( 3 2). ( ) 6 4 y = proj L (x) = x u u u u = ( ) 3 = ( ) 3. 2 L 10 ( ) ( 3 2)

11 Orthogonal Projection onto a Plane Easy example 1 What is the projection of x = 2 onto the xy-plane? 3 Answer: The xy-plane is W = Span{e 1, e 2}, and {e 1, e 2} is an orthogonal basis. 1 x W = proj W 2 = x e1 e 1 + x e2 e 2 = 1 1 e 3 1 e 1 e 2 e e e2 = 2. 0 So this is the same projection as before. x x W x W W [interactive]

12 Orthogonal Projections More complicated example What is the projection of x = 1.4 onto W = Span 0, 1.1? Answer: The basis is orthogonal, so 1.1 x W = proj W 1.4 = x u1 u 1 + x u2 u 2 u u 1 u 2 u 2 = ( 1.1)(1) 1 0 (1.4)(1.1) + (1.45)(.2) (.2) 2 This turns out to be equal to u 2 1.1u x x W xw u 2 u 1 W [interactive]

13 Orthogonal Projections Properties First we restate the property we ve been using all along. Let x be a vector and let x = x W +x W be its orthogonal decomposition with respect to a subspace W. The following vectors are the same: x W proj W (x) The closest vector to x on W We can think of orthogonal projection as a transformation: Theorem Let W be a subspace of R n. proj W : R n R n 1. proj W is a linear transformation. 2. For every x in W, we have proj W (x) = x. 3. For every x in W, we have proj W (x) = 0. x proj W (x). 4. The range of proj W is W and the null space of proj W is W.

14 Poll Let W be a subspace of R n. Poll Let A be the matrix for proj W. What is/are the eigenvalue(s) of A? A. 0 B. 1 C. 1 D. 0, 1 E. 1, 1 F. 0, 1 G. 1, 0, 1 The 1-eigenspace is W. The 0-eigenspace is W. We have dim W + dim W = n, so that gives n linearly independent eigenvectors already. So the answer is D.

15 Orthogonal Projections Matrices What is the matrix for proj W : R 3 R 3, where 1 1 W = Span 0, 1? 1 1 Answer: Recall how to compute the matrix for a linear transformation: A = proj W (e 1) proj W (e 2) proj W (e 3). We compute: proj W (e 1) = proj W (e 2) = proj W (e 3) = e1 u1 e1 u2 u 1 + u 2 = 1 u 1 u 1 u 2 u 2 2 e2 u1 e2 u2 u 1 + u 2 = u 1 u 1 u 2 u 2 3 e3 u1 e3 u2 u 1 + u 2 = 1 u 1 u 1 u 2 u 2 2 5/6 1/3 1/6 Therefore A = 1/3 1/3 1/3. 1/6 1/3 5/6 5/6 1/3 1/ = = 1/3 1 1/ = 1/6 1/ /6

16 Orthogonal Projections Matrix facts Let W be an m-dimensional subspace of R n, let proj W : R n W be the projection, and let A be the matrix for proj L. Fact 1: A is diagonalizable with eigenvalues 0 and 1; it is similar to the diagonal matrix with m ones and n m zeros on the diagonal. Why? Let v 1, v 2,..., v m be a basis for W, and let v m+1, v m+2,..., v n be a basis for W. These are (linearly independent) eigenvectors with eigenvalues 1 and 0, respectively, and they form a basis for R n because there are n of them. Example: If W is a plane in R 3, then A is similar to projection onto the xy-plane: Fact 2: A 2 = A. Why? Projecting twice is the same as projecting once: proj W proj W = proj W = A A = A.

17 Coordinates with respect to Orthogonal Bases Let W be a subspace with orthogonal basis B = {u 1, u 2,..., u m}. For x in W we have proj W (x) = x, so x = proj W (x) = m i=1 x u i u i u i u i = x u1 u 1 u 1 u 1 + x u2 u 2 u 2 u x un u n u n u n. This makes it easy to compute the B-coordinates of x. Corollary Let W be a subspace with orthogonal basis B = {u 1, u 2,..., u m}. Then ( ) x u1 x u 2 x u m [x] B =,,...,. u 1 u 1 u 2 u 2 u m u m [interactive]

18 Coordinates with respect to Orthogonal Bases Example Problem: Find the B-coordinates of x = ( 0 3), where {( ) ( )} 1 4 B =,. 2 2 Old way:( 1 4 ) 0 rref ( 1 0 ) 6/ /20 = [x] B = New way: note B is an orthogonal basis. ( ) ( ) x u1 x u [x] B =, u 2 = u 1 u 1 u 2 u , ( 4) ( ) 6/5. 6/20 = ( ) 6 5, x u u 2 u u 1 [interactive]

19 Orthogonal Projections Distance to a subspace 1 1 What is the distance from e 1 to W = Span 0, 1? 1 1 5/6 Answer: The closest point on W to e 1 is proj W (e 1) = 1/3. 1/6 The distance from e 1 to this point is dist ( e 1, proj W (e ) 1) = (e 1) W 1 5/6 = 0 1/3 0 1/6 1/6 = 1/3 1/6 = (1/6) 2 + ( 1/3) 2 + ( 1/6) 2 = y [interactive] 1 e 1 6 (e 1 ) W W

20 Summary Let W be a subspace of R n. Any vector x in R n can be written in a unique way as x = x W + x W for x W in W and x W in W. This is called its orthogonal decomposition. The vector x W is the closest point to x in W : it is the best approximation. The distance from x to W is x W. If you have an orthogonal basis {u 1, u 2,..., u m} for W, then x W = proj W (x) = x u i u i u i u i = x u1 u 1 u 1 u 1 + x u2 u 2 u 2 u x un u n u n u n. Hence x W = x proj W (x). If you have an orthogonal basis {u 1, u 2,..., u m} for W, then ( ) x u1 x u 2 x u m [x] B =,,...,. u 1 u 1 u 2 u 2 u m u m We can think of proj W : R n R n as a linear transformation. Its null space is W, and its range is W. The matrix A for proj W is diagonalizable with eigenvalues 0 and 1. It is idempotent: A 2 = A.