Tuesday, December 6 ∗∗ Orthogonal bases & Orthogonal projections ∗∗ Solutions 1. A basis B is called an orthonormal basis if it is orthogonal and each basis vector has norm equal to 1. (a) Convert the orthogonal basis 80− 1 0 1 0 19 < 1 1 1 = B = @ 1 A , @ 1 A , @1A : 0 −2 1 ; into an orthonormal basis C . Solution. We just scale each basis vector by its length. The new, orthonormal, basis is 8 0−11 0 1 1 0119 < 1 1 1 = C = p @ 1 A , p @ 1 A , p @1A 2 6 3 : 0 −2 1 ; 011 0−41 (b) Find the coordinates of the vectors v1 = @3A and v2 = @ 2 A in the basis C . 5 7 Solution. The formula is the same as for a general orthogonal basis: writing u1, u2, and u3 for the basis vectors, the formula for each coordinate of v1 in the basis C is u · v i 1 = · 2 ui v1. kuik We compute to find 2 p −6 p 9 p u1 · v1 = p = 2, u2 · v1 = p = − 6, u3 · v1 = p = 3 3, 2 6 3 0 p 1 p2 B C so (v1)C = @−p 6A. Similarly, we get 3 3 r 6 p −16 2 5 u1 · v2 = p = 3 2, u2 · v2 = p = −8 , u3 · v2 = p , 2 6 3 3 0 p 1 3p 2 B C so (v2)C = @−8 p2/3A. 5/ 3 Orthonormal bases are very convenient for calculations. 2. (The Gram-Schmidt process) There is a standard process for converting a basis into an 3 orthogonal basis. Let B = fv1, v2, v3g be a basis for R . In this problem, you will find an orthogonal basis C = fw1, w2, w3g. Start by setting w1 = v1. Then we want w2 to be orthogonal to v1. If we write p for the projection of v2 onto w1, then v2 − p is orthogonal to w1, so we may choose this for w2. In other words, w = v − proj (v ). 2 2 w1 2 Next, we want w3 to be orthogonal to both w1 and w2, so we define w = v − proj (v ) − proj (v ). 3 3 w1 3 w2 3 (a) Use the Gram-Schmidt process to convert 80 1 0 1 0 19 < 1 1 0 = B = @1A , @2A , @2A : 0 1 1 ; into an orthogonal basis C . 011 Solution. We set w1 = @1A. Then 0 011 011 0−1/21 v2 · w1 3 w2 = v2 − w = @2A − @1A = @ 1/2 A . kw k2 1 2 1 1 0 1 We then take 001 011 0−1/21 0−1/31 v3 · w1 v3 · w2 2 2 w3 = v3 − w − w2 = @2A − @1A − @ 1/2 A = @ 1/3 A . kw k2 1 kw k2 2 3/2 1 2 1 0 1 −1/3 (b) Convert this orthogonal basis into an orthonormal basis, and then find the coordi- 071 nates of the vector v = @2A in this orthonormal basis. 1 Solution. The orthonormal basis is 8 011 r 0−1/21 0−1/319 < 1 2 p = C = p @1A , @ 1/2 A , 3 @ 1/3 A . 2 3 : 0 1 −1/3 ; Note that another way to write this same basis is as 8 011 0−11 0−119 < 1 1 1 = C = p @1A , p @ 1 A , p @ 1 A . 2 6 3 : 0 2 −1 ; 0 p 1 9/p 2 B C A computation as in problem 1 gives (v)C = @− p3/2A. −2 3 3. Find the least squares solution to the system of equations 2x + y = 3 −x − y = 2 3x + y = 3. 0 2 1 1 031 Solution. We have A = @−1 −1A and b = @2A. The least squares solution is the 3 1 3 solution to the normal equation AT Ax = ATb. The matrix AT A is 14 6 AT A = 6 3 with inverse 1 3 −6 (AT A)−1 = . 6 −6 14 The least squares solution is therefore given by 1 3 −6 13 1 15 x = (AT A)−1 ATb = = . 6 −6 14 4 6 −22 Note that 0 4/3 1 Ax = @ 7/6 A 6= b 23/6 so the least squares solution x is not an actual solution..