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Notes on Homework 12 1. (a) Consider the function g : V → V given by m g(v) = v − ∑ hv, ei iei ; i =1 we’ll show that it’s a linear transformation. As usual, there are two things to show: First, suppose v1 , v2 ∈ V; we’ll show g(v1 + v2 ) = g(v1 ) + g(v2 ). Well, m g ( v 1 + v 2 ) = v 1 + v 2 − ∑ h v 1 + v 2 , ei i ei i =1 m = v1 + v2 − ∑ (hv1 , ei i + hv2 , ei i)ei additivity of h·, ·i i =1 m = v1 + v2 − ∑ (hv1 , ei iei + hv2 , ei iei ) distributivity of vector mult i =1 m m = v1 + v2 − ∑ (hv1 , ei iei ) − ∑ hv2 , ei iei i =1 m i =1 m = v1 − ∑ (hv1 , ei iei ) + v2 − ∑ hv2 , ei iei i =1 i =1 = g(v1 ) + g(v2 ). Second, suppose v ∈ V, λ ∈ R; we’ll show g(λv) = λg(v), by computing: m g(λv) = λv − ∑ hλv, ei iei i =1 m = λv − ∑ λ hv, ei iei i =1 m = λ (v − ∑ λ hv, ei iei ) i =1 = λg(v). (b) It need not betruethat f (u) = u. Many examples are possible, but, consider the subx 2 space U = { : x ∈ R} ⊂ R2 ; as a basis, try e1 = . (Note that has length 0 0 Professor Dan Bates Colorado State University M369 Linear Algebra Fall 2008 two, and thus {e1 } is not an orthonormal basis for U.) Suppose v = a 0 ∈ U. Then f (v) = hv, e1 ie1 = ( a · 2 + 0 · 0) 4a = 6= v 0 2. 2 0 (a) From class, we know that trying to find a polynomial a0 + a1 x which goes through the four points is the same as solving the equation 4 1 1 1 1 2 a0 = 3 1 3 a1 1 1 4 Ax = b This probably isn’t possible, but we can find the best approximate solution, x̂ = ( A T A)−1 A T b 4 = 7 − 10 (computation omitted); the best-fitting line is y=− 7 x + 4. 10 (b) To find the best-fitting degree two polynomial, we apply the same method, except using the matrix 1 1 12 1 2 22 B= 1 3 32 1 4 42 we calculate ŷ = ( B T B)−1 B T b 21 4 = − 39 20 1 4 Professor Dan Bates Colorado State University M369 Linear Algebra Fall 2008 and the best-fitting quadratic is y= 1 39 21 x− x+ 4 20 4 (c) Now, use 1 1 C= 1 1 1 2 3 4 12 22 32 42 13 23 33 43 we calculate ẑ = (C T C )−1 C T b 21 −27 = 23 2 − 32 and the best-fitting cubic is 3 2 y = − x3 + 3x − 27x + 21 2 2 Professor Dan Bates Colorado State University M369 Linear Algebra Fall 2008 3. 2 1 (a) There are many different ways to do this. For instance, let z1 = 0 , z2 = 2 , 1 0 0 z3 = 1 , and let 1 2 1 0 A = 0 2 1 . 1 0 1 To find numbers x1 , x2 , x3 such that x1 z1 + x2 z2 + x3 z3 = v is to solve the matrix equation x1 1 A x2 = 4 . 7 x3 (Why?) So, solve this equation any way you like. For example 2 1 − 15 5 5 2 2 A−1 = 51 − 5 5 2 1 4 −5 5 5 (calculation omitted) 1 sothatx = A−1 4 7 1 = −1 6 Similarly, to find numbers y1 , y2 , y3 such that y1 z1 + y2 z2 + y3 z3 = w, use 3 −1 3 y=A 2 1 = 1 1 (b) Let B = { z1 , z2 , z3 }. By definition, the Gram matrix is given by Gi j = h zi , z j i. Thus, for example, G1,2 = h z1 , z2 i = 2·1+0·2+1·0 =2 Professor Dan Bates Colorado State University M369 Linear Algebra Fall 2008 Note that this is calculated by multiplying the first row of A T against the second column of A. More generally, G = AT · A 5 2 1 = 2 5 2 1 2 2 (Of course, you can also just compute each h zi , z j i directly and then assemble the information at the end.) (c) From class, we know that hv, wi = [v]BT G [w]B 5 2 1 1 1 = (1 − 1 6) 2 5 2 1 2 2 1 = 29 (d) We could also calculate directly: hv, wi = 1 · 3 + 4 · 3 + 7 · 2 = 29 Professor Dan Bates Colorado State University M369 Linear Algebra Fall 2008