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Math 2270 Spring 2004 Homework 16 Solutions Section 5.5) 8, 10, 16, 20, 23, 24 (8) We define T (v) =< v, w > and take elements v, x ∈ V and a scalar k. Then by properties of inner products, T (v + x) =< v + x, w >=< v, w > + < x, w >= T (v) + T (x) T (kv) =< kv, w >= k < v, w >= kT (v) so T is linear. If w is the zero element of V , then T (v) =< v, w >=< v, 0 >= 0 for all v ∈ V . In this case im(T )= {0} and ker(T )= V . If w 6= 0, then im(T )= ℜ and ker(T ) is the set of all v ∈ V that are orthogonal to w. (10) First, we find the general form for polynomials in P2 that are orthogonal to f (t) = t with respect to the given inner product. 1 < a + bt + ct , t >= 2 2 Z 1 −1 2 3 at + bt + ct 1 1 2 1 3 1 4 dt = at + bt + ct 2 2 3 4 1 −1 1 2 b = 2 3 The previous inner product is equal to zero iff b = 0. Therefore, polynomials orthogonal to f (t) = t are of the form g(t) = a + ct2 . We start with the basis {1, t2 } and use the Gram-Schmidt process to find an orthonormal basis. ||1|| = 1Z 1 dt = 1 2 −1 The first element of our basis is g1 (t) = 1. 1Z 1 2 1 < 1, t >= t dt = 2 −1 3 s s s Z 1 2 1 1 2 4 2 4 2 1 1 2 t − = t4 − t2 + dt = = − + = √ 3 2 −1 3 9 2 5 9 9 45 3 5 √ √ 3 5 2 1 5 2 t − = (3t − 1) g2 (t) = 2 3 2 2 An orthonormal basis is ( √ 5 2 1, (3t − 1) 2 1 ) Math 2270 Spring 2004 Homework 16 Solutions (16) We start with the standard basis {1, t} and use the Gram-Schmidt process to find an orthonormal basis for the given inner product. < 1, 1 >= 1 Z dt = 1 0 < 1, t >= t − 1 2 = s Z 1 0 t2 − t + 1 dt = 4 s 1 1 1 − + = 3 2 4 1 Z t dt = s0 1 2 1 1 = √ 12 2 3 The basis we are looking for is n 1, √ 3 (2t − 1) o Now, we project the function f (t) = t2 onto the linear space defined by the orthonormal basis we found above. 1 t2 dt = 0 √3 Z 1√ √ √ 3 1 1 = − 3(2t3 − t2 ) dt = 3 < t2 , 3(2t − 1) >= 2 3 6 0 √ √ 3 1 1 1 1 projV t2 = + 3(2t − 1) = + t − = t − 3 6 3 2 6 < t , 1 >= Z h i 2 1 (20) < " 1 0 # ,w ~ >= h 1 0 i " 1 2 2 8 # w ~= 1 2 " a b # =0 iff a + 2b e1 are scalar multiples of the " = 0# or a = −2b. Therefore vectors perpendicular to ~ −2 vector . 1 To find an orthonormal basis of ℜ2 with respect to this inner product we start with the basis (" 1 0 # " , −2 1 #) as we already know that these two vectors are orthogonal to one another with respect to this inner product. Therefore, the only work will be to determine their length and normalize 2 Math 2270 Spring 2004 Homework 16 Solutions them if necessary. " " −2 1 #2 1 0 #2 = = h h 1 0 i −2 1 i An orthonormal basis is " 1 2 2 8 #" 1 0 " 1 2 2 8 #" −2 1 (" 1 0 # " , # −1 1/2 h = # = 1 2 h i 0 4 " i 1 0 # " −2 1 =1 # =4 #) (23) We start with the standard basis {1, t} and use the Gram-Schmidt procedure to construct an orthonormal basis. t − 1 2 2 1 < 1, 1 >= (1 + 1) = 1 2 1 1 < 1, t >= (0 + 1) = 2 2 1 −1 1 1 1 −1 = + · = 2 2 2 2 2 4 We have the basis {1, 2t − 1} (24) (a) < f, g + h >=< f, g > + < f, h >= 0 + 8 = 8 (b) ||g + h|| = q < g + h, g + h > = q < g, g > +2 < g, h > + < h, h > = √ 1 + 6 + 50 = √ 57 For part (c) we must first normalize the elements f and √ g. Luckily, they are √ orthogonal to one another since < f, g >= 0. From the table ||f || = 4 = 2 and ||g|| = 1 = 1. The orthonormal basis we are looking for is {f /2, g}. Now, we use the formula for projections projE h =< f /2, h > f f + < g, h > g =< f, h > + < g, h > g = 2f + 3g 2 4 3 Math 2270 Spring 2004 Homework 16 Solutions For part (d), we use our answer from part (c). = ||h − 2f − 3g|| = q q < h − 2f − 3g, h − 2f − 3g > < h, h > +2 < h, −2f > +2 < h, −3g > + < −2f, −2f > +2 < −2f, −3g > + < −3g, −3g > = q 50 − 4(8) − 6(3) + 4(4) + 12(0) + 9(1) = 5 Therefore, an orthonormal basis is 1 1 f, g, (h − 2f − 3g) 2 5 4