PROBLEM 1 (a) We have ||v1|| = sqrt(1*1 + 3*3 + 3*3 + 9*9) = sqrt(100) = 10, so w1 = v1 / 10 = [1/10, 3/10, 3/10, 9/10]. Then w1.v2 = (1*2 + 3*1 + 3*6 + 9*3) / 10 = 5, so the projection of v2 to the orthogonal complement of the span of w1 is v2 - (w1.v2) w1 = [3/2, -1/2, -9/2, -3/2] (as a column vector). We calculate that this has norm sqrt(25)=5, so w2 is obtained by dividing it by 5 to obtain [3/10, -1/10, 9/10, -3/10]. Then w1.w2 = 3/100 - 3/100 + 27/100 - 27/100 = 0 as it should be. (b) Either by doing the computation in (a) in reverse, or by forming the dot products with w1 and w2 to obtain the coefficients of v1 and v2, we find that v1 = 10 w1 and v2 = 5 w1 + 5 w2. That is, the coordinates are [10] [5] v1 = [ ] v2 = [ ] [ 0] [5]. (c) The dot product of e1 with any vector w is the first component of w. Thus e1.w1 = 1/10 and e1.w2 = 3/10. Therefore the projection of e1 onto V is e1|| = (1/10)w1 + (3/10)w2 = [1/10, 0, 3/10, 0]. [A popular but harder route was the formula Q * Q-transpose (5.3.10 on p.216) for the projection to a subspace V with a known orthonormal basis.] The part of e1 perpendicular to V is then e1 - e1|| = [9/10, 0, -3/10, 0], which indeed is orthogonal to w1 (because (9/10)(1/10) - (3/10)(3/10) = 0) and to w2 (because (9/10)(3/10) - (3/10)(9/10) = 0).