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Unformatted text preview: Answers to Exercises in: Vector Spaces and Projections Stat 640 Answer 1 (a) If we write a 1 v 1 + a 2 v 2 , we get ( a 1 ,a 1 ,a 1 ,a 2 ,a 2 ,a 2 ) , so we can describe this space as all vectors in IR 6 for which the first three elements are the same, and the last three elements are the same. (b) If we write a 1 v 1 + a 2 v 2 , we get ( a 1 ,a 1 ,a 1 ,a 1 + a 2 ,a 1 + a 2 ,a 1 + a 2 ) . Note that we can again describe this space as all vectors in IR 6 for which the first three elements are the same, and the last three elements are the same, so that this linear space is identical to that in part (a). (c) If we write a 1 v 1 + a 2 v 2 + a 3 v 3 , we get ( a 1 + a 3 ,a 1 + a 3 ,a 1 + a 3 ,a 2 + a 3 ,a 2 + a 3 ,a 2 + a 3 ) . Note that we can again describe this space as all vectors in IR 6 for which the first three elements are the same, and the last three elements are the same, so that this linear space is identical to that in part (a). (d) If we write a 1 v 1 + a 2 v 2 , we get ( a 1 ,a 1 , , ,a 2 ,a 2 ) , so this space is all vectors in IR 6 for which the first two elements are the same, the second two elements are zero, and the last two elements are the same, (e) If we write a 1 v 1 + a 2 v 2 , we get ( a 1 + a 2 ,a 1 + 2 a 2 ,a 1 + 3 a 2 ,a 1 + 4 a 2 ,a 1 + 5 a 2 ) . This is all vectors in IR 5 so that the difference between subsequent elements is the same. Examples are ( 3 , 5 , 7 , 9 , 11) or (0 ,. 1 ,. 2 ,. 3 ,. 4) . Note that if we plot any vector in this space against the vector v 2 = (1 , 2 , 3 , 4 , 5) , the five points lie on a straight line. Answer 2 All but (e). Answer 3 (a) These are linearly independent, because the space spanned by any pair will not contain the third vector. (b) These are not linearly independent, because the third vector is the sum of the first two. (c) These are not linearly independent, because v 3 = 3 v 1 v 2 . Answer 4 A simple basis for this twodimensional subspace of IR 6 is v 1 = (1 , 1 , 1 , , , 0) and v 1 = (0 , , , 1 , 1 , 1) . Answer 5 k y a x k 2 = n X i =1 ( y i ax i ) 2 = S, say . Now take the derivative of S with respect to a : dS da = 2 n X i =1 ( y i ax i ) x i , which is zero when a = n i =1 x i y i n i =1 x 2 i . The second derivative of S with respect to a is positive, so the zero is a minimum. Answer 6 (a) ( y  x ) = h x , y i k x k 2 x = 11 10 (2 , 2 , 1 , 1) = (2 . 2 , 2 . 2 , 1 . 1 , 1 . 1) . (b) (5 , 5 , , 0) (c) (4 . 5 , 4 . 5 , 4 . 5 , 4 . 5) (d) (44 / 3 , 88 / 3 , 132 / 3 , 176 / 3) Answer 7 h y ( y  x ) , x i = h y h x , y i k x k 2 x , x i = h y , x i  h x , y i k x k 2 h x , x i = h y , x i  h x , y i = 0 , because h x , x i = k x k 2 ....
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 Spring '08
 MaryMeyer

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