Introduction to Linear algebra-Strang-Solutions-Manual_ver13

68 c 86 2 6 c 8 c 2 8 c 6 d 1 true 96

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Unformatted text preview: C v2 w2 =kvkkwk D v  w=kvkkwk. This is cos  because ˇ ˛ D  . 1 2 2 .u2 C U1 / and ju2 jjU2 j  1 .u2 C U2 /. The whole line 1 2 2 2 1 1 2 2 2 2 becomes :96  .:6/.:8/ C .:8/.:6/  2 .:6 C :8 / C 2 .:8 C :6 / D 1. True: :96 < 1. p The cosine of  is x= x 2 C y 2 , near side over hypotenuse. Then j cos  j2 is not greater than 1: x 2 =.x 2 C y 2 /  1. The vectors w D .x; y/ with .1; 2/  w D x C 2y D 5 lie on a line in the xy plane. p The shortest w on that line is .1; 2/. (The Schwarz inequality kwk  v  w=kvk D 5 p is an equality when cos  D 0 and w D .1; 2/ and kwk D 5.) The length kv wk is between 2 and 8 (triangle inequality when kvk D 5 and kwk D 3). The dot product v  w is between 15 and 15 by the Schwarz inequality. Three vectors in the plane could make angles greater than 90ı with each other: for example .1; 0/; . 1; 4/; . 1; 4/. Four vectors could not do this (360ı total angle). How many can do this in R3 or Rn ? Ben Harris and Greg Marks showed me that the answer is n C 1: The vectors from the center of a regular simplex in Rn to its n C 1 vertices all have negative dot products. If n C 2 vectors in Rn had negative dot products, project them onto the plane orthogonal to the last one. Now you have n C 1 vectors in Rn 1 with negative dot products. Keep going to 4 vectors in R2 : no way! For a specific example, pick v D p 2; 3/ and then w D . 3; 1; 2/. In this example .1; p cos  D v  w=kvkkwk D 7= 14 14 D 1=2 and  D 120ı . This always happens when x C y C z D 0: 1 12 v  w D xz C xy C yz D .x C y C z/2 .x C y 2 C z 2 / 2 2 1 1 This is the same as v  w D 0 kvkkwk: Then cos  D : 2 2 p Wikipedia gives this proof of geometric mean G D 3 xyz  arithmetic mean A D .x C y C z/=3. First there is equality in case x D y D z . Otherwise A is somewhere between the three positive numbers, say for example z < A < y . Use the known inequality g  a for the two positive numbers x and y C z A. Their 1 1 mean a D 2 .x C y C z A/ is 2 .3A A/ D same as A! So a  g says that A3  g 2 A D x.y C z A/A. But .y C z A/A D .y A/.A z / C yz > yz . Substitute to find A3 > xyz D G 3 as we wanted to prove. Not easy! 24 Example 6 gives ju1 jjU1 j  25 26 27 28 29 30 There are many proofs of G D .x1 x2    xn /1=n  A D .x1 C x2 C    C xn /=n. In calculus you are maximizing G on the plane x1 C x2 C    C xn D n. The maximum occurs when all x ’s are equal. 31 The columns of the 4 by 4 “Hadamard matrix” (times 1 ) are perpendicular unit 2 vectors: 2 3 1 1 1 1 1 61 1 1 1 17 HD 4 : 1 1 1 15 2 2 1 1 1 1 32 The commands V D randn .3; 30/I D D sqrt .diag .V 0  V //I U D V \D I will give 30 random unit vectors in the columns of U . Then u 0  U is a row matrix of 30 dot products whose average absolute value may be close to 2= . Solutions to Exercises 6 Problem Set 1.3, page 29 1 2s1 C 3s2 C 4s3 D .2; 5; 9/. The same vector b comes from S times x D .2; 3; 4/: " 100 110 111 #" # " # "# .row 1/  x 2 2 3 D .row 2/  x D 5 : 4 .row 2/  x 9 2 The solutions are y1 D 1, y2 D 0, y3 D 0 (right side D column 1) and y1 D 1, y2 D 3, y3 D 5. That second example illustrat...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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