Products whose average absolute value may be close to 2 /π. ģ2 The commandsV=randn(3,30) D=sqrt(diag(Vģ0 random unit vectors in the columns ofU. The maximumģ1 The columns of the 4 by 4 “Hadamard matrix” (times xn)Ĭalculus you are maximizingGon the planex 1 +x 2 +.Use the known inequalityg≤afor thetwopositive numbersxandy+z−A. Somewhere between the three positive numbers, say for examplez < A < y. In this exampleģ0 Wikipedia gives this proof of geometric mean G =Ī= (x+y+z)/ 3. Keep going to 4 vectors inRĢ9 For a specific example, pickv= (1, 2 ,−3)and thenw= (− 3, 1 ,2). Project them onto the plane orthogonal to the last one. ? Ben Harris and Greg Marks showed me that theĪnswer isn+ 1.The vectors from the center of a regular simplex inR wis between− 15 and 15 by the Schwarz inequality.Ģ8 Three vectors in the plane could make angles greater than 90Įxample(1,0),(− 1 ,4),(− 1 ,−4).)Ģ7 The length‖v−w‖is between 2 and 8 (triangle inequality when‖v‖= 5and‖w‖=ģ ). Subtractingcvis the key to constructingĢ6 The vectorsw= (x, y)with(1,2) thesew’s fill half of 3 -dimensional space.ġ2 (1,1)perpendicular to(1,5)−c(1,1)if(1,1) w= 0: perpendicular!ġ0 Slopes 2 / 1 and− 1 / 2 multiply to give− 1 : thenv.ĩ Ifv 2 w 2 /v 1 w 1 =− 1 thenv 2 w 2 =−v 1 w 1 orv 1 w 1 +v 2 w 2 =v.Orπ/ 2 radians (c) cosθ = 2/(2)(2) = 1/ 2 soθ = 60Ĩ (a) False:vandware any vectors in the plane perpendicular tou (b) True:u All vectors perpendicular to(1, 1 ,1)and(1, 2 ,3) Whole plane of vectors perpendicular tou 2, and a whole circle of unit vectors in thatĦ All vectorsw= (c, 2 c)are perpendicular tov. Perpendicular tou 1 (and so is(− 3 ,1)/ √ Four vectors whose combinations fill 4 -dimensional space: one example is the This is true even ifu= 0 theģ0 The combinations ofvandwfill the planeunlessvandwlie on the same line through (and infinitely many) combinations will produceb. Three lie on a line that does not containb.Yes, if one combination producesbthen two V.No, three vectorsu,v,win thex-yplane could fail to producebif all Add to find 2 v= (6, 10 ,14)Ģ9 Two combinations out of infinitely many that produceb= (0,1)are− 2 u+vand The six equations come from theĬomponents ofv+w= (4, 5 ,6)andv−w= (2, 5 ,8). = 16corners and 2 ♴ = 8three-dimensional facesĪnd 24 two-dimensional faces and 32 edges in Worked Example2.4 A.Ģ8 There are 6 unknown numbersv 1, v 2, v 3, w 1, w 2, w 3. So the sum of twelve vectorsĢ6 Two equations come from the two components:c+ 3d= 14and 2 c+d= 8.The 2 : 00 is 30ġ4 Moving the origin to 6 : 00 addsj= (0,1)to every vector. The center point is(ġ2 The combinations ofi= (1, 0 ,0)andi+j= (1, 1 ,0)fill thexyplane inxyzspace.ġ3 Sum=zero vector. Three possible parallelograms!ġ0 i−j= (1, 1 ,0)is in the base (x-yplane).i+j+k= (1, 1 ,1)is the opposite cornerįrom(0, 0 ,0). Adding diagonals gives 2 v(or 2 w).ĩ The fourth corner can be(4,4)or(4,0)or(− 2 ,2). If we took all whole numberscandd, the lattice would lie over the whole plane.Ĩ The other diagonal isv−w(or elsew−v). There is no solution tocv+dw= (3, 3 ,6)ħ The nine combinationsc(2,1) +d(0,1)withc= 0, 1, 2 andd= (0, 1 ,2)will lie on a Stated another way:u=−v−wis in the plane ofvandw.Ħ The components of everycv+dwadd to zero because the components ofvand ofwĪdd to zero.c= 3andd= 9give(3, 3 ,−6). The vectorsu,v,ware in the same plane because a combination gives
video lectures: /∼gs email: Wellesley - Cambridge Press Box 812060 Wellesley, Massachusetts 02482 Problem Set 1.1, page 8Ģ v+w= (2,3)andv−w= (6,−1)will be the diagonals of the parallelogram withģ This problem gives the diagonalsv+wandv−wof the parallelogram and asks for
#Linear algebra strang 5th pdf manual
INTRODUCTION TO LINEAR ALGEBRA Fifth Edition MANUAL FOR INSTRUCTORS Gilbert Strang Massachusetts Institute of Technology /linearalgebra /18.
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