Linear Algebra Problem 1 Math 310 Summer II 2016 Prof Bayly

Math 310 Summer II 2016 (Prof. Bayly) Homework 2 Please make your sketches on centimeter graph paper, which you can print out from d21 if you Remember that we now use the row-column formula iin to express the dot product of two vectors. Also Pythagoras\'s formula for the length of vector llri- vri T. Let i (1 2) (4,2) (a) Sketch ii and ii in your sketch? (b) Calculate llrll, lull, llii TII, and llii (c) Check that till s llum linl (soo-called \"Triangle Inequalities (d) Check that lliI2 (Pythagoras\'s Theorem since ii and are legs of a right triangle with hypotenuse u r.) (2) Let ii (1 2, 3, -4) ,E (2,1,4,3 Repeat question (1), except for the sketches of course (a) Find an explicit formula for f(z) (it should be quadratic). (b) Find the value z that minimizes f(r). Is Trmin) i a (c) sketch a,D, armin, and Trmin). Does rozmin) look (5) Let a (1,2,3)T,6 (3,2,1)T,E a (2,1,3) and I, y real variables. (a Find an explicit formula for f(r, y (it should be quadratic. (b) Find the value ymin. that minimizes z,y). Is r(zmin, ymin) a, b? What theorems in geometry are illustrated by the calculations in problems (3-5)?

Solution

u= (1,-2)^{T ,} x = (4,2)^T

u+x = (1+4 , -2+2)^T = (5,0)^T

U-X = (1-4, -2-2)^T = (-3,-4)^T

U^T X = (1,-2) . (4,2)^T = 4 -4 = 0 => U is perpendicular to x.

Also , ||u|| = sqrt{1+4} = sqrt{5 }

||x|| = sqrt{16+4} = sqrt{20} = 2 sqrt{5}

||u+x|| = sqrt{25+0} = 5

||u-x|| = sqrt{9+16} = sqrt{25} = 5

Also, ||u+x|| <= ||u|| + ||x|| [as 5 < 3sqrt{5}]

||u||²+ ||x||² = ||u-x||²   [ As 5+20 = 25 ]

Hence proved.