Probs1 - Physics 316, Problem Set 1, due Tuesday October 4...

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Physics 316, Problem Set 1, due Tuesday October 4 in class 1. Schwarz’s inequality: A vector space V has a norm, with the properties listed in the Glossary (see materials in web site). Show that its natural scalar product ( x , y ) satisfies ( x , y ) 2 ( x , x )( y , y ). This is known as the Schwartz inequality. Solution: ( x , y )= 1 2 || x + y 2 -| | x 2 | y 2 · We observe that ( x , x x 2 for any x . According to the triangle inequality, x + y | |≤| | x + y x + y 2 ≤| | x 2 + y 2 +2 x || || y Expressing x + y using the definition of ( x , y ), 2( x , y )+ x 2 + x 2 = x + y 2 | x 2 + y 2 x y 2( x , y ) 2 x y dividing by 2 and squaring. .. ( x , y ) 2 ( x , x )( y , y ) QED 2. Affine transformations and linear transformations: A certain affine space is also a vector space, and a line in this vector space is given by y ( η ) b + η x . a) Find the analogous expression for a line passing through the origin ( 0 ) of the vector space that is parallel to the given line . b) A transformation f : x R 2 y R 2 in a two-dimensional space R 2 is defined by ( y 1 ,y 2 ab cd ( x 1 ,x 2 )+( g 1 ,g 2 ) , where a...d , g 1 and g 2 are constants. Show that the image of the generic line defined in a) is also a line. Solution: a) Let’s call the desired line passing through the origin z ( η ). If z = b 0 + η x 0 includes the origin, then b 0 must be 0 . To make z be parallel to y we can take x 0 = x : there is no value of η,η 0 such that y ( η z ( η 0 ) thus the two lines don’t intersect. The same is true if x 0 is a scalar multiple of x . b) Since x ( η ) is a line, it can be expressed x ( η b + η u . Using the component language x =( x 1 2 ) , this means ( x 1 2 )( η )=( b 1 + ηu 1 ,b 2 + 2 ). We need to show that ( y 1 2 )( η ) can be expressed in the form B + η U , or in component language, ( y 1 2 B 1 + ηU 1 ,B 2 + 2 ). On the other hand the formula says ( y 1 2 ( b 1
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Probs1 - Physics 316, Problem Set 1, due Tuesday October 4...

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