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WINTER 2007 assign6-sol

# WINTER 2007 assign6-sol - Math 235 Assignment 6 Due 9:15am...

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Math 235 Assignment 6 Due 9:15am, Wednesday March 7, 2007. 1. From the Text § 6.2 #10. Show that { u 1 , u 2 , u 3 } is orthogonal basis for R 3 . Then express x as a linear combination of the u ’s. u 1 = 3 - 3 0 , u 2 = 2 2 - 1 , u 3 = 1 1 4 , x = 5 - 3 1 . Solution. u 1 · u 2 = u 1 T u 2 = 3 * 2 + ( - 3) * 2 + 0 * ( - 1) = 0 u 1 · u 3 = u 1 T u 3 = 3 * 1 + ( - 3) * 1 + 0 * 4 = 0 u 2 · u 3 = u 2 T u 3 = 2 * 1 + 2 * 1 + ( - 1) * 4 = 0 Thus { u 1 , u 2 , u 3 } is an orthogonal set. Since the set consists of linearly independent vectors, the three vectors form a basis for R 3 . x = x · u 1 u 1 · u 1 u 1 + x · u 2 u 2 · u 2 u 2 + x · u 3 u 3 · u 3 u 3 = 5 * 3 + ( - 3) * ( - 3) + 1 * 0 3 * 3 + ( - 3) * ( - 3) + 0 * 0 u 1 + 5 * 2 + ( - 3) * 2 + 1 * ( - 1) 2 * 2 + 2 * 2 + ( - 1) * ( - 1) u 2 + 5 * 1 + ( - 3) * 1 + 1 * 4 1 * 1 + 1 * 1 + 4 * 4 u 3 = 4 3 u 1 + 1 3 u 2 + 2 3 u 3 . #34. Given u = 0 in R n , let L =Span { u } . For y in R n , the reflection of y in L is the point r ( y ) defined by r ( y ) = 2 · proj L y - y . Show that the mapping y -→ r ( y ) is a linear transformation. 1

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Solution. We need to show that r ( c y + d x ) = c * r ( y ) + d * r ( x ) , where c and d are scalars.
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