assign5_soln

assign5_soln - Math 136 Assignment 5 Solutions 1. For each...

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Math 136 Assignment 5 Solutions 1. For each of the following mappings: (i) Prove that it is linear. (ii) Find the standard matrix of the mapping (iii) Find a spanning set for the nullspace. (iv) Find a spanning set for the range. a) f ( x 1 , x 2 , x 3 ) = ( x 1 - x 2 , x 1 - 2 x 3 ). (i) Let ~x, ~ y R 3 and s R , then f ( s~x + ~ y ) = f sx 1 + y 1 sx 2 + y 2 sx 3 + y 3 = ± ( sx 1 + y 1 ) - ( sx 2 + y 2 ) ( sx 1 + y 1 ) - 2( sx 3 + y 3 ) ² = s ± x 1 - x 2 x 1 - 2 x 3 ² + ± y 1 - y 2 y 1 - 2 y 3 ² = sf ( ~x ) + f ( ~ y ) (ii) f (1 , 0 , 0) = ± 1 1 ² , f (0 , 1 , 0) = ± - 1 0 ² , f (0 , 0 , 1) = ± 0 - 2 ² . Hence [ f ] = ± 1 - 1 0 1 0 - 2 ² . (iii) If f ( x 1 , x 2 , x 3 ) = (0 , 0), then we must have x 1 - x 2 = 0 and x 1 - 2 x 3 = 0. So, x 2 = x 1 and x 3 = 1 2 x 1 . Thus, if ~x Null( f ), then ~x = x 1 x 2 x 3 = x 1 x 1 1 2 x 1 = x 1 1 1 1 / 2 So Null( f ) = span 1 1 1 / 2 . (iv) For any ~x R 3 , we have f ( x 1 , x 2 , x 3 ) = ± x 1 - x 2 x 1 - 2 x 3 ² = x 1 ± 1 1 ² + x 2 ± - 1 0 ² + x 3 ± 0 - 2 ² . Thus Range( f ) = span ³± 1 1 ² , ± - 1 0 ² , ± 0 - 2 ²´ = ³± 1 1 ² , ± - 1 0 ²´ NOTE: Since Range( f ) = R 2 , we could pick any other basis for R 2 (for example the standard basis).
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2 b) f ( x 1 , x 2 ) = ( x 2 , 0 , x 1 - x 2 , x 1 ). (i) Let ~x, ~ y R 2 and s R , then f ( s~x + ~ y ) = f ±² sx 1 + y 1 sx 2 + y 2 ³´ = sx 2 + y 2 0 ( sx 1 + y 1 ) - ( sx 2 + y 2 ) sx 1 + y 1 = s x 2 0 x 1 - x 2 x 1 + y 2 0 y 1 - y 2 y 1 = sf ( ~x ) + f ( ~ y ) (ii)
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This note was uploaded on 11/25/2010 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.

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assign5_soln - Math 136 Assignment 5 Solutions 1. For each...

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