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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 ( sx + 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 ) = s x 1 - x 2 x 1 - 2 x 3 + y 1 - y 2 y 1 - 2 y 3 = sf ( x ) + f ( y ) (ii) f (1 , 0 , 0) = 1 1 , f (0 , 1 , 0) = - 1 0 , f (0 , 0 , 1) = 0 - 2 . Hence [ f ] = 1 - 1 0 1 0 - 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 = x 1 1 1 + x 2 - 1 0 + x 3 0 - 2 . Thus Range( f ) = span 1 1 , - 1 0 , 0 - 2 = 1 1 , - 1 0 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 ( sx + y ) = f sx 1 + y 1 sx 2 + y 2 = 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) f (1 , 0) = 0 0 1 1 , f (0 , 1) = 1 0 - 1 0 . Hence [ f ] = 0 1 0 0 1 - 1 1 0 .
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