A0_soln - Math 235 Assignment 0 Solutions 1 Determine projv...

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Math 235 Assignment 0 Solutions 1. Determine proj v x and perp v x where a) v = 2 3 - 2 and x = 4 - 1 3 . Solution: proj x v = x · v v 2 v = - 1 17 2 3 - 2 = - 2 / 17 - 3 / 17 2 / 17 b) v = - 1 2 1 - 3 and x = 2 - 1 2 1 .. Solution: proj x v = x · v v 2 2 v = - 1 15 - 1 2 1 - 3 = 1 / 15 - 2 / 15 - 1 / 15 3 / 15 . 2. Prove algebraically that proj x ( v ) and perp x v are orthogonal. Solution: We have proj x ( v ) · perp x v = v · x x 2 x · v - v · x x 2 x = v · x x 2 ( x · v ) - v · x x 2 2 ( x · x ) = ( v · x ) 2 x 2 - ( v · x ) 2 x 4 x 2 = ( v · x ) 2 x 2 - ( v · x ) 2 x 2 = 0 . Hence, they are orthogonal. 3. Solve the system z 1 - z 2 + iz 3 = 2 i (1 + i ) z 1 - iz 2 + iz 3 = - 2 + i (1 - i ) z 1 + ( - 1 + 2 i ) z 2 + (1 + 2 i ) z 3 = 3 + 2 i
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2 Solution: Making an augmented matrix and row-reducing we get 1 - 1 i 2 i 1 + i - i i - 2 + i 1 - i - 1 + 2 i 1 + 2 i 3 + 2 i 1 0 1 + i i 0 1 1 - i 0 0 0 0 x 3 does not have a leading one so let x 3 = t C . Then we have x 1 = i - (1 + i ) t x 2 = - i - t x 3 = t So the general solution is x = i - i 0 + t - 1 - i - 1 1 . 4. Prove each of the following mappings are linear and find the standard matrix of each. a) proj (2 , 2 , - 1) . Solution: Let n = 2 2 - 1 and let x, y R 3 and k R . Then by using properties of the dot product we get proj (2 , 2 , - 1) ( kx + y ) = ( kx + y ) · n ) n 2 n = k x · n n 2 n + y · n n 2 2 n = k proj n x + proj n y Hence, proj (2 , 2 , - 1) is linear. We have proj (2 , 2 , - 1) e 1 = e · n ) n 2 n = 2 9 2 2 - 1 = 4 / 9 4 / 9 - 2 / 9 proj (2 , 2 , - 1) e 2 = e · n ) n 2 n = 2 9 2 2 - 1 = 4 / 9 4 / 9 - 2 / 9 proj (2 , 2 , - 1) e 3 = e · n ) n 2 n = - 1 9 2 2 - 1 = - 2 / 9 - 2 / 9 1 / 9
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3 Hence [proj (2 , 2 , - 1) ] = 4 / 9 4 / 9 - 2 / 9 4 / 9 4 / 9 - 2 / 9 - 2 / 9 - 2 / 9 1 / 9 . b) L ( ax 2 + bx + c ) = a b + c 0 a + b . Solution: Let x = ax 2 + bx + c and y = a 1 x 2 + b 1 x + c 1 and k R , then L ( kx + y ) = L ( kax 2 + kbx + kc + a 1 x 2 + b 1 x + c 1 ) = L (( ka + a 1 ) x 2 + ( kb + b 1 ) x + ( kc + c 1 )) = ( ka + a 1 ) ( kb + b 1 ) + ( kc + c 1 ) 0 ( ka + a 1 ) + ( kb + b 1 ) = k a b + c 0 a + b + a 1 b 1 + c 1 0 a 1 + b 1 = kL ( x ) + L ( y ) Hence, L is linear.
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