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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 235 Assignment 0 Solutions 1. Determine proj ~x ~v and perp ~x ~v where a) ~x = (2 , 3 ,- 2) and ~v = (4 ,- 1 , 3). Solution: proj ~x ~v = (4 ,- 1 , 3) (2 , 3 ,- 2) k (2 , 3 ,- 2) k 2 (2 , 3 ,- 2) =- 1 17 (2 , 3 ,- 2) = (- 2 17 ,- 3 17 , 2 17 ). b) ~x = (- 1 , 2 , 1 ,- 3) and ~v = (2 ,- 1 , 2 , 1). Solution: proj ~x ~v = (2 ,- 1 , 2 , 1) (- 1 , 2 , 1 ,- 3) k (- 1 , 2 , 1 ,- 3) k 2 (- 1 , 2 , 1 ,- 3) =- 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 k ~x k 2 ~x ~v- ~v ~x k ~x k 2 ~x = ~v ~x k ~x k 2 ( ~x ~v )- ~v ~x k ~x k 2 2 ( ~x ~x ) = ( ~v ~x ) 2 k ~x k 2- ( ~v ~x ) 2 k ~x k 4 k ~x k 2 = ( ~v ~x ) 2 k ~x k 2- ( ~v ~x ) 2 k ~x k 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 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 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 2 So the general solution is ~x = i- i + 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 ~x,~ y R 3 and k R . Then by using properties of the dot product we get proj (2 , 2 ,- 1) ( k~x + ~ y ) = ( k~x + ~ y ) (2 , 2 ,- 1) k (2 , 2 ,- 1) k 2 (2 , 2 ,- 1) = k ~x (2 , 2 ,- 1) k (2 , 2 ,- 1) k 2 (2 , 2 ,- 1) + ~ y (2 , 2 ,- 1) k (2 , 2 ,- 1) k 2 (2 , 2 ,- 1) = k proj (2 , 2 ,- 1) ~x + proj ( , 2 ,- 1) ~ y Hence, proj (2 , 2 ,- 1) is linear. We have proj (2 , 2 ,- 1) (1 , , 0) = (1 , , 0) (2 , 2 ,- 1) k (2 , 2 ,- 1) k 2 (2 , 2 ,- 1) = 2 9 (2 , 2 ,- 1) = ( 4 9 , 4 9 ,- 2 9 ) proj (2 , 2 ,- 1) (0 , 1 , 0) = (0 , 1 , 0) (2 , 2 ,- 1) k (2 , 2 ,- 1) k 2 (2 , 2 ,- 1) = 2 9 (2 , 2 ,- 1) = ( 4 9 , 4 9 ,- 2 9 ) proj (2 , 2 ,- 1) (0 , , 1) = (0 , , 1) (2 , 2 ,- 1) k (2 , 2 ,- 1) k 2 (2 , 2 ,- 1) =- 1 9 (2 , 2 ,- 1) = (- 2 9 ,- 2 9 , 1 9 ) 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 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 ( k~x + ~ 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 ) ( ka + a 1 ) + ( kb + b 1 ) = k...
View Full Document

Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online