A10_soln - Math 235 Assignment 10 Solutions 1. Let L : C 2...

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Unformatted text preview: Math 235 Assignment 10 Solutions 1. Let L : C 2 C 2 be defined by L ( z 1 ,z 2 ) = z 1 + (1 + i ) z 2 (1 + i ) z 1 + 2 iz 2 . (a) Prove that L is linear. Solution: Let ~w,~ z C 2 and , C . Then L ( ~ z + ~w ) = L ( z 1 + w 1 ,z 2 + w 2 ) = z 1 + w 1 + (1 + i )( z 2 + w 2 ) (1 + i )( z 1 + w 1 ) + 2 i ( z 2 + w 2 ) = z 1 + (1 + i ) z 2 (1 + i ) z 1 + 2 iz 2 + w 1 + (1 + i ) w 2 (1 + i ) w 1 + 2 iw 2 = L ( ~ z ) + L ( ~w ) Hence, L is linear. (b) Determine a basis for the range and kernel of L . Solution: If ~ z = z 1 z 2 ker( L ), then = L ( ~ z ) = z 1 + (1 + i ) z 2 (1 + i ) z 1 + 2 iz 2 Hence, we need to solve the homogeneous system of equations z 1 + (1 + i ) z 2 = 0 (1 + i ) z 1 + 2 iz 2 = 0 Row reducing the corresponding coefficient matrix gives 1 1 + i 1 + i 2 i 1 1 + i Hence, a basis for ker( L ) is- 1- i 1 . Every vector in the range of L has the form z 1 + (1 + i ) z 2 (1 + i ) z 1 + 2 iz 2 = z 1 1 1 + i + z 2 1 + i 2 i = z 1 1 1 + i + (1 + i ) z 2 1 1 + i = z 1 + (1 + i ) z 2 1 1 + i Therefore, a basis for Range( L ) is 1 1 + i . 2 (c) Use the result of part (b) to define a basis B for C 2 such that [ L ] B is diagonal. Solution: If ~ z ker( L ), then L ( ~ z ) = ~ 0 = 0 ~ z . Consequently, we pick ~ z 1 =- 1- 1 1 . In part (b) we showed that L ( ~ z ) = z...
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A10_soln - Math 235 Assignment 10 Solutions 1. Let L : C 2...

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