A10_soln

# A10_soln - Math 235 Assignment 10 Solutions z1(1 i)z2(1...

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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 α, β C . Then L ( α~ z + β ~w ) = L ( αz 1 + βw 1 , αz 2 + βw 2 ) = αz 1 + βw 1 + (1 + i )( αz 2 + βw 2 ) (1 + i )( αz 1 + βw 1 ) + 2 i ( αz 2 + βw 2 ) = α z 1 + (1 + i ) z 2 (1 + i ) z 1 + 2 iz 2 + β w 1 + (1 + i ) w 2 (1 + i ) w 1 + 2 iw 2 = αL ( ~ z ) + βL ( ~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 0 0 = 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 0 0 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 .

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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 1 + (1 + i ) z 2 1 1 + i . Hence, if we take ~ z 2 = 1 1 + i , we get L ( ~ z 2 ) = 1 + (1 + i )(1 + i ) 1 1 + i = (1 + 2 i ) 1 1 + i Thus, if we take B = { ~ z 1 , ~ z 2 } we get L ( ~ z 1 ) = ~ 0 = 0 ~ z 1 + 0 ~ z 2 L ( ~ z 2 ) = (1 + 2 i ) ~ z 2 = 0 ~ z 1 + (1 + 2 i ) ~ z 2 Hence, [ L ] B = 0 0 0 1 + 2 i 2. Let Z
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