practiceprob10_soln

practiceprob10_soln - Math 235 Assignment 10 Practice...

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Unformatted text preview: Math 235 Assignment 10 Practice Problems 1. Suppose that a real 2 × 2 matrix A has 2 + i as an eigenvalue with a corresponding eigenvector 1 + i i . Determine A . Solution: Since A is real we know that the other eigenvalue of A is 2- i . Moreover, by the result of Assignment 10 #6, we get that an eigenvector corresponding to 2- i is the conjugate of the eigenvector corresponding to 2 + i . Hence, A is diagonalized by P = 1 + i 1- i i- i to D = 2 + i 2- i . Then P- 1 AP = D , so A = PDP- 1 = 1 + i 1- i i- i 2 + i 2- i 1- 2 i- i- 1 + i- i 1 + i = 1 2- 1 3 2. Let L : C 2 → C 2 be defined by L ( z 1 ,z 2 ) =- iz 1 + (1 + i ) z 2 (- 1 + i ) z 1- 2 iz 2 . a) Prove that L is linear and find the standard matrix of L . Solution: Let ~ z = z 1 z 2 and ~ y = y 1 y 2 and α,β ∈ C . Then L ( α~ z + β~ y ) = L αz 1 + βy 1 αz 2 + βy 2 =- i ( αz 1 + βy 1 ) + (1 + i )( αz 2 + βy 2 (- 1 + i )( αz 1 + βy 1 )- 2 i ( αz 2 + βy 2 ) = α- iz 1 + (1 + i ) z 2 (- 1 + i ) z 1- 2 iz 2 + β- iy 1 + (1 + i ) y 2 (- 1 + i ) y 1- 2 iy 2 = αL ( ~ z ) + βL ( ~ y ) Thus L is linear. We have L (1 , 0) =- i- 1 + i and L (0 , 1) = 1 + i- 2 i . Hence [ L ] =- i 1 + i- 1 + i- 2 i . b) Determine the range and kernel of L . Solution: Row reducing [ L ] we get- i 1 + i- 1 + i- 2 i ∼ 1- 1 + i A basis for ker L is the general solution of [ L ] ~x = ~ 0, hence a basis is 1- i 1 . The range of L is equal to the columnspace of [ L ]. Thus, a basis for the range of L is- i- 1 + i . 2 3. For each of the following matrices, either diagonalize the matrix over C , or show that it is not diagonalizable....
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This note was uploaded on 01/21/2012 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

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practiceprob10_soln - Math 235 Assignment 10 Practice...

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