This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 01/21/2012 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
 Spring '08
 CELMIN
 Math

Click to edit the document details