practiceprob10_soln

# practiceprob10_soln - Math 235 Assignment 10 Practice...

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

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 0 0 2 - i . Then P - 1 AP = D , so A = PDP - 1 = 1 + i 1 - i i - i 2 + i 0 0 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 0 0 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 .

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

View Full Document
2 3. For each of the following matrices, either diagonalize the matrix over C , or show that it is not diagonalizable. (a) 2 2 - 1 - 4 1 2 2 2 - 1 Solution: We have C ( λ ) = 2 - λ 2 - 1 - 4 1 - λ 2 2 2 - 1 - λ = - λ 0 λ - 4 1 - λ 2 2 2 - 1 - λ = 0 0 λ - 2 1 - λ 2 1 - λ 2 - 1 - λ = - λ ( λ 2 - 2 λ + 5) Hence, the eigenvalues are λ 1 = 0 and the roots of λ 2 - 2 λ + 5. By the quadratic formula,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern