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Question 08 (b) 

Let be the linear map given by rotation clockwise by 5 degrees. Find a basis of consisting of eigenvalues of T, or explain why this is not possible. 
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Hint 

Science:Math Exam Resources/Courses/MATH221/December 2007/Question 08 (b)/Hint 1 
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Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. There is no basis of consisting of eigenvectors of T. This is because T has no eigenvectors. Why? If v were an eigenvector of T, then for some . As T is a rotation, it does not change the length of v, so the only possible values for are . This implies T(v) lies either parallel to v (if ) or antiparallel (if ). But this is impossible, as T(v) is a clockwise rotation by 5 degrees. Therefore T has no eigenvectors. 