Science:Math Exam Resources/Courses/MATH221/December 2007/Question 08 (a)
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Question 08 (a) 

Let be the linear map given by reflection across the line . 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 (a)/Hint 1 
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Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. Since T is a reflection, it does not change the length of any vector. Therefore, 1 and 1 are the only possible eigenvalues for T. The eigenvalue 1 corresponds to a vector which remains invariant under T, and therefore these vectors must lie on the line of reflection . In particular, we see that is an eigenvector for T corresponding to the eigenvalue 1, as it lies on the line . The eigenvalue 1 corresponds to a 180 degree reversal in orientation; these vectors must therefore lie on the line perpendicular to the line of reflection, that is, any vector lying on the line . Since lies on the line , we conclude that it is an eigenvector corresponding to the eigenvalue 1. We therefore have two (linearly independent) eigenvectors for T, and , which form a basis for . 