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

MATH221 December 2007

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Question 08 (b)
Let $T:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be the linear map given by rotation clockwise by 5 degrees. Find a basis of $\mathbb {R} ^{2}$ consisting of eigenvalues of T, or explain why this is not possible.

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. There is no basis of $\mathbb {R} ^{2}$ consisting of eigenvectors of T. This is because T has no eigenvectors. Why? If v were an eigenvector of T, then $T(v)=\lambda v$ for some $\lambda \in \mathbb {R}$. As T is a rotation, it does not change the length of v, so the only possible values for $\lambda$ are $\pm 1$. This implies T(v) lies either parallel to v (if $\lambda =1$) or anti-parallel (if $\lambda =-1$). But this is impossible, as T(v) is a clockwise rotation by 5 degrees. Therefore T has no eigenvectors.

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