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

MATH221 December 2007

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Question 08 (a)
Let $T:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be the linear map given by reflection across the line $y={\frac {1}{2}}x$. 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. 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 $y={\frac {1}{2}}x$. In particular, we see that ${\begin{pmatrix}2\\1\end{pmatrix}}$ is an eigenvector for T corresponding to the eigenvalue 1, as it lies on the line $y={\frac {1}{2}}x$. 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 $y=-2x$. Since ${\begin{pmatrix}1\\-2\end{pmatrix}}$ lies on the line $y=-2x$, we conclude that it is an eigenvector corresponding to the eigenvalue -1. We therefore have two (linearly independent) eigenvectors for T, ${\begin{pmatrix}2\\1\end{pmatrix}}$ and ${\begin{pmatrix}1\\-2\end{pmatrix}}$, which form a basis for $\mathbb {R} ^{2}$.

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