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

MATH221 December 2007

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Question 08 (d)
Let $T:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be the linear map whose matrix is given by ${\begin{pmatrix}1&0\\2/3&1\end{pmatrix}}$ (a shear in the y-direction). 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. No such basis exists. Note that 1 is the only eigenvalue for T, as the characteristic polynomial is $(\lambda -1)^{2}$. To determine the eigenvectors corresponding to the eigenvalue 1, we solve the corresponding equation ${\begin{pmatrix}0&0\\2/3&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}.$ Thus ${\begin{pmatrix}x\\y\end{pmatrix}}$ is an eigenvector for T if and only if x = 0. This means that there is only one linearly independent eigenvector, namely ${\begin{pmatrix}0\\1\end{pmatrix}}$. As a basis of $\mathbb {R} ^{2}$ consists of two linearly independent vectors, there exists no basis consisting of eigenvectors of T.

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