(I) Let $T_\theta$ be a linear transformation $T_\theta : \Bbb R^2 → \Bbb R^2$ which rotates the plane $\theta$ degree anti-clockwise. For which values of $\theta$ does $T_\theta$ have at least an eigenvalue?

(II) In those cases, please find the eigenvalues and the dimension of the associated eigenspaces.

Part (I) of the question I think I've already solved correctly, but I've included it here just in case I missed something. I believe I'm only stuck on the very last bit of finding eigenspaces. Any help is greatly appreciated!

My attempt (I): So, the rotation matrix is $$A=\begin{bmatrix}\cos \theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$$ and in particular, given that $T\mathbf{v} = \lambda\mathbf{v}$ , we have $\det(A-\lambda I) = 0$ where $I$ is the identity matrix.

This shows that $$det\bigg(\begin{bmatrix}\cos\theta-\lambda & -\sin\theta\\ \sin\theta & \cos\theta-\lambda\end{bmatrix}\bigg)=(\cos\theta-\lambda)^2 + \sin^2\theta=0,$$ which solving for $\theta$ yields $\theta=0$ and $\theta=\pi$ that have at least one eigenvalue.

(II): Plugging in $\theta=0$ and $\theta=\pi$ to the above equation yields $$(\cos 0-\lambda)^2+ \sin^20=0 \qquad\implies\qquad \lambda=1$$ and $$(\cos\pi-\lambda)^2+\sin^2\pi=0 \qquad\implies\qquad \lambda=-1.$$

Now, when i try to solve for the eigenspaces, I have $T\mathbf{v} = \lambda\mathbf{v}$, so for $\theta=0$ and $\lambda = 1$, $$\begin{bmatrix}1-\lambda & 0\\0 & 1-\lambda\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$

A similar situation happens for when $\theta=\pi$ and $\lambda=-1$ with the zero vector. This doesn't make any sense to me as an answer. What have I done wrong, or am I just not understanding what this answer is saying?

  • $\begingroup$ Your computation of $(cosθ-λ)^2 + sin^2θ=0$ should give a second degree equation in $\lambda$, from which you should conclude $\lambda=\pm e^{i\theta}$. Remember that $\theta$ is fixed and you want the $\lambda$s as function of $\theta$. $\endgroup$ – Jean Marie Feb 25 '16 at 11:14
  • $\begingroup$ @Apple I tried to improve a bit the Mathjax in your post. Please check that I did not introduce any mistake. $\endgroup$ – Surb Feb 25 '16 at 11:17
  • 1
    $\begingroup$ Ah I did the same. Is there a way to cancel my edit? $\endgroup$ – Shuri2060 Feb 25 '16 at 11:20
  • $\begingroup$ @QuestionAsker Done (sorry I rejected it). But I don't think that you can cancel your suggested edit. $\endgroup$ – Surb Feb 25 '16 at 11:21


You have just found that a rotation of $\theta =0$ ( or $\theta= 2n\pi$) is an identity, so all vectors are eigenvectors. And for $\theta=\pi +2n\pi$ the rotation is an inversion (with respect to the origin) of the vectors, so....


A vector is "Eigen" when its image is parallel to itself.

This only occurs for the null rotation ($\theta=0, Rv=v$) or the reflection ($\theta=\pi, Rv=-v$). In these cases, any vector is "Eigen".

  • $\begingroup$ So am I right in thinking that the eigenspaces are simply \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} for θ=0, λ=1 and \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix} for θ=π, λ=-1? $\endgroup$ – Apple Bottom Jeans Feb 25 '16 at 11:30
  • $\begingroup$ @AppleBottomJeans: no, both Eigenspaces are $\mathbb R^2$. $\endgroup$ – Yves Daoust Feb 25 '16 at 11:34
  • $\begingroup$ Oh duh. Of course. So it's just \begin{bmatrix}1 \\ 0\end{bmatrix} and \begin{bmatrix}0 \\ 1\end{bmatrix}? $\endgroup$ – Apple Bottom Jeans Feb 25 '16 at 11:36
  • $\begingroup$ @AppleBottomJeans: even less. Both are $\mathbb R^2$. $\endgroup$ – Yves Daoust Feb 25 '16 at 11:37
  • $\begingroup$ @AppleBottomJeans Did you mean eigenvector? $\endgroup$ – Shuri2060 Feb 25 '16 at 11:37

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