# How to show eigenvalues of any rotation matrix of $3 \times 3$ are $\{ 1, e^{i\theta}, e^{-i\theta}\}$ [duplicate]

Eigenvalues of any rotation matrix of $3 \times 3$ are $\{ 1, e^{i\theta}, e^{-i\theta}\}$ where $\theta$ is the rotation angle, and corresponding eigenvectors are ${a,I,J}$ which are rotation axis and circular points for the plane orthogonal to $a$ respectively. In a 2-dimensional projective plane $\Bbb P^2$, $I,J$ can be written as $$I=[1,i,0]^T,J=[1,-i,0]^T$$ To my understanding, $Ra=a$ accounts for eigenvalue 1 and corresponding eigenvector $a$, but I cannot figure out other cases. Can anyone give a brief proof?

Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision at page 628.

## marked as duplicate by John Hughes, amd, Community♦Dec 25 '17 at 2:50

• Hint: What is the matrix for your rotation map with respect to the basis $\{a,I,J\}$? – Ted Shifrin Dec 18 '17 at 3:42
• In fact,I can figure out that in simple case, such as rotating about z-axis in 3d space. Here rotation matrix is like Givens rotation $R_z(\theta)$.But I cannot prove the general case. – Finley Dec 18 '17 at 3:51
• Hint: what Ted said: "with respect to the basis $\{a, I, J\}$." – John Hughes Dec 18 '17 at 4:11
• Further hint: being able to write the matrix of a transformation wrt a given basis is a really good skill for anyone doing vision or graphics...so if you can't yet do that, now's a great time to grab a linear algebra text and learn how. It'll be in the chapter on linear transformations, or one or two after that. – John Hughes Dec 18 '17 at 4:21
• @JohnHughes thanks for your patience and informative advice :) – Finley Dec 18 '17 at 4:30

Decomposing a matrix into its eigenvectors and eigenvalues is hard. But for the special case of a rotation around the $z$ axis, you already have the eigenvectors and eigenvalues, so you merely have to multiply them to confirm this is the rotation you're after:
$$Q\Lambda Q^{-1}=\\ \begin{pmatrix}1&1&0\\-i&i&0\\0&0&1\end{pmatrix} \begin{pmatrix}e^{i\theta}&0&0\\0&e^{-i\theta}&0\\0&0&1\end{pmatrix} \left(\begin{pmatrix}1&1&0\\-i&i&0\\0&0&1\end{pmatrix}^{-1}\right)=\\ \frac12\begin{pmatrix}1&1&0\\-i&i&0\\0&0&1\end{pmatrix} \begin{pmatrix}e^{i\theta}&0&0\\0&e^{-i\theta}&0\\0&0&1\end{pmatrix} \begin{pmatrix}1&i&0\\1&-i&0\\0&0&2\end{pmatrix}=\\ \begin{pmatrix} \frac{e^{i\theta}+e^{-i\theta}}{2} & -\frac{e^{i\theta}-e^{-i\theta}}{2i} & 0 \\ \frac{e^{i\theta}-e^{-i\theta}}{2i} & \frac{e^{i\theta}+e^{-i\theta}}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}= \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} = R_z(\theta)$$
The columns of the first matrix are the eigenvectors, the second matrix has the corresponding eigenvalues, and the third is the inverse of the first. It came as something of a surprise to me that the positive angle is associated with $J$ not $I$, but unless I messed up some sign somewhere this is the case. Multiplying values and taking the complex definitions of triginometric functions into account, you get the rotation as expected.
Quick sanity check: try $\theta=\frac\pi2$ for the first eigenvector:
$$\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix} \begin{pmatrix}1\\-i\\0\end{pmatrix}= \begin{pmatrix}i\\1\\0\end{pmatrix}= i\begin{pmatrix}1\\-i\\0\end{pmatrix}= e^{i\pi/2}\begin{pmatrix}1\\-i\\0\end{pmatrix}$$