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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.

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

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    $\begingroup$ Hint: What is the matrix for your rotation map with respect to the basis $\{a,I,J\}$? $\endgroup$ – Ted Shifrin Dec 18 '17 at 3:42
  • $\begingroup$ 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. $\endgroup$ – Finley Dec 18 '17 at 3:51
  • $\begingroup$ Hint: what Ted said: "with respect to the basis $\{a, I, J\}$." $\endgroup$ – John Hughes Dec 18 '17 at 4:11
  • $\begingroup$ 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. $\endgroup$ – John Hughes Dec 18 '17 at 4:21
  • $\begingroup$ @JohnHughes thanks for your patience and informative advice :) $\endgroup$ – Finley Dec 18 '17 at 4:30
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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.

And how about a rotation around some other axis? Well, the concept of eigenvectors and eigenvalues does not depend on the choice of coordinate system, as long as it's basis is orthonormal. So the above check is generic enough to cover all other situations, too.

Quick sanity check: try $\theta=\frac\pi2$ for the first eigenvector:

$$\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix} \begin{pmatrix}1\\-i\\0\end{pmatrix}= \begin{pmatrix}i\\1\\0\end{pmatrix}= i\begin{pmatrix}1\\-i\\0\end{pmatrix}= e^{i\pi/2}\begin{pmatrix}1\\-i\\0\end{pmatrix}$$

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