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I'm trying to solve for the eigenvalues of and the eigenvectors of a rotation matrix (about the z-axis):

$$A = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Using $\det (\lambda I - A) = 0$, I determined that $\lambda \in \{1, \cos\theta\}$ (though I'm not certain that that is 100% correct). To solve for the eigenvectors, I decided that it would be easiest to put the following matrix in reduced row echelon form.

$$\begin{bmatrix} 1 - \cos\theta & \sin\theta & 0 \\ -\sin\theta & 1 - \cos\theta & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

My TI-89 suggests that this matrix in reduced row echelon form is as follows:

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

But I'm not entirely sure which trigonometric identities that it's using to figure that out. Any ideas?

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The characteristic polynomial is $$ \det(\lambda I-A)= \det\begin{bmatrix} \lambda-\cos\theta & \sin\theta & 0 \\ -\sin\theta & \lambda-\cos\theta & 0 \\ 0 & 0 & \lambda-1 \end{bmatrix} = (\lambda-1) \det\begin{bmatrix} \lambda-\cos\theta & \sin\theta\\ -\sin\theta & \lambda-\cos\theta \end{bmatrix} $$ and, finally, $$ (\lambda-1)(\lambda^2-2\lambda\cos\theta+1) $$ and the eigenvalues are $1$, $\cos\theta+i\sin\theta$ and $\cos\theta-i\sin\theta$.

An eigenvector for $1$ is readily computable as $[0\ 0\ 1]^T$, but if you're interested in finding a row echelon form of $$ \begin{bmatrix} 1-\cos\theta & \sin\theta & 0 \\ -\sin\theta & 1-\cos\theta & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ you can do as follows; set $\theta=2\varphi$, so $1-\cos\theta=2\sin^2\varphi$ and $\sin\theta=2\sin\varphi\cos\varphi$.

The case in which $\sin\varphi=0$ must be dealt with separately; assume $\sin\varphi\ne0$: \begin{align} \begin{bmatrix} 2\sin^2\varphi & 2\sin\varphi\cos\varphi & 0 \\ -2\sin\varphi\cos\varphi & 2\sin^2\varphi & 0 \\ 0 & 0 & 0 \end{bmatrix} &\to \begin{bmatrix} 1 & \cot\varphi & 0 \\ -2\sin\varphi\cos\varphi & 2\sin^2\varphi & 0 \\ 0 & 0 & 0 \end{bmatrix} &&R_1\gets\frac{1}{2\sin^2\varphi}R_1 \\[6px]&\to \begin{bmatrix} 1 & \cot\varphi & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} &&R_2\gets R_2+2R_1\sin\varphi\cos\varphi \\[6px]&\to \begin{bmatrix} 1 & \cot\varphi & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} &&R_2\gets \frac{1}{2}R_2 \\[6px]&\to \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} &&R_1\gets R_1-R_2\cot\varphi \end{align}

The case $\sin\varphi=0$ corresponds to $\theta=2k\pi$, when the original matrix is the identity.

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first of all, I believe your rotation matrix is incorrect: your rotation matrix is simply collapsing the $z$-axis, while it should leave the axis unchanged. The correct matrix would be

$$ R(\theta) = \begin{bmatrix} cos(\theta) & \sin(\theta)& 0 \\ -sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Notice that the last column is $$ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$

which means it leaves the last coordinate of your vector unchanged.

Now, as for the eigenvalues and eigenvectors, in this case, there's a super intuitive way to look at it without the mathematical machinery that I will describe first.

What does an eigenvector $\vec{v}$ do? It scales itself when acted on by $R(\theta)$. That is

$$R(\theta) \vec v = \lambda \vec v, \lambda \in \mathbb{R}$$

However, when you rotate the plane, no vector gets scaled - every vector moves around (rotates). Hence, the general rotation matrix has no eigenvectors.

However, when $\theta = 0$, the rotation matrix $R(\theta)$ is equal to the identity matrix $I$, where every vector is an eigenvector. We know that the identity matrix has eigenvalue $1$, so

$$R(0^\circ) \ \text{has eigenvalue} \ 1$$

You Another special case is when $\theta = 180^circ$, then this rotation corresponds to simply negating any vector $\vec v$. This has an eigenvector of $-1$, since every vector simply gets negated when you rotate the plane by $180^\circ$.

Hence,

$$R(180^\circ) \ \text{has eigenvalue} \ -1$$

Rotations where $\theta \neq 0^\circ, \neq 180^\circ$ have no eigenvalues

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  • $\begingroup$ Ah. Yeah, that was a typo. Fixed. “What does an eigenvector $\vec{v}$ do? It scales itself when acted on by $R(\theta)$. […] However, when you rotate the plane, no vector gets scaled - every vector moves around (rotates). Hence, the general rotation matrix has no eigenvectors.” Would it be possible for the eigenvector to be a function of $\theta$? $\endgroup$ – Tyler Crompton May 4 '16 at 9:50
  • $\begingroup$ Yes, they are "functions of $\theta$" but probably not in the way you are thinking. $R(\theta)$ is a function that creates matrices. For every $\theta_0 \in \mathbb{R}$, there is a matrix $M = R(\theta_0)$, and this matrix $M$ has eigenvectors (or may not). Hence, in some sense, the eigenvectors are a "function of theta", but they are constant for a fixed $\theta_0$. In this case, the only $\theta_0$ that yield eigenvectors are $\theta_0 = 0^\circ, 180^\circ$ $\endgroup$ – Siddharth Bhat May 4 '16 at 9:53
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First, your eigenvalues are wrong

$det(xI - A) = - x( (\cos(\theta) - x)^2 + \sin(\theta)^2) = - x(1 + x^2 -2x\cos(\theta)) = -x(x-e^{i\theta})(x-e^{-i\theta}) $

Which gives as eigenvalues $0,e^{i\theta},e^{-i\theta} $.

For the eigenvectors, if there are no obvious solutions, calculate a basis of $\ker(A-\lambda I)$

PS : Please, in the future, avoid stuff like "RREF", we are not supposed to guess what it means. Besides, not all people on this forum are native english speakers

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  • $\begingroup$ Do you have a resource for further reading on how you got those particular eigenvalues? The Wikipedia briefly mentions it but doesn't even go into as much detail as you did. $\endgroup$ – Tyler Crompton May 4 '16 at 10:03

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