Reduced row echelon form of matrix with trigonometric expressions 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?
 A: 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
A: 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.
A: 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
