Isometry without eigenvalues So I'm tasked with finding an $R^2$ isometry that has no eigenvalues in $\mathbb{R}$. Trouble is, well, I don't really "get" isometries as they pertain to eigenvalues and transformation matrices. I could trivially come up with...say the following matrix:
$$\begin{bmatrix}1 & -1\\
\frac{1}{4} & 1\end{bmatrix}$$
Now the eigenvalue would be:
\begin{align}
(1 - \lambda)^2 - (-\frac{1}{2}) &= 0\\
\frac{5}{4} - 2\lambda + \lambda^2 &= 0\\
\lambda &= \frac{2\pm\sqrt{(-2)^2-4\cdot\frac{5}{4}}}{2}\\
\lambda &= \frac{2\pm\sqrt{4-5}}{2}
\end{align}
That would have a complex eigenvalue (of $1 \pm \frac{i}{2}$), but that's very certainly not of absolute value 1. Since $a$ in the equation is always 1, I guess I'd need to come up with numbers that result in $\sqrt{2} + \sqrt{2}i$ so that the absolute value would indeed be 1.
However, that doesn't really have much to do with the definition of an isometry, which is defined through inner product space: I can't really find the link between the eigenvalue and the inner product space. I guess my question is, how do I go about formulating an isometry that also fills all these conditions?
 A: Let $f\colon\Bbb R^2\longrightarrow\Bbb R^2$ be an isometry. Then both $f(1,0)$ and $f(0,1)$ have norm $1$. Also, since isometries preserve angles, $f(1,0)$ and $f(0,1)$ are orthogonal. So, the matrix of $f$ with respect to the standard basis is either of the form$$\begin{bmatrix}a&b\\b&-a\end{bmatrix}\tag1$$or of the form$$\begin{bmatrix}a&-b\\b&a\end{bmatrix},\tag2$$with $a^2+b^2=1$. If it is of the form $(1)$, then $f$ has real eigenvalues: $\pm\sqrt{a^2+b^2}$. On the other hand, if it has the form $(2)$, then its eigenvalues are $a\pm bi$, which are real if and only if $b=0$ (that is, if and only if $f=\pm\operatorname{Id}_2$).
A: Rotation by $90^{0}$ is an isometry which has no real eigen values.  If $x$ is vector such that its rotation by $90^{0}$ is parallel to $x$ then $x$ must be $0$. This Linear transformation is of the form $T(x,y)=(-y,x)$. The matrix is $A =\begin{bmatrix} 0 \,-1\\ 1\,\, \, \, 0 \end{bmatrix}  $
A: You might be interested in the three reflections theorem$^*$, which states that every isometry is a product of at most three reflections, and which can be used to completely classify isometries of the plane:

An isometry is either a reflection or glide reflection, a translation, or a rotation.

Hence, every isometry has one of the following:

*

*A line of fixed points.

*No fixed points and a single invariant line.

*No fixed points and a parallel family of invariant lines.

*A single fixed point.

From above, it is clear rotations about the origin, through an angle $\theta\neq 0\pmod \pi$, cannot have real eigenvalues. To prove this, let $$r_{\theta}=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} $$ be a rotation through $0\lt\theta\leq 2\pi$ radians, then $\lambda$ an eigenvalue means it satisfies the characteristic equation $$\det (r_{\theta}-\lambda\mathbf{1})=0\implies (\cos\theta - \lambda)^2 +\sin^2\theta=0$$ $$\implies \cos\theta - \lambda=\pm i\sin\theta$$$$\implies \lambda=cos\theta\pm i\sin\theta=e^{\pm i\theta},$$ which is real precisely when $\theta$ equals $\pi$ or $2\pi$.
It is natural to ask whether any other isometries have no real eigenvalues. Reflections and translations both have visible real eigenvalues, so we are only left with glide reflections. Appealing to the characteristic equation again, it turns out, if $$\overline{g}=\begin{bmatrix} a & b \\ c & d\end{bmatrix}$$ is the matrix of a glide reflection$,^{**}$ then the eigenvalues of $\overline{g}$ are $$\frac{a+d\pm\sqrt{(a+d)^2+4}}{2},$$ which are clearly real. Thus

The only isometries of the plane with no real eigenvalues are rotations about $O$ through an angle $\theta\neq 0\pmod{\pi}$.

$*$Check out e.g. Geometry of Surfaces by John Stillwell.
$**$ In order to express isometries like $\overline{g}$ as a single $2\times 2$ matrix, we need to "extend" the plane by adding a point at infinity to form what is called the extended or projective plane $\mathbb{CP}^1=\mathbb{C}\cup\{\infty\}$. This means the matrix $\overline{g}$ is actually an element of the projective special linear group $PSL(2,\mathbb{R})$.
