Isometry of space/Rotation axis/Complement/Unitary/Section

Theorem

Let

$\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3}$

be a linear isometry. Then there exists an eigenvector with eigenvalue

${}1$ or ${}-1$.

Proof

The characteristic polynomial ${}P$ of ${}\varphi$ is a normed polynomial of degree three. For ${}t\to +\infty$, we have ${}P(t)\to +\infty$, and for ${}t\to -\infty$, we have ${}P(t)\to -\infty$. Due to the intermediate value theorem, ${}P$ has a real zero. Such a zero is an eigenvalue of ${}\varphi$. Because of fact, this eigenvalue equals ${}1$ or ${}-1$.

$\Box$

A proper linear isometry of the space transforms the unit sphere into itself. One might imagine an isometry as a rotation of a ball lying in a suitable bowl.

Theorem

A proper isometry

$\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3}$

has an

eigenvector with eigenvalue ${}1$, that is, there exists a line (through the origin)

that is a fixed line for ${}\varphi$.

Proof

We consider the characteristic polynomial of ${}\varphi$, that is,

${}P(\lambda )=\det {\left(\lambda E_{3}-\varphi \right)}\,.$

This is a normed real polynomial of degree three. For ${}\lambda =0$, we have

${}P(0)=\det {\left(-\varphi \right)}=-\det \varphi =-1\,.$

As the polynomial ${}P(\lambda )\to \infty$ as ${}\lambda \to \infty$, there must be a positive ${}\lambda$ that is a zero of ${}P$. Due to fact, we must have ${}\lambda =1$.

$\Box$

Theorem

Let

$\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3}$

be a proper isometry. Then ${}\varphi$ is a rotation around a fixed axis. This means that ${}\varphi$ is described, with respect to a suitable orthonormal basis, by a matrix of the form

${\begin{pmatrix}1&0&0\\0&\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\0&\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}.$

Proof

Due to fact, there exists an eigenvector ${}u$ for the eigenvalue ${}1$. Let ${}U=\mathbb {R} u$ denote the corresponding line. This line is fixed and, in particular, invariant under ${}\varphi$. Because of fact, also the orthogonal complement ${}U^{\perp }$ is invariant under ${}\varphi$. That is, there exists a linear isometry

$\varphi _{2}\colon U^{\perp }\longrightarrow U^{\perp }$

that coincides on ${}U^{\perp }$ with ${}\varphi$. Here, ${}\varphi _{2}$ is proper. Therefore, by fact, ${}\varphi _{2}$ is a plane rotation. If we choose a vector of length one in ${}U$, and an orthonormal basis of ${}U^{\perp }$, then ${}\varphi$ has with respect to this basis the given matrix form.

$\Box$