# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 34/refcontrol

    Diagonalizability of isometries over the complex numbers

## LemmaLemma 34.1 change

Let $\varphi \colon V\rightarrow V$ be a linear isometryMDLD/linear isometry (K) on a finite-dimensionalMDLD/finite-dimensional (vs) ${\mathbb {K} }$-vector spaceMDLD/vector space $V$, endowed with an inner product.MDLD/inner product Let $U\subseteq V$ denote an invariant linear subspace.MDLD/invariant linear subspace Then also the orthogonal complementMDLD/orthogonal complement $U^{\perp }$ is invariant. In particular, $\varphi$ can be written as a direct sumMDLD/direct sum
    $\varphi =\varphi _{U}\oplus \varphi _{U^{\perp }}\,,$
where the restrictions $\varphi _{U}$ and $\varphi _{U^{\perp }}$

### Proof

We have
    $U^{\perp }={\left\{v\in V\mid \left\langle v,u\right\rangle =0{\text{ for all }}u\in U\right\}}\,.$
For such a $v\in U^{\perp }$ and an arbitrary $u\in U$, we have
    $\left\langle \varphi (v),u\right\rangle =\left\langle \varphi ^{-1}(\varphi (v)),\varphi ^{-1}(u)\right\rangle =\left\langle v,u'\right\rangle =0\,,$
since $u'=\varphi ^{-1}(u)\in U$ due to the invariance of $U$. Therefore, $\varphi (v)\in U^{\perp }$.
$\Box$
The following statement is called spectral theorem or, more specific, spectral theorem for complex isometries. In this course, we will get to know further spectral theorems; see Theorem 41.11
and Theorem 42.9 .

## TheoremTheorem 34.2 change

Let $V$ be a finite-dimensionalMDLD/finite-dimensional $\mathbb {C}$-vector space,MDLD/vector space endowed with an inner product,MDLD/inner product and let
    $\varphi \colon V\longrightarrow V$
be an isometry.MDLD/isometry (C) Then $V$ has an orthonormal basisMDLD/orthonormal basis of eigenvectorsMDLD/eigenvectors of $\varphi$. In particular, $\varphi$ is

### Proof

We do induction on the dimensionMDLD/dimension (vs) of $V$. The statement is clear in the one-dimensional case. Due to the Fundamental theorem of algebra and Theorem 23.2 , $\varphi$ has an eigenvalueMDLD/eigenvalue and an eigenvector, which we can normalize. Let $E\subseteq V$ be the corresponding eigenline.MDLD/eigenline Since we have an isometry, the orthogonal complementMDLD/orthogonal complement $E^{\perp }$ is also, due to Lemma 34.1 , $\varphi$-invariant,MDLD/invariant (linear subspace) and the restriction
    $\varphi {|}_{E^{\perp }}\colon E^{\perp }\longrightarrow E^{\perp }$
is again an isometry. By the induction hypothesis, there exists on $E^{\perp }$ an orthonormal basis consisting of eigenvectors. Together with the first eigenvector, these form an orthonormal basis of eigenvectors of $V$.
$\Box$
    Angles

## Remark

For vectors $v$ and $w$, different from $0$, in a euklidean vector spaceMDLD/euklidean vector space $V$, the inequality of Cauchy-Schwarz implies that
    $-1\leq {\frac {\left\langle v,w\right\rangle }{\Vert {v}\Vert \cdot \Vert {w}\Vert }}\leq 1\,$
holds. Using the trigonometric function cosine (as a bijective mapping $[0,\pi ]\rightarrow [-1,1]$) and its inverse function, the angle between the two vectors can be defined, by setting
    $\angle (v,w):=\arccos {\frac {\left\langle v,w\right\rangle }{\Vert {v}\Vert \cdot \Vert {w}\Vert }}\,.$
The angle is a real number between $0$ and $\pi$. The equation above can be read as
    $\left\langle v,w\right\rangle =\Vert {v}\Vert \cdot \Vert {w}\Vert \cdot \cos {\left(\angle (v,w)\right)}\,.$
This provides the possibility to define the inner product in this way. However, then we have to find an independent definition for the angle. This approach might look a bit more intuitive but is has computationally and in terms of the proofs many disadvantages.
For an affine spaceMDLD/affine space $E$ over a Euclidean vector space $V$, and three given points $P,Q,R\in E$ (a triangle) with $Q,R\neq P$, the angle $\angle (Q,P,R)$ of the triangle at $P$ is the angle $\angle ({\overrightarrow {PQ}},{\overrightarrow {PR}})$.
    Plane isometries

## TheoremTheorem 34.4 change

Let
    $\varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2}$
be a proper linear isometry.MDLD/proper linear isometry Then $\varphi$ is a rotation, and the describing matrix with respect to the standard basis has the form
    $D(\theta )={\begin{pmatrix}\operatorname {cos} \,\theta &-\operatorname {sin} \,\theta \\\operatorname {sin} \,\theta &\operatorname {cos} \,\theta \end{pmatrix}}\,$
with a uniquely determined rotation angle
$\theta \in [0,2\pi [$

### Proof

Let $(x,y)$ and $(u,v)$ be the images of the standard vectors $(1,0)$ and $(0,1)$. An isometry preserves the length of a vector; therefore,
    $\Vert {\begin{pmatrix}x\\y\end{pmatrix}}\Vert ={\sqrt {x^{2}+y^{2}}}=1\,.$
Hence, $x$ is a real number between $-1$ and $+1$, and $y=\pm {\sqrt {1-x^{2}}}$, that is, $(x,y)$ is a point on the real unit circle. The unit circle is parametrized by the trigonometric functions, that is, there exists a uniquely determined angle $\theta$, $0\leq \theta <2\pi$, such that
    $(x,y)=(\operatorname {cos} \,\theta ,\operatorname {sin} \,\theta )\,.$
Since an isometry preserves orthogonality, we have
    $\left\langle {\begin{pmatrix}x\\y\end{pmatrix}},{\begin{pmatrix}u\\v\end{pmatrix}}\right\rangle =xu+yv=0\,.$
In case $y=0$, this implies (because of $x=\pm 1$) $u=0$. Then $v=\pm 1$, and, due to the properness assumption, the sign must be that of $x$.
So let $y\neq 0$. Then
    ${\begin{pmatrix}-v\\u\end{pmatrix}}={\frac {u}{y}}{\begin{pmatrix}x\\y\end{pmatrix}}\,.$
Since both vectors have the length $1$, the modulus of the scalar factor $u/y$ is $1$. In case $u=y$, we had $v=-x$ and the determinant were $-1$. Therefore, $u=-y$ and $v=x$, yielding the claim.
$\Box$
The composition of two rotations ${\begin{pmatrix}\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}$ and ${\begin{pmatrix}\operatorname {cos} \,\beta &-\operatorname {sin} \,\beta \\\operatorname {sin} \,\beta &\operatorname {cos} \,\beta \end{pmatrix}}$ is ${\begin{pmatrix}\operatorname {cos} \,(\alpha +\beta )&-\operatorname {sin} \,(\alpha +\beta )\\\operatorname {sin} \,(\alpha +\beta )&\operatorname {cos} \,(\alpha +\beta )\end{pmatrix}}$. This property is intuitively clear, using intuitive properties of plane rotations. Using the addition theorems for the trigonometric functions, we can proof them. Moreover, the addition theorems follow from this property, see exercise *****. Therefore, the group of plane rotations is commutative.

## TheoremTheorem 34.5 change

Let
    $\varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2}$
be an improperMDLD/improper (isometry) linear isometry.MDLD/linear isometry Then $\varphi$ is an axis reflection, and its describing matrix, with respect to the standard basis, has the form
    ${\begin{pmatrix}\cos \alpha &\sin \alpha \\\sin \alpha &-\cos \alpha \end{pmatrix}}$
with a uniquely determined angle
$\alpha \in [0,2\pi [$

### Proof

We consider
    $\varphi \circ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}},$
which is by the multiplication theorem for the determinant a proper isometry.MDLD/proper isometry Because of Theorem 34.4 , there exists a uniquely determined angle $\alpha \in [0,\pi [$ such that
    $\varphi \circ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}}={\begin{pmatrix}\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}\,.$
Therefore,
    $\varphi ={\begin{pmatrix}\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}\circ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}}={\begin{pmatrix}\cos \alpha &\sin \alpha \\\sin \alpha &-\cos \alpha \end{pmatrix}}\,.$
$\Box$
If such an axis rotation is given, then ${\begin{pmatrix}-\sin \alpha \\\cos \alpha -1\end{pmatrix}}$ is an eigenvector for the eigenvalue $1$, and the axis of reflection is $\mathbb {R} {\begin{pmatrix}-\sin \alpha \\\cos \alpha -1\end{pmatrix}}$, see exercise *****. An axis reflection is described, with respect to the basis consisting of a vector $\neq 0$ of the reflecting axis and a corresponding orthogonal vector $\neq 0$, by the matrix ${\begin{pmatrix}1&0\\0&-1\end{pmatrix}}$. This means that the description given in Theorem 34.5
(with respect to the standard basis) can be improved drastically.
    Isometries of space

## TheoremCreate referencenumber

Let
    $\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3}$
be a linear isometry.MDLD/linear isometry (R) Then there exists an eigenvectorMDLD/eigenvector with eigenvalueMDLD/eigenvalue
$1$
$-1$

### Proof

The characteristic polynomialMDLD/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 eigenvalueMDLD/eigenvalue of $\varphi$. Because of Theorem 33.12 , 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.

## TheoremTheorem 34.7 change

A proper isometryMDLD/proper isometry
    $\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3}$
eigenvectorMDLD/eigenvector with eigenvalueMDLD/eigenvalue $1$, that is, there exists a line (through the origin)
$\varphi$

### Proof

We consider the characteristic polynomialMDLD/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 Theorem 33.12 , we must have $\lambda =1$.
$\Box$

## TheoremTheorem 34.8 change

Let
    $\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3}$
be a proper isometry.MDLD/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 Theorem 34.7 , 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 Lemma 34.1 , 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 Theorem 34.4 , $\varphi _{2}$ 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.
$\Box$

## CorollaryCreate referencenumber

At the beginning of a football game, the football lies on the center spot. After a goal is scored, the ball is again put on the center spot. In this situation, the following holds: There exists at least two (antipodal) points on the ball (its surface)

### Proof

The total movement is a linear isometry;MDLD/linear isometry (R) therefore, the statement follows from Theorem 34.8 .
$\Box$
    The decomposition theorem for isometries

## LemmaLemma 34.10 change

Let $V\neq 0$ be a realMDLD/real finite-dimensionalMDLD/finite-dimensional (fgvs) vector space,MDLD/vector space and let
    $\varphi \colon V\longrightarrow V$
denote an endomorphism.MDLD/endomorphism Then there exists a $\varphi$-invariantMDLD/invariant (endomorphism) linear subspaceMDLD/linear subspace $U\subseteq V$ of dimension
$1$
$2$

### Proof

We may assume $V=\mathbb {R} ^{n}$, and that $\varphi$ is described by the matrix $M$ with respect to the standard basis. If $\varphi$ has an eigenvalue, then we are done. Otherwise, we consider the corresponding complex mapping, that is,
    $\varphi \colon \mathbb {C} ^{n}\longrightarrow \mathbb {C} ^{n},$
which is given by the same matrix $M$. This matrix has a complex eigenvalue $a+b{\mathrm {i} }$, and a complex eigenvector $v\in \mathbb {C} ^{n}$. In particular, we have
    $Mv=(a+b{\mathrm {i} })v\,.$
Writing
    $v=v_{1}+{\mathrm {i} }v_{2}\,$
with $v_{1},v_{2}\in \mathbb {R} ^{n}$, this means
    $Mv_{1}+{\mathrm {i} }Mv_{2}=Mv=(a+b{\mathrm {i} }){\left(v_{1}+{\mathrm {i} }v_{2}\right)}=av_{1}-bv_{2}+{\mathrm {i} }(av_{2}+bv_{1})\,.$
Comparing the real part and the imaginary part we can deduce that $Mv_{1},Mv_{2}\in \langle v_{1},v_{2}\rangle$. Therefore, the real linear subspace $\langle v_{1},v_{2}\rangle$ is invariant.
$\Box$

## TheoremCreate referencenumber

Let
    $\varphi \colon V\longrightarrow V$
be an isometryMDLD/isometry on the euclidean vector spaceMDLD/euclidean vector space $V$. Then $V$ is a orthogonal direct sumMDLD/orthogonal direct sum
    $V=G_{1}\oplus \cdots \oplus G_{p}\oplus H_{1}\oplus \cdots \oplus H_{q}\oplus E_{1}\oplus \cdots \oplus E_{r}\,$
of $\varphi$-invariantMDLD/invariant (endomorphism) linear subspaces,MDLD/linear subspaces
$G_{i},H_{j}$
$E_{k}$
$\varphi$
$G_{i}$
$H_{j}$
$E_{k}$

### Proof

We do induction over the dimensionMDLD/dimension (vs) $n$ of $V$. The one-dimensional case is clear, due to Theorem 33.12 . Let $n=2$. The determinant has, because of Lemma 33.13
either the value $1$ or $-1$. In case it is $-1$, the characteristic polynomial has two distinct zeroes, and these zeroes are, due to Theorem 33.12 , $1$ and $-1$. Then we have a reflection at an axis, and
    $V=G\oplus H\,.$
If the determinant is $1$, then we are in the situation of Theorem 34.4 , and we have a rotation. If the angle of rotation is $0$, then we have the identity, and we can decompose $V=G_{1}\oplus G_{2}$. If the angle of rotation is $\pi$, then we have a point reflection $-\operatorname {Id}$; therefore, we have a decomposition $V=H_{1}\oplus H_{2}$. For all other angles, there is no eigenvector.
Let now $n\geq 1$ be arbitrary, and suppose that the statement is proven for smaller dimensions. Because of Lemma 34.10 , there exists a $\varphi$-invariant linear subspace $U$ of dimension $1$ or $2$, and, because of Lemma 34.1 , there exists an invariant orthogonal complement,MDLD/orthogonal complement that is,
    $V=U\oplus W\,.$
The induction hypothesis, applied to $W$, yields the result.
$\Box$
In this decomposition, $G_{1}\oplus \cdots \oplus G_{p}$ is the eigenspace for the eigenvalue $1$, and $H_{1}\oplus \cdots \oplus H_{q}$ is the eigenspace for the eigenvalue $-1$; the decompositions are not unique. The Isometry is proper if and only if $q$ is even.