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

Diagonalizability of isometries over the complex numbers

Lemma

Let ${}\varphi \colon V\rightarrow V$ be a linear isometry on a finite-dimensional ${}{\mathbb {K} }$-vector space ${}V$, endowed with an inner product. Let ${}U\subseteq V$ denote an invariant linear subspace. Then also the orthogonal complement ${}U^{\perp }$ is invariant. In particular, ${}\varphi$ can be written as a direct sum

${}\varphi =\varphi _{U}\oplus \varphi _{U^{\perp }}\,,$

where the restrictions ${}\varphi _{U}$ and ${}\varphi _{U^{\perp }}$

are also isometries.

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 Fact ***** and Fact *****.

Theorem

Let ${}V$ be a finite-dimensional ${}\mathbb {C}$-vector space, endowed with an inner product, and let

$\varphi \colon V\longrightarrow V$

be an isometry. Then ${}V$ has an orthonormal basis of eigenvectors of ${}\varphi$. In particular, ${}\varphi$ is

diagonalizable.

Proof

We do induction on the dimension 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 eigenvalue and an eigenvector, which we can normalize. Let ${}E\subseteq V$ be the corresponding eigenline. Since we have an isometry, the orthogonal complement ${}E^{\perp }$ is also, due to Fact *****, ${}\varphi$-invariant, 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 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 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

Theorem

Let

$\varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2}$

be a 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.

Theorem

Let

$\varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2}$

be an improper 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. Because of Fact *****, 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 Fact ***** (with respect to the standard basis) can be improved drastically.

Isometries of space

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.

$\Box$

Corollary

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)

that are at the new kick-off exactly at the position where they were at the first kick-off. The total movement of the ball is a rotation around an axis.

Proof

The total movement is a linear isometry; therefore, the statement follows from Fact *****.

$\Box$

The decomposition theorem for isometries

Lemma

Let ${}V\neq 0$ be a real finite-dimensional vector space, and let

$\varphi \colon V\longrightarrow V$

denote an endomorphism. Then there exists a ${}\varphi$-invariant linear subspace ${}U\subseteq V$ of dimension

${}1$ or ${}2$.

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.

$\Box$

Theorem

Let

$\varphi \colon V\longrightarrow V$

be an isometry on the euclidean vector space ${}V$. Then ${}V$ is a 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 ${}\varphi$-invariant linear subspaces,

where the ${}G_{i},H_{j}$ are one-dimensional, and the ${}E_{k}$ are two-dimensional. The restriction of ${}\varphi$ to the ${}G_{i}$ is the identity, the restriction to ${}H_{j}$ is the negative identity, and the restriction to ${}E_{k}$ is a rotation without eigenvalue.

Proof

We do induction over the dimension ${}n$ of ${}V$. The one-dimensional case is clear, due to Fact *****. Let ${}n=2$. The determinant has, because of Fact ***** either the value ${}1$ or ${}-1$. In case it is ${}-1$, the characteristic polynomial has two distinct zeroes, and these zeroes are, due to Fact *****, ${}1$ 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 Fact *****, 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 Fact *****, there exists a ${}\varphi$-invariant linear subspace ${}U$ of dimension ${}1$ or ${}2$, and, because of Fact *****, there exists an invariant 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.