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

We have

${\displaystyle {}U^{\perp }={\left\{v\in V\mid \left\langle v,u\right\rangle =0{\text{ for all }}u\in U\right\}}\,.}$

For such a ${\displaystyle {}v\in U^{\perp }}$ and an arbitrary ${\displaystyle {}u\in U}$, we have

${\displaystyle {}\left\langle \varphi (v),u\right\rangle =\left\langle \varphi ^{-1}(\varphi (v)),\varphi ^{-1}(u)\right\rangle =\left\langle v,u'\right\rangle =0\,,}$

since ${\displaystyle {}u'=\varphi ^{-1}(u)\in U}$ due to the invariance of ${\displaystyle {}U}$. Therefore, ${\displaystyle {}\varphi (v)\in U^{\perp }}$.

We do induction on the dimension of ${\displaystyle {}V}$. The statement is clear in the one-dimensional case. Due to the Fundamental theorem of algebra and Theorem 23.2 , ${\displaystyle {}\varphi }$ has an eigenvalue and an eigenvector, which we can normalize. Let ${\displaystyle {}E\subseteq V}$ be the corresponding eigenline. Since we have an isometry, the orthogonal complement ${\displaystyle {}E^{\perp }}$ is also, due to Lemma 34.1 , ${\displaystyle {}\varphi }$-invariant, and the restriction

${\displaystyle \varphi {|}_{E^{\perp }}\colon E^{\perp }\longrightarrow E^{\perp }}$

is again an isometry. By the induction hypothesis, there exists on ${\displaystyle {}E^{\perp }}$ an orthonormal basis consisting of eigenvectors. Together with the first eigenvector, these form an orthonormal basis of eigenvectors of ${\displaystyle {}V}$.

Let ${\displaystyle {}{\begin{pmatrix}x\\y\end{pmatrix}}}$ and ${\displaystyle {}{\begin{pmatrix}u\\v\end{pmatrix}}}$ be the images of the standard vectors ${\displaystyle {}{\begin{pmatrix}1\\0\end{pmatrix}}}$ and ${\displaystyle {}{\begin{pmatrix}0\\1\end{pmatrix}}}$. An isometry preserves the length of a vector; therefore,

${\displaystyle {}\Vert {\begin{pmatrix}x\\y\end{pmatrix}}\Vert ={\sqrt {x^{2}+y^{2}}}=1\,.}$

Since an isometry preserves orthogonality, we have

${\displaystyle {}\left\langle {\begin{pmatrix}x\\y\end{pmatrix}},{\begin{pmatrix}u\\v\end{pmatrix}}\right\rangle =xu+yv=0\,.}$

In case ${\displaystyle {}y=0}$, this implies (because of ${\displaystyle {}x=\pm 1}$) ${\displaystyle {}u=0}$. Then ${\displaystyle {}v=\pm 1}$, and, due to the properness assumption, the sign must be that of ${\displaystyle {}x}$.

So let ${\displaystyle {}y\neq 0}$. Then

${\displaystyle {}{\begin{pmatrix}-v\\u\end{pmatrix}}={\frac {u}{y}}{\begin{pmatrix}x\\y\end{pmatrix}}\,.}$

Since both vectors have the length ${\displaystyle {}1}$, the modulus of the scalar factor ${\displaystyle {}u/y}$ is ${\displaystyle {}1}$. In case ${\displaystyle {}u=y}$, we had ${\displaystyle {}v=-x}$ and the determinant was ${\displaystyle {}-1}$. Therefore, ${\displaystyle {}u=-y}$ and ${\displaystyle {}v=x}$. Hence, the describing matrix of ${\displaystyle {}\varphi }$ with respect to the standard basis is

${\displaystyle {\begin{pmatrix}x&-y\\y&x\end{pmatrix}}{\text{ where }}x^{2}+y^{2}=1.}$

In particular, ${\displaystyle {}x}$ is a real number between ${\displaystyle {}-1}$ and ${\displaystyle {}+1}$, and ${\displaystyle {}y=\pm {\sqrt {1-x^{2}}}}$, that is, ${\displaystyle {}(x,y)}$ 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 ${\displaystyle {}\theta }$, ${\displaystyle {}0\leq \theta <2\pi }$, such that

${\displaystyle {}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}\,.}$

We consider

${\displaystyle \varphi \circ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}},}$

which is by the multiplication theorem for the determinant a proper isometry. Because of Theorem 34.4 , there exists a uniquely determined angle ${\displaystyle {}\alpha \in [0,\pi [}$ such that

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

Therefore,

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

The characteristic polynomial ${\displaystyle {}P}$ of ${\displaystyle {}\varphi }$ is a normed polynomial of degree three. For ${\displaystyle {}t\to +\infty }$, we have ${\displaystyle {}P(t)\to +\infty }$, and for ${\displaystyle {}t\to -\infty }$, we have ${\displaystyle {}P(t)\to -\infty }$. Due to the intermediate value theorem, ${\displaystyle {}P}$ has a real zero. Such a zero is, according to Theorem 23.2 , an eigenvalue of ${\displaystyle {}\varphi }$. Because of Theorem 33.12 , this eigenvalue equals ${\displaystyle {}1}$ or ${\displaystyle {}-1}$.

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

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

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

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

As the polynomial ${\displaystyle {}P(\lambda )\to \infty }$ as ${\displaystyle {}\lambda \to \infty }$, there must be a positive ${\displaystyle {}\lambda }$ that is a zero of ${\displaystyle {}P}$. Due to Theorem 33.12 , we must have ${\displaystyle {}\lambda =1}$.

Due to Theorem 34.7 , there exists an eigenvector ${\displaystyle {}u}$ for the eigenvalue ${\displaystyle {}1}$. Let ${\displaystyle {}U=\mathbb {R} u}$ denote the corresponding line. This line is fixed and, in particular, invariant under ${\displaystyle {}\varphi }$. Because of Lemma 34.1 , also the orthogonal complement ${\displaystyle {}U^{\perp }}$ is invariant under ${\displaystyle {}\varphi }$. That is, there exists a linear isometry

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

that coincides on ${\displaystyle {}U^{\perp }}$ with ${\displaystyle {}\varphi }$. Here, ${\displaystyle {}\varphi _{2}}$ is proper. Therefore, by Theorem 34.4 , ${\displaystyle {}\varphi _{2}}$ is a plane rotation. If we choose a vector of length one in ${\displaystyle {}U}$, and an orthonormal basis of ${\displaystyle {}U^{\perp }}$, then ${\displaystyle {}\varphi }$ has with respect to this basis the given matrix form.

The total movement is a linear isometry; therefore, the statement follows from Theorem 34.8 .

We may assume ${\displaystyle {}V=\mathbb {R} ^{n}}$, and that ${\displaystyle {}\varphi }$ is described by the matrix ${\displaystyle {}M}$ with respect to the standard basis. If ${\displaystyle {}\varphi }$ has an eigenvalue, then we are done. Otherwise, we consider the corresponding complex mapping, that is,

${\displaystyle \varphi \colon \mathbb {C} ^{n}\longrightarrow \mathbb {C} ^{n},}$

which is given by the same matrix ${\displaystyle {}M}$. This matrix has a complex eigenvalue ${\displaystyle {}a+b{\mathrm {i} }}$, and a complex eigenvector ${\displaystyle {}v\in \mathbb {C} ^{n}}$. In particular, we have

${\displaystyle {}Mv=(a+b{\mathrm {i} })v\,.}$

Writing

${\displaystyle {}v=v_{1}+{\mathrm {i} }v_{2}\,}$

with ${\displaystyle {}v_{1},v_{2}\in \mathbb {R} ^{n}}$, this means

${\displaystyle {}Mv_{1}+{\mathrm {i} }Mv_{2}=Mv=(a+b{\mathrm {i} }){\left(v_{1}+{\mathrm {i} }v_{2}\right)}=av_{1}-bv_{2}+{\mathrm {i} }(av_{2}+bv_{1})\,.}$

Comparing the real part and the imaginary part we can deduce that ${\displaystyle {}Mv_{1},Mv_{2}\in \langle v_{1},v_{2}\rangle }$. Therefore, the real linear subspace ${\displaystyle {}\langle v_{1},v_{2}\rangle }$ is invariant.

We do induction over the dimension ${\displaystyle {}n}$ of ${\displaystyle {}V}$. The one-dimensional case is clear, due to Theorem 33.12 . Let ${\displaystyle {}n=2}$. The determinant has, because of Lemma 33.13 either the value ${\displaystyle {}1}$ or ${\displaystyle {}-1}$. In case it is ${\displaystyle {}-1}$, the characteristic polynomial has two distinct zeroes, and these zeroes are, due to Theorem 33.12 , ${\displaystyle {}1}$ and ${\displaystyle {}-1}$. Then we have a reflection at an axis, and

${\displaystyle {}V=G\oplus H\,.}$

If the determinant is ${\displaystyle {}1}$, then we are in the situation of Theorem 34.4 , and we have a rotation. If the angle of rotation is ${\displaystyle {}0}$, then we have the identity, and we can decompose ${\displaystyle {}V=G_{1}\oplus G_{2}}$. If the angle of rotation is ${\displaystyle {}\pi }$, then we have a point reflection ${\displaystyle {}-\operatorname {Id} }$; therefore, we have a decomposition ${\displaystyle {}V=H_{1}\oplus H_{2}}$. For all other angles, there is no eigenvector.

Let now ${\displaystyle {}n\geq 1}$ be arbitrary, and suppose that the statement is proven for smaller dimensions. Because of Lemma 34.10 , there exists a ${\displaystyle {}\varphi }$-invariant linear subspace ${\displaystyle {}U}$ of dimension ${\displaystyle {}1}$ or ${\displaystyle {}2}$, and, because of Lemma 34.1 , there exists an invariant orthogonal complement, that is,

${\displaystyle {}V=U\oplus W\,.}$

The induction hypothesis, applied to ${\displaystyle {}W}$, yields the result.