Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 35

Angle-preserving mappings

Definition

\varphi \colon V\longrightarrow W

between Euclidean vector spaces ${}V$ and ${}W$ is called angle-preserving, if for any two vectors ${}u,v\in V$ (different from ${}0$ ) the relation

{}\angle (\varphi (u),\varphi (v))=\angle (u,v)\,

holds.

Angles are only defined for vectors different from ${}0$ . Therefore, angle-preserving mappings are injective. An isometry is always angle-preserving, because the norm and the angles are defined with respect to an inner product, and are not changed under an isometry. Further examples of angle-preserving mappings are homotheties by a scalar different from ${}0$ , see Exercise 35.1 . An angle-preserving mapping maps orthogonal vectors to orthogonal vectors.

Example

Let

\varphi \colon \mathbb {C} \longrightarrow \mathbb {C}

be a ${}\mathbb {C}$ -linear mapping that is given by the multiplication with the complex number

{}w=a+b{\mathrm {i} }\neq 0\,.

With respect to the real basis ${}1,{\mathrm {i} }$ of ${}\mathbb {C} =\mathbb {R} ^{2}$ , this mapping is described by the real ${}2\times 2$ -matrix

{\begin{pmatrix}a&-b\\b&a\end{pmatrix}}.

We write this matrix as

{}{\begin{pmatrix}a&-b\\b&a\end{pmatrix}}={\begin{pmatrix}{\sqrt {a^{2}+b^{2}}}&0\\0&{\sqrt {a^{2}+b^{2}}}\end{pmatrix}}{\begin{pmatrix}{\frac {a}{\sqrt {a^{2}+b^{2}}}}&-{\frac {b}{\sqrt {a^{2}+b^{2}}}}\\{\frac {b}{\sqrt {a^{2}+b^{2}}}}&{\frac {a}{\sqrt {a^{2}+b^{2}}}}\end{pmatrix}}\,.

Therefore, we have a composition of an isometry (a plane rotation) and of a homothety with the scalar factor

{}\vert {w}\vert ={\sqrt {a^{2}+b^{2}}}\,;

in particular, this is an angle-preserving mapping.

Theorem

Let

\varphi \colon V\longrightarrow V

denote an angle-preserving linear mapping on the Euclidean vector space ${}V$ . Then there exists an isometry

\psi \colon V\longrightarrow V

and a homothety

\sigma \colon V\longrightarrow V

such that

{}\varphi =\sigma \circ \psi \,.

Proof

Let

{}r=\det \varphi \,,

and set

{}s:={\sqrt[{n}]{\vert {r}\vert }}\,,

where ${}n$ denotes the dimension of ${}V$ . Let ${}\sigma$ be the homothety with factor ${}s$ . We consider the mapping

{\displaystyle {}\psi

This mapping is still angle-preserving, and its determinant is ${}1$ or ${}-1$ . Because of Exercise 33.18 , ${}\psi$ is an isometry.

\Box

Remark

For an angle-preserving mapping

\varphi \colon V\longrightarrow W

between Euclidean vector spaces ${}V$ and ${}W$ , not only the angles at the origin, but all angles are preserved. For points ${}P,Q,R\in V$ , the angle of the triangle at ${}Q$ coincides with the angle at ${}\varphi (Q)$ of the image triangle ${}\varphi (P),\varphi (Q),\varphi (R)$ , because of

{}\angle (P,Q,R)=\angle ({\overrightarrow {QP}},{\overrightarrow {QR}})=\angle (\varphi ({\overrightarrow {QP}}),\varphi ({\overrightarrow {QR}}))=\angle ({\overrightarrow {\varphi (Q)\varphi (P)}},{\overrightarrow {\varphi (Q)\varphi (R)}})=\angle (\varphi (P),\varphi (Q),\varphi (R))\,.

Distances between sets

Definition

For two nonempty subsets ${}A,B\subseteq M$ in a metric space ${}M$ ,

{}d(A,B):=\inf {\left(d(P,Q),\,P\in A,\,Q\in B\right)}\,

is called the distance of the subsets

{}A

and

{}B

.

We will apply this concept for normed vector spaces and for Euclidean vector spaces. For two points ${}P,Q\in V$ , the distance between the sets ${}\{P\}$ and ${}\{Q\}$ equals ${}d(P,Q)$ .

We will mainly work in situations where the infimum is obtained, that is, the infimum is also a minimum. This is the typical behavior for linear objects.

Lemma

Let ${}V$ be a Euclidean vector space, ${}U\subseteq V$ a linear subspace, and ${}v\in V$ . Then ${}p_{U}(v)$ is the point on ${}U$ that has, among all points of ${}U$ , the minimal distance to ${}v$ . In particular, we have

{}d(v,U)=d(v,p_{U}(v))\,.

Proof

For ${}u\in U$ , we have, due to the Pythagorean theorem,

{}d(v,u)^{2}=d(v,p_{U}(v))^{2}+d(p_{U}(v),u)^{2}\,,

as ${}p_{U}(v)-u\in U$ and ${}v-p_{U}(v)\in U^{\perp }$ are perpendicular to each other. This expression is minimal if and only if ${}d(p_{U}(v),u)=0$ , and this holds if and only if

{}p_{U}(v)=u\,.

\Box

In this context, ${}p_{U}(v)$ is also called the dropped perpendicular foot of ${}v$ on ${}U$ .

Corollary

Let ${}V$ be a Euclidean vector space, ${}U\subseteq V$ a linear subspace, and ${}v\in V$ . Let ${}u_{1},\ldots ,u_{m}$ denote an orthonormal basis of ${}U$ . Then

{}d(v,U)^{2}=\Vert {v}\Vert ^{2}-\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle ^{2}\,.

Proof

Because of Lemma 35.6 , we have

{}d(v,U)=d(v,p_{U}(v))\,,

and, according to Lemma 32.14 , we have

{}p_{U}(v)=\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}\,.

The vectors ${}p_{U}(v)$ and ${}v-p_{U}(v)$ are orthogonal to each other. Therefore, using the Pythagorean theorem, we have

{}{\begin{aligned}d(v,U)^{2}&=\Vert {v-p_{U}(v)}\Vert ^{2}\\&=\Vert {v}\Vert ^{2}-\Vert {p_{U}(v)}\Vert ^{2}\\&=\Vert {v}\Vert ^{2}-\Vert {\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}}\Vert ^{2}\\&=\Vert {v}\Vert ^{2}-\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle ^{2}.\end{aligned}}

\Box

Example

Let ${}J\subseteq \{1,\ldots ,n\}$ , and let ${}U=\langle e_{i},\,i\in J\rangle \subseteq \mathbb {R} ^{n}$ denote the linear subspace spanned by this choice of standard vectors. Let

{}v={\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}\,.

Then, the distance of ${}v$ to ${}U$ equals

{}d(v,U)={\sqrt {\sum _{i\not \in J}v_{i}^{2}}}\,.

The dropped perpendicular foot of ${}v$ on ${}U$ is

{}p_{U}(v)={\begin{pmatrix}w_{1}\\\vdots \\w_{n}\end{pmatrix}}\,,

where

{}w_{i}={\begin{cases}v_{i},\,{\text{ if }}i\in J,\,\\0,\,{\text{ if }}i\notin J\,.\end{cases}}\,

Corollary

Let ${}a\in \mathbb {R} ^{n}$ be a vector with ${}\Vert {a}\Vert =1$ , and let

{}U={\left\{x\in \mathbb {R} ^{n}\mid a_{1}x_{1}+\cdots +a_{n}x_{n}=0\right\}}=(\mathbb {R} a)^{\perp }\,

denote the linear subspace defined by ${}a$ as normal vector. Then, for a vector ${}v\in \mathbb {R} ^{n}$ , the distance to ${}U$ equals

{}d(U,v)=\vert {\left\langle a,v\right\rangle }\vert \,.

Proof

Let ${}a,u_{2},\ldots ,u_{n}$ be an orthonormal basis of ${}\mathbb {R} ^{n}$ , and write

{}v=\lambda a+\sum _{i=2}^{n}c_{i}u_{i}\,.

Then

{}p_{U}(v)=\sum _{i=2}^{n}c_{i}u_{i}\,,

and, due to Lemma 35.6 , we have

{}d(U,v)=\Vert {v-p_{U}(v)}\Vert =\Vert {\lambda a}\Vert =\vert {\lambda }\vert \Vert {a}\Vert =\vert {\lambda }\vert \,.

In conjunction with

{}\left\langle a,v\right\rangle =\left\langle a,\lambda a+\sum _{i=2}^{n}c_{i}u_{i}\right\rangle =\left\langle a,\lambda a\right\rangle =\lambda \,

this yields the result.

\Box

These considerations do also hold for affine subspaces.

Example

Let ${}E$ be a real affine space over the Euclidean vector space ${}V$ , let ${}P\in E$ be a point, and let ${}F\subseteq E$ denote an affine subspace. In case ${}P\in F$ , the distance of ${}P$ to ${}F$ equals ${}0$ . In general, we write

{}F=Q+U\,

with a point ${}Q\in F$ and with a linear subspace ${}U\subseteq V$ . We determine the orthogonal complement ${}W=U^{\perp }$ of ${}U$ in ${}V$ . If ${}u_{1},\ldots ,u_{m}$ is a basis of ${}U$ , and ${}w_{1},\ldots ,w_{k}$ is a basis of ${}W$ , then there exists a unique representation

{}{\overrightarrow {PQ}}=\sum _{i=1}^{m}a_{i}u_{i}+\sum _{j=1}^{k}b_{j}w_{j}\,.

In this case,

{}L=P+\sum _{j=1}^{k}b_{j}w_{j}=Q-\sum _{i=1}^{m}a_{i}u_{i}\,

is the dropped perpendicular foot of ${}P$ on ${}F$ , and the distance of ${}P$ to ${}L$ is

{}d{\left(P,F\right)}=d{\left(P,L\right)}=\Vert {\sum _{j=1}^{k}b_{j}w_{j}}\Vert \,.

If the ${}w_{j}$ form an orthonormal basis of ${}U$ , then this equals ${}{\sqrt {\sum _{j=1}^{k}b_{j}^{2}}}$ .

Example

We want to determine in the Euclidean plane the distance between the point ${}P={\begin{pmatrix}4\\5\end{pmatrix}}$ and the line ${}G$ given by ${}2x-3y=7$ . The line has the form

{}G={\left\{{\begin{pmatrix}{\frac {7}{2}}\\0\end{pmatrix}}+t{\begin{pmatrix}3\\2\end{pmatrix}}\mid t\in \mathbb {R} \right\}}\,,

and ${}{\begin{pmatrix}2\\-3\end{pmatrix}}$ is a vector perpendicular to ${}G$ . We have

{}{\begin{pmatrix}{\frac {7}{2}}\\0\end{pmatrix}}-{\begin{pmatrix}4\\5\end{pmatrix}}={\begin{pmatrix}-{\frac {1}{2}}\\-5\end{pmatrix}}=-{\frac {23}{26}}{\begin{pmatrix}3\\2\end{pmatrix}}+{\frac {14}{13}}{\begin{pmatrix}2\\-3\end{pmatrix}}\,.

Therefore, the dropped perpendicular foot is

{}{\begin{pmatrix}4\\5\end{pmatrix}}+{\frac {14}{13}}{\begin{pmatrix}2\\-3\end{pmatrix}}={\begin{pmatrix}{\frac {80}{13}}\\{\frac {23}{13}}\end{pmatrix}}\,,

and the distance is

{\frac {14}{13}}{\sqrt {13}}.

Lemma

Let ${}V$ be a Euclidean vector space, and let ${}E_{1}=P_{1}+U_{1}$ and ${}E_{2}=P_{2}+U_{2}$ denote nonempty affine subspaces with the linear subspaces ${}U_{1},U_{2}\subseteq V$ . Let

{}P_{1}-P_{2}=u_{1}+u_{2}+u\,

with ${}u_{1}\in U_{1}$ , ${}u_{2}\in U_{2}$ , and ${}u\in {\left(U_{1}+U_{2}\right)}^{\perp }$ . Then the distance ${}d(E_{1},E_{2})$ equals ${}\Vert {u}\Vert$ ; it is obtained in the points ${}P_{1}-u_{1}\in E_{1}$ and

{}P_{2}+u_{2}\in E_{2}

. In particular, the connecting vector of the points, where the minimal distance is obtained, is perpendicular to

{}E_{1}

and to

{}E_{2}

.

Proof

We write ${}P_{1}-P_{2}=u_{1}+u_{2}+u$ with ${}u_{1}\in U_{1}$ , ${}u_{2}\in U_{2}$ , and ${}u\in {\left(U_{1}+U_{2}\right)}^{\perp }$ ; such a decomposition does always exist, ${}u_{1},u_{2}$ are not uniquely determined (in case ${}U_{1}\cap U_{2}\neq 0$ ), but ${}u$ is uniquely determined. We have

{}P_{1}-u_{1}=P_{2}+u_{2}+u\,,

and ${}Q_{1}:=P_{1}-u_{1}\in E_{1}$ , and ${}Q_{2}:=P_{2}+u_{2}\in E_{2}$ . The distance between ${}Q_{1}$ and ${}Q_{2}$ is ${}\Vert {u}\Vert$ . For arbitrary points ${}R_{1}=Q_{1}+v_{1}\in E_{1}$ and ${}R_{2}=Q_{2}+v_{2}\in E_{2}$ fulfilling ${}v_{1}\in U_{1}$ and ${}v_{2}\in U_{2}$ , we have

{}{\begin{aligned}d(R_{1},R_{2})^{2}&=\Vert {R_{1}-R_{2}}\Vert ^{2}\\&=\Vert {v_{1}-v_{2}+u}\Vert ^{2}\\&=\left\langle v_{1}-v_{2}+u,v_{1}-v_{2}+u\right\rangle \\&=\left\langle v_{1}-v_{2},v_{1}-v_{2}\right\rangle +\left\langle u,u\right\rangle \\&\geq \left\langle u,u\right\rangle ,\end{aligned}}

that is,

{}d(R_{1},R_{2})\geq \Vert {u}\Vert \,.

\Box

In the previous statement, the points where the minimum is obtained, are not unique determined; for example, think about two parallel lines in the plane. If the intersection of the linear spaces corresponding to ${}E_{1},E_{2}$ equals ${}0$ , then we have uniqueness. This is true in the case of skew lines.

Example

Two (affine) lines ${}G,H\subseteq \mathbb {R} ^{3}$ are called skew if they do not have any point in common and if they are also not parallel, meaning their vectors are linearly independent. Then, these vectors generate a plane; there exists a vector ${}u$ perpendicular to this plane. We can compute one such vector, the normal vector, with the cross product. Let

{}G=P+\mathbb {R} v\,,

and

{}H=Q+\mathbb {R} w\,.

The linear system of equations

{}P-Q=av+bw+cu\,

has a unique solution ${}(a,b,c)\in \mathbb {R} ^{3}$ . Here, ${}P-av\in G$ and ${}Q+bw\in H$ are the dropped perpendicular feet, where the distance of the lines is obtained, according to Lemma 35.12 . This distance is ${}\Vert {cu}\Vert$ .

Corollary

Let

{}G=P+\mathbb {R} v\,

and

{}H=Q+\mathbb {R} w\,

be skew lines in ${}\mathbb {R} ^{3}$ , with vectors ${}v,w\in \mathbb {R} ^{3}$ . Let ${}u$ be a normed vector that is perpendicular to ${}v$ and ${}w$ . Then

{}d(G,H)=\vert {\left\langle P-Q,u\right\rangle }\vert \,.

Proof

We use Example 35.13 , and consider

{}P-Q=av+bw+cu\,.

According to Cramer's rule, we obtain, using Lemma 33.3 (5), and the property that ${}u$ is a linear multiple of ${}v\times w$ ,

{}{\begin{aligned}c&={\frac {\det {\begin{pmatrix}v_{1}&w_{1}&P_{1}-Q_{1}\\v_{2}&w_{2}&P_{2}-Q_{2}\\v_{3}&w_{3}&P_{3}-Q_{3}\end{pmatrix}}}{\det {\begin{pmatrix}v_{1}&w_{1}&u_{1}\\v_{2}&w_{2}&u_{2}\\v_{3}&w_{3}&u_{3}\end{pmatrix}}}}\\&={\frac {\left\langle v\times w,P-Q\right\rangle }{\left\langle v\times w,u\right\rangle }}\\&={\frac {\left\langle u,P-Q\right\rangle }{\left\langle u,u\right\rangle }}\\&=\left\langle u,P-Q\right\rangle .\end{aligned}}

\Box

Example

Let

{}G=P+\mathbb {R} v\,

and

{}H=Q+\mathbb {R} w\,

be skew lines. We want to understand the distance problem between the two lines as an extremal problem in the sense of higher-dimensional analysis. Let

{}P={\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}\,

and

{}Q={\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\end{pmatrix}}\,.

The square of the distance between the two points

{}P'={\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}+s{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\,

and

{}Q'={\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\end{pmatrix}}+t{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}\,

is (setting ${}c_{i}=a_{i}-b_{i}$ )

{}{\begin{aligned}d(P',Q')^{2}&={\left(c_{1}+sv_{1}-tw_{1}\right)}^{2}+{\left(c_{2}+sv_{2}-tw_{2}\right)}^{2}+{\left(c_{3}+sv_{3}-tw_{3}\right)}^{2}\\&=c_{1}^{2}+s^{2}v_{1}^{2}+t^{2}w_{1}^{2}+2sc_{1}v_{1}-2tc_{1}w_{1}-2stv_{1}w_{1}+c_{2}^{2}+s^{2}v_{2}^{2}+t^{2}w_{2}^{2}+2sa_{2}v_{2}-2tc_{2}w_{2}-2stv_{2}w_{2}+c_{3}^{2}+s^{2}v_{3}^{2}+t^{2}w_{3}^{2}+2sc_{3}v_{3}-2tc_{3}w_{3}-2stv_{3}w_{3}\\&=c_{1}^{2}+c_{2}^{2}+c_{3}^{2}+2s{\left(c_{1}v_{1}+c_{2}v_{2}+c_{3}v_{3}\right)}-2t{\left(c_{1}w_{1}+c_{2}w_{2}+c_{3}w_{3}\right)}+s^{2}{\left(v_{1}^{2}+v_{2}^{2}+v_{3}^{2}\right)}+t^{2}{\left(w_{1}^{2}+w_{2}^{2}+w_{3}^{2}\right)}-2st{\left(v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3}\right)}.\end{aligned}}

We interpret this expression with methods of Analysis 2. We consider the data given by the lines as fixed parameters, so that we have a real-valued expression ${}f(s,t)$ in the two real variables ${}s$ and ${}t$ , and we want to determine its extrema. The partial derivatives are

{\frac {\partial f}{\partial s}}=2{\left(c_{1}v_{1}+c_{2}v_{2}+c_{3}v_{3}\right)}+2s{\left(v_{1}^{2}+v_{2}^{2}+v_{3}^{2}\right)}-2t{\left(v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3}\right)}\,

and

{\frac {\partial f}{\partial t}}=2{\left(c_{1}w_{1}+c_{2}w_{2}+c_{3}w_{3}\right)}+2t{\left(w_{1}^{2}+w_{2}^{2}+w_{3}^{2}\right)}-2s{\left(v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3}\right)}\,.

If we equate this with ${}0$ , then we obtain an inhomogeneous linear system of equations with two equations in the variables ${}s$ and ${}t$ . Using Cramer's rule, we get

{}s={\frac {\det {\begin{pmatrix}-c_{1}v_{1}-c_{2}v_{2}-c_{3}v_{3}&-v_{1}w_{1}-v_{2}w_{2}-v_{3}w_{3}\\-c_{1}w_{1}-c_{2}w_{2}-c_{3}w_{3}&w_{1}^{2}+w_{2}^{2}+w_{3}^{2}\end{pmatrix}}}{\det {\begin{pmatrix}v_{1}^{2}+v_{2}^{2}+v_{3}^{2}&-v_{1}w_{1}-v_{2}w_{2}-v_{3}w_{3}\\-v_{1}w_{1}-v_{2}w_{2}-v_{3}w_{3}&w_{1}^{2}+w_{2}^{2}+w_{3}^{2}\end{pmatrix}}}}\,

and

{}t={\frac {\det {\begin{pmatrix}v_{1}^{2}+v_{2}^{2}+v_{3}^{2}&-c_{1}v_{1}-c_{2}v_{2}-c_{3}v_{3}\\-v_{1}w_{1}-v_{2}w_{2}-v_{3}w_{3}&-c_{1}w_{1}-c_{2}w_{2}-c_{3}w_{3}\end{pmatrix}}}{\det {\begin{pmatrix}v_{1}^{2}+v_{2}^{2}+v_{3}^{2}&-v_{1}w_{1}-v_{2}w_{2}-v_{3}w_{3}\\-v_{1}w_{1}-v_{2}w_{2}-v_{3}w_{3}&w_{1}^{2}+w_{2}^{2}+w_{3}^{2}\end{pmatrix}}}}\,.

If ${}v$ and ${}w$ are normed, then these expressions can be simplified to

{}s={\frac {-\left\langle P-Q,v\right\rangle -\left\langle P-Q,w\right\rangle \left\langle v,w\right\rangle }{1-\left\langle v,w\right\rangle ^{2}}}\,

and

{}t={\frac {-\left\langle P-Q,w\right\rangle -\left\langle P-Q,v\right\rangle \left\langle v,w\right\rangle }{1-\left\langle v,w\right\rangle ^{2}}}\,.

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