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

Angle-preserving mappings

Definition

A linear mapping

$\phi :V⟶W$

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

$\mathrm{\angle }(\phi (u),\phi (v))=\mathrm{\angle }(u,v)$

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

$\phi :ℂ⟶ℂ$

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

$w=a+b\mathrm{i}\ne 0.$

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

$\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right).$

We write this matrix as

$\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)=\left(\begin{array}{cc}\sqrt{a^2+b^2}& 0\\ 0& \sqrt{a^2+b^2}\end{array}\right)\left(\begin{array}{cc}\frac{a}{\sqrt{a^2+b^2}}& -\frac{b}{\sqrt{a^2+b^2}}\\ \frac{b}{\sqrt{a^2+b^2}}& \frac{a}{\sqrt{a^2+b^2}}\end{array}\right).$

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

$|w|=\sqrt{a^2+b^2};$

in particular, this is an angle-preserving mapping.

Theorem

Let

$\phi :V⟶V$

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

$\psi :V⟶V$

and a homothety

$\sigma :V⟶V$

such that

$\phi =\sigma \circ \psi .$

Proof

Let

$r=det\phi ,$

and set

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

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

$\psi :={\sigma }^{-1}\circ \phi .$

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

$◻$

Remark

For an angle-preserving mapping

$\phi :V⟶W$

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 $\phi (Q)$ of the image triangle $\phi (P),\phi (Q),\phi (R)$, because of

$\mathrm{\angle }(P,Q,R)=\mathrm{\angle }(\overrightarrow{QP},\overrightarrow{QR})=\mathrm{\angle }(\phi (\overrightarrow{QP}),\phi (\overrightarrow{QR}))=\mathrm{\angle }(\overrightarrow{\phi (Q)\phi (P)},\overrightarrow{\phi (Q)\phi (R)})=\mathrm{\angle }(\phi (P),\phi (Q),\phi (R)).$

Distances between sets

Definition

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

$d(A,B):=inf(d(P,Q),P\in A,Q\in B)$

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

$p_U(v)=u.$

$◻$

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 $u_1,\dots ,u_m$ denote an orthonormal basis of $U$. Then

$d(v,U{)}^2=\Vert v{\Vert }^2-\sum _{i=1}^m{⟨v,u_i⟩}^2.$

Proof

Because of Lemma 35.6 , we have

$d(v,U)=d(v,p_U(v)),$

and, according to Lemma 32.14 , we have

$p_U(v)=\sum _{i=1}^m⟨v,u_i⟩u_i.$

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

$\begin{array}{cc}d(v,U{)}^2& =\Vert v-p_U(v){\Vert }^2\\ & =\Vert v{\Vert }^2-\Vert p_U(v){\Vert }^2\\ & =\Vert v{\Vert }^2-\Vert \sum _{i=1}^m⟨v,u_i⟩u_i{\Vert }^2\\ & =\Vert v{\Vert }^2-\sum _{i=1}^m{⟨v,u_i⟩}^2.\end{array}$

$◻$

Example

Let $J\subseteq \{1,\dots ,n\}$, and let $U=⟨e_i,i\in J⟩\subseteq {ℝ}^n$ denote the linear subspace spanned by this choice of standard vectors. Let

$v=\left(\begin{array}{c}{v}_1\\ ⋮\\ v_n\end{array}\right).$

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

$d(v,U)=\sqrt{\sum _{i\in ̸J}{v}_i^2}.$

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

$p_U(v)=\left(\begin{array}{c}{w}_1\\ ⋮\\ w_n\end{array}\right),$

where

$w_i=\{\begin{array}{c}{v}_i,\text{ if }i\in J,\\ 0,\text{ if }i\notin J.\end{array}$

Corollary

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

$U=\{x\in {ℝ}^n\mid a_1{x}_1+\cdots +a_n{x}_n=0\}=(ℝa{)}^{\perp }$

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

$d(U,v)=|⟨a,v⟩|.$

Proof

Let $a,u_2,\dots ,u_n$ be an orthonormal basis of ${ℝ}^n$, 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

$d(U,v)=\Vert v-p_U(v)\Vert =\Vert \lambda a\Vert =|\lambda |\Vert a\Vert =|\lambda |.$

In conjunction with

$⟨a,v⟩=⟨a,\lambda a+\sum _{i=2}^n{c}_i{u}_i⟩=⟨a,\lambda a⟩=\lambda$

this yields the result.

$◻$

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 $u_1,\dots ,u_m$ is a basis of $U$, and $w_1,\dots ,w_k$ 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

$d(P,F)=d(P,L)=\Vert \sum _{j=1}^k{b}_j{w}_j\Vert .$

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 $P=\left(\begin{array}{c}4\\ 5\end{array}\right)$ and the line $G$ given by $2x-3y=7$. The line has the form

$G=\{\left(\begin{array}{c}\frac{7}{2}\\ 0\end{array}\right)+t\left(\begin{array}{c}3\\ 2\end{array}\right)\mid t\in ℝ\},$

and $\left(\begin{array}{c}2\\ -3\end{array}\right)$ is a vector perpendicular to $G$. We have

$\left(\begin{array}{c}\frac{7}{2}\\ 0\end{array}\right)-\left(\begin{array}{c}4\\ 5\end{array}\right)=\left(\begin{array}{c}-\frac{1}{2}\\ -5\end{array}\right)=-\frac{23}{26}\left(\begin{array}{c}3\\ 2\end{array}\right)+\frac{14}{13}\left(\begin{array}{c}2\\ -3\end{array}\right).$

Therefore, the dropped perpendicular foot is

$\left(\begin{array}{c}4\\ 5\end{array}\right)+\frac{14}{13}\left(\begin{array}{c}2\\ -3\end{array}\right)=\left(\begin{array}{c}\frac{80}{13}\\ \frac{23}{13}\end{array}\right),$

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 $u\in {(U_1+U_2)}^{\perp }$. 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 $u\in {(U_1+U_2)}^{\perp }$; such a decomposition does always exist, $u_1,u_2$ are not uniquely determined (in case $U_1\cap U_2\ne 0$), 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

$\begin{array}{cc}d(R_1,R_2{)}^2& =\Vert R_1-R_2{\Vert }^2\\ & =\Vert v_1-v_2+u{\Vert }^2\\ & =⟨v_1-v_2+u,v_1-v_2+u⟩\\ & =⟨v_1-v_2,v_1-v_2⟩+⟨u,u⟩\\ & \ge ⟨u,u⟩,\end{array}$

that is,

$d(R_1,R_2)\ge \Vert u\Vert .$

$◻$

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 $G,H\subseteq {ℝ}^3$ 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

$G=P+ℝv,$

and

$H=Q+ℝw.$

The linear system of equations

$P-Q=av+bw+cu$

has a unique solution $(a,b,c)\in {ℝ}^3$. 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

$G=P+ℝv$

and

$H=Q+ℝw$

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

$d(G,H)=|⟨P-Q,u⟩|.$

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 $v\times w$,

$\begin{array}{cc}c& =\frac{det\left(\begin{array}{ccc}{v}_1& w_1& P_1-Q_1\\ v_2& w_2& P_2-Q_2\\ v_3& w_3& P_3-Q_3\end{array}\right)}{det\left(\begin{array}{ccc}{v}_1& w_1& u_1\\ v_2& w_2& u_2\\ v_3& w_3& u_3\end{array}\right)}\\ & =\frac{⟨v\times w,P-Q⟩}{⟨v\times w,u⟩}\\ & =\frac{⟨u,P-Q⟩}{⟨u,u⟩}\\ & =⟨u,P-Q⟩.\end{array}$

$◻$

Example

Let

$G=P+ℝv$

and

$H=Q+ℝw$

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

$P=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)$

and

$Q=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right).$

The square of the distance between the two points

$P^{\prime }=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)+s\left(\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right)$

and

$Q^{\prime }=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)+t\left(\begin{array}{c}{w}_1\\ w_2\\ w_3\end{array}\right)$

is (setting $c_i=a_i-b_i$)

$\begin{array}{cc}d(P^{\prime },Q^{\prime }{)}^2& ={(c_1+s{v}_1-t{w}_1)}^2+{(c_2+s{v}_2-t{w}_2)}^2+{(c_3+s{v}_3-t{w}_3)}^2\\ & =c_1^2+s^2{v}_1^2+t^2{w}_1^2+2s{c}_1{v}_1-2t{c}_1{w}_1-2st{v}_1{w}_1+c_2^2+s^2{v}_2^2+t^2{w}_2^2+2s{a}_2{v}_2-2t{c}_2{w}_2-2st{v}_2{w}_2+c_3^2+s^2{v}_3^2+t^2{w}_3^2+2s{c}_3{v}_3-2t{c}_3{w}_3-2st{v}_3{w}_3\\ & =c_1^2+c_2^2+c_3^2+2s(c_1{v}_1+c_2{v}_2+c_3{v}_3)-2t(c_1{w}_1+c_2{w}_2+c_3{w}_3)+s^2(v_1^2+v_2^2+v_3^2)+t^2(w_1^2+w_2^2+w_3^2)-2st(v_1{w}_1+v_2{w}_2+v_3{w}_3).\end{array}$

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

$\frac{\partial f}{\partial s}=2(c_1{v}_1+c_2{v}_2+c_3{v}_3)+2s(v_1^2+v_2^2+v_3^2)-2t(v_1{w}_1+v_2{w}_2+v_3{w}_3)$

and

$\frac{\partial f}{\partial t}=2(c_1{w}_1+c_2{w}_2+c_3{w}_3)+2t(w_1^2+w_2^2+w_3^2)-2s(v_1{w}_1+v_2{w}_2+v_3{w}_3).$

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

$s=\frac{det\left(\begin{array}{cc}-c_1{v}_1-c_2{v}_2-c_3{v}_3& -v_1{w}_1-v_2{w}_2-v_3{w}_3\\ -c_1{w}_1-c_2{w}_2-c_3{w}_3& w_1^2+w_2^2+w_3^2\end{array}\right)}{det\left(\begin{array}{cc}{v}_1^2+v_2^2+v_3^2& -v_1{w}_1-v_2{w}_2-v_3{w}_3\\ -v_1{w}_1-v_2{w}_2-v_3{w}_3& w_1^2+w_2^2+w_3^2\end{array}\right)}$

and

$t=\frac{det\left(\begin{array}{cc}{v}_1^2+v_2^2+v_3^2& -c_1{v}_1-c_2{v}_2-c_3{v}_3\\ -v_1{w}_1-v_2{w}_2-v_3{w}_3& -c_1{w}_1-c_2{w}_2-c_3{w}_3\end{array}\right)}{det\left(\begin{array}{cc}{v}_1^2+v_2^2+v_3^2& -v_1{w}_1-v_2{w}_2-v_3{w}_3\\ -v_1{w}_1-v_2{w}_2-v_3{w}_3& w_1^2+w_2^2+w_3^2\end{array}\right)}.$

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

$s=\frac{-⟨P-Q,v⟩-⟨P-Q,w⟩⟨v,w⟩}{1-{⟨v,w⟩}^2}$

and

$t=\frac{-⟨P-Q,w⟩-⟨P-Q,v⟩⟨v,w⟩}{1-{⟨v,w⟩}^2}.$

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