# Linear Algebra/Topic: Orthonormal Matrices/Solutions

## Solutions

Problem 1

Decide if each of these is an orthonormal matrix.

1. $\begin{pmatrix} 1/\sqrt{2} &-1/\sqrt{2} \\ -1/\sqrt{2} &-1/\sqrt{2} \end{pmatrix}$
2. $\begin{pmatrix} 1/\sqrt{3} &-1/\sqrt{3} \\ -1/\sqrt{3} &-1/\sqrt{3} \end{pmatrix}$
3. $\begin{pmatrix} 1/\sqrt{3} &-\sqrt{2}/\sqrt{3} \\ -\sqrt{2}/\sqrt{3} &-1/\sqrt{3} \end{pmatrix}$
1. Yes.
2. No, the columns do not have length one.
3. Yes.
Problem 2

Write down the formula for each of these distance-preserving maps.

1. the map that rotates $\pi/6$ radians, and then translates by $\vec{e}_2$
2. the map that reflects about the line $y=2x$
3. the map that reflects about $y=-2x$ and translates over $1$ and up $1$

Some of these are nonlinear, because they involve a nontrivial translation.

1. $\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x\cdot\cos(\pi/6)-y\cdot\sin(\pi/6) \\ x\cdot\sin(\pi/6)+y\cdot\cos(\pi/6) \end{pmatrix} +\begin{pmatrix} 0 \\ 1 \end{pmatrix} =\begin{pmatrix} x\cdot(\sqrt{3}/2)-y\cdot(1/2)+0 \\ x\cdot(1/2)+y\cdot\cos(\sqrt{3}/2)+1 \end{pmatrix}$
2. The line $y=2x$ makes an angle of $\arctan(2/1)$ with the $x$-axis. Thus $\sin\theta=2/\sqrt{5}$ and $\cos\theta=1/\sqrt{5}$.
$\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x\cdot(1/\sqrt{5})-y\cdot(2/\sqrt{5}) \\ x\cdot(2/\sqrt{5})+y\cdot(1/\sqrt{5}) \end{pmatrix}$
3. $\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x\cdot(1/\sqrt{5})-y\cdot(-2/\sqrt{5}) \\ x\cdot(-2/\sqrt{5})+y\cdot(1/\sqrt{5}) \end{pmatrix} +\begin{pmatrix} 1 \\ 1 \end{pmatrix} =\begin{pmatrix} x/\sqrt{5}+2y/\sqrt{5}+1 \\ -2x/\sqrt{5}+y/\sqrt{5}+1 \end{pmatrix}$
Problem 3
1. The proof that a map that is distance-preserving and sends the zero vector to itself incidentally shows that such a map is one-to-one and onto (the point in the domain determined by $d_0$, $d_1$, and $d_2$ corresponds to the point in the codomain determined by those three). Therefore any distance-preserving map has an inverse. Show that the inverse is also distance-preserving.
2. Prove that congruence is an equivalence relation between plane figures.
1. Let $f$ be distance-preserving and consider $f^{-1}$. Any two points in the codomain can be written as $f(P_1)$ and $f(P_2)$. Because $f$ is distance-preserving, the distance from $f(P_1)$ to $f(P_2)$ equals the distance from $P_1$ to $P_2$. But this is exactly what is required for $f^{-1}$ to be distance-preserving.
2. Any plane figure $F$ is congruent to itself via the identity map $\mbox{id}:\mathbb{R}^2\to \mathbb{R}^2$, which is obviously distance-preserving. If $F_1$ is congruent to $F_2$ (via some $f$) then $F_2$ is congruent to $F_1$ via $f^{-1}$, which is distance-preserving by the prior item. Finally, if $F_1$ is congruent to $F_2$ (via some $f$) and $F_2$ is congruent to $F_3$ (via some $g$) then $F_1$ is congruent to $F_3$ via $g\circ f$, which is easily checked to be distance-preserving.
Problem 4

In practice the matrix for the distance-preserving linear transformation and the translation are often combined into one. Check that these two computations yield the same first two components.

$\begin{pmatrix} a &c \\ b &d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} +\begin{pmatrix} e \\ f \end{pmatrix} \qquad \begin{pmatrix} a &c &e \\ b &d &f \\ 0 &0 &1 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}$

(These are homogeneous coordinates; see the Topic on Projective Geometry).

The first two components of each are $ax+cy+e$ and $bx+dy+f$.
1. The Pythagorean Theorem gives that three points are colinear if and only if (for some ordering of them into $P_1$, $P_2$, and $P_3$), $\text{dist}\,(P_1,P_2)+\text{dist}\,(P_2,P_3)=\text{dist}\,(P_1,P_3)$. Of course, where $f$ is distance-preserving, this holds if and only if $\text{dist}\,(f(P_1),f(P_2))+\text{dist}\,(f(P_2),f(P_3))=\text{dist}\,(f(P_1),f(P_3))$, which, again by Pythagoras, is true if and only if $f(P_1)$, $f(P_2)$, and $f(P_3)$ are colinear. The argument for betweeness is similar (above, $P_2$ is between $P_1$ and $P_3$). If the figure $F$ is a triangle then it is the union of three line segments $P_1P_2$, $P_2P_3$, and $P_1P_3$. The prior two paragraphs together show that the property of being a line segment is invariant. So $f(F)$ is the union of three line segments, and so is a triangle. A circle $C$ centered at $P$ and of radius $r$ is the set of all points $Q$ such that $\text{dist}\,(P,Q)=r$. Applying the distance-preserving map $f$ gives that the image $f(C)$ is the set of all $f(Q)$ subject to the condition that $\text{dist}\,(P,Q)=r$. Since $\text{dist}\,(P,Q)=\text{dist}\,(f(P),f(Q))$, the set $f(C)$ is also a circle, with center $f(P)$ and radius $r$.
3. One that was mentioned in the section is the "sense" of a figure. A triangle whose vertices read clockwise as $P_1$, $P_2$, $P_3$ may, under a distance-preserving map, be sent to a triangle read $P_1$, $P_2$, $P_3$ counterclockwise.