# 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 $P_{1}P_{2}$ , $P_{2}P_{3}$ , and $P_{1}P_{3}$ . 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.