# Linear Algebra/Topic: Orthonormal Matrices/Solutions

## Solutions

Problem 1

Decide if each of these is an orthonormal matrix.

1. ${\displaystyle {\begin{pmatrix}1/{\sqrt {2}}&-1/{\sqrt {2}}\\-1/{\sqrt {2}}&-1/{\sqrt {2}}\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}1/{\sqrt {3}}&-1/{\sqrt {3}}\\-1/{\sqrt {3}}&-1/{\sqrt {3}}\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}1/{\sqrt {3}}&-{\sqrt {2}}/{\sqrt {3}}\\-{\sqrt {2}}/{\sqrt {3}}&-1/{\sqrt {3}}\end{pmatrix}}}$
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 ${\displaystyle \pi /6}$  radians, and then translates by ${\displaystyle {\vec {e}}_{2}}$
2. the map that reflects about the line ${\displaystyle y=2x}$
3. the map that reflects about ${\displaystyle y=-2x}$  and translates over ${\displaystyle 1}$  and up ${\displaystyle 1}$

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

1. ${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x\cdot \cos(\pi /6)-y\cdot \sin(\pi /6)\\x\cdot \sin(\pi /6)+y\cdot \cos(\pi /6)\end{pmatrix}}+{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}x\cdot ({\sqrt {3}}/2)-y\cdot (1/2)+0\\x\cdot (1/2)+y\cdot \cos({\sqrt {3}}/2)+1\end{pmatrix}}}$
2. The line ${\displaystyle y=2x}$  makes an angle of ${\displaystyle \arctan(2/1)}$  with the ${\displaystyle x}$ -axis. Thus ${\displaystyle \sin \theta =2/{\sqrt {5}}}$  and ${\displaystyle \cos \theta =1/{\sqrt {5}}}$ .
${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x\cdot (1/{\sqrt {5}})-y\cdot (2/{\sqrt {5}})\\x\cdot (2/{\sqrt {5}})+y\cdot (1/{\sqrt {5}})\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x\cdot (1/{\sqrt {5}})-y\cdot (-2/{\sqrt {5}})\\x\cdot (-2/{\sqrt {5}})+y\cdot (1/{\sqrt {5}})\end{pmatrix}}+{\begin{pmatrix}1\\1\end{pmatrix}}={\begin{pmatrix}x/{\sqrt {5}}+2y/{\sqrt {5}}+1\\-2x/{\sqrt {5}}+y/{\sqrt {5}}+1\end{pmatrix}}}$
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 ${\displaystyle d_{0}}$ , ${\displaystyle d_{1}}$ , and ${\displaystyle d_{2}}$  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 ${\displaystyle f}$  be distance-preserving and consider ${\displaystyle f^{-1}}$ . Any two points in the codomain can be written as ${\displaystyle f(P_{1})}$  and ${\displaystyle f(P_{2})}$ . Because ${\displaystyle f}$  is distance-preserving, the distance from ${\displaystyle f(P_{1})}$  to ${\displaystyle f(P_{2})}$  equals the distance from ${\displaystyle P_{1}}$  to ${\displaystyle P_{2}}$ . But this is exactly what is required for ${\displaystyle f^{-1}}$  to be distance-preserving.
2. Any plane figure ${\displaystyle F}$  is congruent to itself via the identity map ${\displaystyle {\mbox{id}}:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}$ , which is obviously distance-preserving. If ${\displaystyle F_{1}}$  is congruent to ${\displaystyle F_{2}}$  (via some ${\displaystyle f}$ ) then ${\displaystyle F_{2}}$  is congruent to ${\displaystyle F_{1}}$  via ${\displaystyle f^{-1}}$ , which is distance-preserving by the prior item. Finally, if ${\displaystyle F_{1}}$  is congruent to ${\displaystyle F_{2}}$  (via some ${\displaystyle f}$ ) and ${\displaystyle F_{2}}$  is congruent to ${\displaystyle F_{3}}$  (via some ${\displaystyle g}$ ) then ${\displaystyle F_{1}}$  is congruent to ${\displaystyle F_{3}}$  via ${\displaystyle g\circ f}$ , 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.

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

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

The first two components of each are ${\displaystyle ax+cy+e}$  and ${\displaystyle bx+dy+f}$ .
1. The Pythagorean Theorem gives that three points are colinear if and only if (for some ordering of them into ${\displaystyle P_{1}}$ , ${\displaystyle P_{2}}$ , and ${\displaystyle P_{3}}$ ), ${\displaystyle {\text{dist}}\,(P_{1},P_{2})+{\text{dist}}\,(P_{2},P_{3})={\text{dist}}\,(P_{1},P_{3})}$ . Of course, where ${\displaystyle f}$  is distance-preserving, this holds if and only if ${\displaystyle {\text{dist}}\,(f(P_{1}),f(P_{2}))+{\text{dist}}\,(f(P_{2}),f(P_{3}))={\text{dist}}\,(f(P_{1}),f(P_{3}))}$ , which, again by Pythagoras, is true if and only if ${\displaystyle f(P_{1})}$ , ${\displaystyle f(P_{2})}$ , and ${\displaystyle f(P_{3})}$  are colinear. The argument for betweeness is similar (above, ${\displaystyle P_{2}}$  is between ${\displaystyle P_{1}}$  and ${\displaystyle P_{3}}$ ). If the figure ${\displaystyle F}$  is a triangle then it is the union of three line segments ${\displaystyle P_{1}P_{2}}$ , ${\displaystyle P_{2}P_{3}}$ , and ${\displaystyle P_{1}P_{3}}$ . The prior two paragraphs together show that the property of being a line segment is invariant. So ${\displaystyle f(F)}$  is the union of three line segments, and so is a triangle. A circle ${\displaystyle C}$  centered at ${\displaystyle P}$  and of radius ${\displaystyle r}$  is the set of all points ${\displaystyle Q}$  such that ${\displaystyle {\text{dist}}\,(P,Q)=r}$ . Applying the distance-preserving map ${\displaystyle f}$  gives that the image ${\displaystyle f(C)}$  is the set of all ${\displaystyle f(Q)}$  subject to the condition that ${\displaystyle {\text{dist}}\,(P,Q)=r}$ . Since ${\displaystyle {\text{dist}}\,(P,Q)={\text{dist}}\,(f(P),f(Q))}$ , the set ${\displaystyle f(C)}$  is also a circle, with center ${\displaystyle f(P)}$  and radius ${\displaystyle r}$ .
3. One that was mentioned in the section is the "sense" of a figure. A triangle whose vertices read clockwise as ${\displaystyle P_{1}}$ , ${\displaystyle P_{2}}$ , ${\displaystyle P_{3}}$  may, under a distance-preserving map, be sent to a triangle read ${\displaystyle P_{1}}$ , ${\displaystyle P_{2}}$ , ${\displaystyle P_{3}}$  counterclockwise.