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 Jim Hefferon committed Dec 05, 2011 1 % Chapter 3, Topic _Linear Algebra_ Jim Hefferon  Jim Hefferon committed Oct 28, 2013 2 % http://joshua.smcvt.edu/linearalgebra  Jim Hefferon committed Dec 05, 2011 3 4 5 6 7 8 % 2001-Jun-12 \topic{Orthonormal Matrices} \index{orthonormal matrix|(} \index{matrix!orthonormal|(} In \emph{The Elements},\index{Euclid} Euclid considers two figures to be the same if they have the same size and shape.  Jim Hefferon committed Jan 09, 2012 9 10 That is, while the triangles below are not equal because they are not the same set of points,  Jim Hefferon committed Nov 27, 2013 11 they are, for Euclid's purposes, essentially  Jim Hefferon committed Dec 05, 2011 12 indistinguishable  Jim Hefferon committed Nov 27, 2013 13 because we can imagine  Jim Hefferon committed Dec 05, 2011 14 15 16 17 18 19 20 21 22 23 picking the plane up, sliding it over and rotating it a bit, although not warping or stretching it, and then putting it back down, to superimpose the first figure on the second. (Euclid never explicitly states this principle but he uses it often \cite{Casey}.) \begin{center} \includegraphics{ch3.61} \end{center}  Jim Hefferon committed Oct 28, 2013 24 25 In modern terms picking the plane up~\ldots'' is taking a  Jim Hefferon committed Dec 05, 2011 26 map from the plane to itself.  Jim Hefferon committed Oct 28, 2013 27 Euclid considers only transformations  Jim Hefferon committed Jan 09, 2012 28 that may slide or turn the plane but not bend or stretch it.  Jim Hefferon committed Oct 28, 2013 29 Accordingly, define a map $\map{f}{\Re^2}{\Re^2}$ to be  Jim Hefferon committed Nov 27, 2013 30 31 \definend{distance-preserving}\index{distance-preserving map}% \index{map!distance-preserving}\index{function!distance-preserving}  Jim Hefferon committed Dec 05, 2011 32 or a \definend{rigid motion}\index{rigid motion} or an  Jim Hefferon committed Oct 28, 2013 33 \definend{isometry}\index{isometry}  Jim Hefferon committed Dec 05, 2011 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 if for all points $P_1,P_2\in\Re^2$, the distance from $f(P_1)$ to $f(P_2)$ equals the distance from $P_1$ to $P_2$. We also define a plane \definend{figure}\index{plane figure} to be a set of points in the plane and we say that two figures are \definend{congruent}\index{congruent plane figures}% \index{plane figure!congruence} if there is a distance-preserving map from the plane to itself that carries one figure onto the other. Many statements from Euclidean geometry follow easily from these definitions. Some are: (i)~collinearity is invariant under any distance-preserving map (that is, if $P_1$, $P_2$, and $P_3$ are collinear then so are $f(P_1)$, $f(P_2)$, and $f(P_3)$), (ii)~betweeness is invariant under any distance-preserving map (if $P_2$ is between $P_1$ and $P_3$ then so is $f(P_2)$ between $f(P_1)$ and $f(P_3)$), (iii)~the property of being a triangle is invariant under any distance-preserving map (if a figure is a triangle then the image of that figure is also a triangle), (iv)~and the property of being a circle is invariant under any distance-preserving map.  Jim Hefferon committed Jan 09, 2012 58 59 In 1872, F.~Klein\index{Klein, F.} suggested that we can define Euclidean geometry as the study of properties that  Jim Hefferon committed Dec 05, 2011 60 61 are invariant under these maps. (This forms part of Klein's Erlanger\index{Erlanger Program} Program,  Jim Hefferon committed Jan 09, 2012 62 63 which proposes the organizing principle that we can describe each kind of geometry\Dash Euclidean, projective, etc.\Dash  Jim Hefferon committed Dec 05, 2011 64 65 as the study of the properties that are invariant under some group of transformations.  Jim Hefferon committed Oct 28, 2013 66 The word group' here means more than just collection'  Jim Hefferon committed Dec 05, 2011 67 68 69 70 71 but that lies outside of our scope.) We can use linear algebra to characterize the distance-preserving maps of the plane.  Jim Hefferon committed Oct 28, 2013 72 To begin, observe that  Jim Hefferon committed Dec 05, 2011 73 74 75 76 77 there are distance-preserving transformations of the plane that are not linear. The obvious example is this \emph{translation}.\index{translation} \begin{equation*} \colvec{x \\ y} \quad\mapsto\quad  Jim Hefferon committed Jan 08, 2012 78  \colvec{x \\ y}+\colvec[r]{1 \\ 0}=\colvec{x+1 \\ y}  Jim Hefferon committed Dec 05, 2011 79 80 \end{equation*} However,  Jim Hefferon committed Oct 28, 2013 81 82 this example turns out to be the only one, in that if $f$ is distance-preserving and sends $\zero$ to $\vec{v}_0$  Jim Hefferon committed Dec 05, 2011 83 84 85 then the map $\vec{v}\mapsto f(\vec{v})-\vec{v}_0$ is linear. That will follow immediately from this statement:~a map $t$ that is distance-preserving and sends $\zero$ to itself is linear.  Jim Hefferon committed Oct 28, 2013 86 To prove this equivalent statement, consider the standard basis and suppose that  Jim Hefferon committed Dec 05, 2011 87 88 89 90 91 92 \begin{equation*} t(\vec{e}_1)=\colvec{a \\ b} \qquad t(\vec{e}_2)=\colvec{c \\ d} \end{equation*} for some $a,b,c,d\in\Re$.  Jim Hefferon committed Oct 28, 2013 93 To show that $t$ is linear we can show that  Jim Hefferon committed Dec 05, 2011 94 95 96 97 98 99 100 101 it can be represented by a matrix, that is, that $t$ acts in this way for all $x,y\in\Re$. \begin{equation*} \vec{v}=\colvec{x \\ y} \mapsunder{t} \colvec{ax+cy \\ bx+dy} \tag*{\text{($*$)}}\end{equation*} Recall that if we fix three non-collinear points  Jim Hefferon committed Jan 09, 2012 102 103 104 105 106 107 then we can determine any point by giving its distance from those three. So we can determine any point $\vec{v}$ in the domain by its distance from $\zero$, $\vec{e}_1$, and $\vec{e}_2$. Similarly, we can determine any point $t(\vec{v})$ in the codomain by its distance from  Jim Hefferon committed Dec 05, 2011 108 109 110 111 the three fixed points $t(\zero)$, $t(\vec{e}_1)$, and $t(\vec{e}_2)$ (these three are not collinear because, as mentioned above, collinearity is invariant and $\zero$, $\vec{e}_1$, and $\vec{e}_2$ are not collinear).  Jim Hefferon committed Oct 28, 2013 112 Because $t$ is distance-preserving we can say more:~for the  Jim Hefferon committed Dec 05, 2011 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 point $\vec{v}$ in the plane that is determined by being the distance $d_0$ from $\zero$, the distance $d_1$ from $\vec{e}_1$, and the distance $d_2$ from $\vec{e}_2$, its image $t(\vec{v})$ must be the unique point in the codomain that is determined by being $d_0$ from $t(\zero)$, $d_1$ from $t(\vec{e}_1)$, and $d_2$ from $t(\vec{e}_2)$. Because of the uniqueness, checking that the action in ($*$) works in the $d_0$, $d_1$, and $d_2$ cases \begin{equation*} \dist(\colvec{x \\ y},\zero) =\dist(t(\colvec{x \\ y}),t(\zero)) =\dist(\colvec{ax+cy \\ bx+dy},\zero) \end{equation*}  Jim Hefferon committed Jan 09, 2012 128 (we assumed that $t$ maps $\zero$ to itself)  Jim Hefferon committed Dec 05, 2011 129 130 131 132 133 134 135 136 137 138 139 140 141 142 \begin{equation*} \dist(\colvec{x \\ y},\vec{e}_1) =\dist(t(\colvec{x \\ y}),t(\vec{e}_1)) =\dist(\colvec{ax+cy \\ bx+dy},\colvec{a \\ b}) \end{equation*} and \begin{equation*} \dist(\colvec{x \\ y},\vec{e}_2) =\dist(t(\colvec{x \\ y}),t(\vec{e}_2)) =\dist(\colvec{ax+cy \\ bx+dy},\colvec{c \\ d}) \end{equation*} suffices to show that ($*$) describes $t$. Those checks are routine.  Jim Hefferon committed Oct 28, 2013 143 144 Thus any distance-preserving $\map{f}{\Re^2}{\Re^2}$ is a linear map plus a translation,  Jim Hefferon committed Dec 05, 2011 145 146 $f(\vec{v})=t(\vec{v})+\vec{v}_0$ for some constant vector $\vec{v}_0$ and linear map $t$ that is distance-preserving.  Jim Hefferon committed Oct 28, 2013 147 So in order to understand distance-preserving maps what remains is to  Jim Hefferon committed Jan 09, 2012 148 understand distance-preserving linear maps.  Jim Hefferon committed Dec 05, 2011 149   Jim Hefferon committed Jan 09, 2012 150 151 Not every linear map is distance-preserving. For example  Jim Hefferon committed Dec 05, 2011 152 $\vec{v}\mapsto 2\vec{v}$ does not preserve distances.  Jim Hefferon committed Jan 09, 2012 153   Jim Hefferon committed Dec 05, 2011 154 155 But there is a neat characterization:~a linear transformation $t$ of the plane is distance-preserving if and only if both  Jim Hefferon committed Jan 09, 2012 156 $\norm{t(\vec{e}_1)}=\norm{t(\vec{e}_2)}=1$,  Jim Hefferon committed Dec 05, 2011 157 158 and $t(\vec{e}_1)$ is orthogonal to $t(\vec{e}_2)$. The only if' half of that statement is easy\Dash because $t$ is  Jim Hefferon committed Jan 09, 2012 159 distance-preserving it must preserve the lengths of vectors  Jim Hefferon committed Dec 05, 2011 160 161 and because $t$ is distance-preserving the Pythagorean theorem shows that it must preserve orthogonality.  Jim Hefferon committed Jan 09, 2012 162 163 To show the if' half we can check that the map preserves lengths of vectors because then for all  Jim Hefferon committed Dec 05, 2011 164 165 166 $\vec{p}$ and $\vec{q}$ the distance between the two is preserved $\norm{t(\vec{p}-\vec{q}\,)}=\norm{t(\vec{p})-t(\vec{q}\,)} =\norm{\vec{p}-\vec{q}\,}$.  Jim Hefferon committed Jan 09, 2012 167 For that check let  Jim Hefferon committed Dec 05, 2011 168 169 170 171 172 173 174 \begin{equation*} \vec{v}=\colvec{x \\ y} \quad t(\vec{e}_1)=\colvec{a \\ b} \quad t(\vec{e}_2)=\colvec{c \\ d} \end{equation*}  Jim Hefferon committed Jan 09, 2012 175 and with the if' assumptions that  Jim Hefferon committed Dec 05, 2011 176 177 178 179 180 181 182 183 184 185 186 187 $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$ we have this. \begin{align*} \norm{t(\vec{v}\,)}^2 &= (ax+cy)^2+(bx+dy)^2 \\ &= a^2x^2+2acxy+c^2y^2+b^2x^2+2bdxy+d^2y^2 \\ &= x^2(a^2+b^2)+y^2(c^2+d^2)+2xy(ac+bd) \\ &= x^2 + y^2 \\ &= \norm{\vec{v}\,}^2 \end{align*} One thing that is neat about this characterization is that we can easily recognize  Jim Hefferon committed Jan 09, 2012 188 189 matrices that represent such a map with respect to the standard bases: the columns  Jim Hefferon committed Dec 05, 2011 190 are of length one and are mutually orthogonal.  Jim Hefferon committed Jan 09, 2012 191 This is an  Jim Hefferon committed Dec 05, 2011 192 193 \definend{orthonormal matrix}\index{orthonormal matrix}% \index{matrix!orthonormal}  Jim Hefferon committed Oct 28, 2013 194 (or, more informally,  Jim Hefferon committed Dec 05, 2011 195 \definend{orthogonal matrix}\index{orthogonal matrix}\index{matrix!orthogonal}  Jim Hefferon committed Oct 28, 2013 196 since people often use this term to mean not just that the columns are  Jim Hefferon committed Jan 09, 2012 197 orthogonal but also that they have length one).  Jim Hefferon committed Dec 05, 2011 198   Jim Hefferon committed Oct 28, 2013 199 We can leverage this characterization to  Jim Hefferon committed Jan 09, 2012 200 understand the geometric actions of  Jim Hefferon committed Dec 05, 2011 201 distance-preserving maps.  Jim Hefferon committed Jan 09, 2012 202 203 Because $\norm{t(\vec{v}\,)}=\norm{\vec{v}\,}$, the map~$t$ sends any $\vec{v}$ somewhere on the circle about the origin that has  Jim Hefferon committed Dec 05, 2011 204 205 radius equal to the length of $\vec{v}$. In particular,  Jim Hefferon committed Jan 09, 2012 206 $\vec{e}_1$ and $\vec{e}_2$ map to the unit circle.  Jim Hefferon committed Dec 05, 2011 207 208 209 What's more, %because of the orthogonality restriction, once we fix the unit vector $\vec{e}_1$ as mapped to the vector with components $a$ and $b$ then there are only two places  Jim Hefferon committed Jan 09, 2012 210 211 where $\vec{e}_2$ can go if its image is to be perpendicular to the first vector's image:~it can map either to one  Jim Hefferon committed Dec 05, 2011 212 213 214 215 216 217 where $\vec{e}_2$ maintains its position a quarter circle clockwise from $\vec{e}_1$ \begin{center} \begin{tabular}{@{}c@{}}\includegraphics{ch3.62}\end{tabular} \qquad $\rep{t}{\stdbasis_2,\stdbasis_2}=  Jim Hefferon committed Jan 08, 2012 218  \begin{mat}  Jim Hefferon committed Dec 05, 2011 219 220  a &-b \\ b &a  Jim Hefferon committed Jan 08, 2012 221  \end{mat}$  Jim Hefferon committed Dec 05, 2011 222 \end{center}  Jim Hefferon committed Jan 09, 2012 223 or to one where it goes a quarter circle counterclockwise.  Jim Hefferon committed Dec 05, 2011 224 225 226 227 \begin{center} \begin{tabular}{@{}c@{}}\includegraphics{ch3.63}\end{tabular} \qquad $\rep{t}{\stdbasis_2,\stdbasis_2}=  Jim Hefferon committed Jan 08, 2012 228  \begin{mat}  Jim Hefferon committed Dec 05, 2011 229 230  a &b \\ b &-a  Jim Hefferon committed Jan 08, 2012 231  \end{mat}$  Jim Hefferon committed Dec 05, 2011 232 233 \end{center}  Jim Hefferon committed Oct 28, 2013 234 The geometric description of these two cases is easy.  Jim Hefferon committed Jan 09, 2012 235 236 Let $\theta$ be the counterclockwise angle between the $x$-axis and the image of $\vec{e}_1$.  Jim Hefferon committed Dec 05, 2011 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 The first matrix above represents, with respect to the standard bases, a \definend{rotation}\index{rotation}\index{linear map!rotation} of the plane by $\theta$ radians. \begin{center} \begin{tabular}{@{}c@{}}\includegraphics{ch3.62}\end{tabular} \qquad $\colvec{x \\ y} \mapsunder{t} \colvec{x\cos\theta-y\sin\theta \\ x\sin\theta+y\cos\theta}$ \end{center} The second matrix above represents a \definend{reflection}\index{reflection}\index{linear map!reflection} of the plane through the line bisecting the angle between $\vec{e}_1$ and $t(\vec{e}_1)$. \begin{center} \begin{tabular}{@{}c@{}}\includegraphics{ch3.64}\end{tabular} \qquad $\colvec{x \\ y}\mapsunder{t} \colvec{x\cos\theta+y\sin\theta \\ x\sin\theta-y\cos\theta}$ \end{center} (This picture shows $\vec{e}_1$ reflected up into the first quadrant and $\vec{e}_2$ reflected down into the fourth quadrant.)  Jim Hefferon committed Jan 09, 2012 260 Note: in the domain  Jim Hefferon committed Dec 05, 2011 261 262 263 the angle between $\vec{e}_1$ and $\vec{e}_2$ runs counterclockwise, and in the first map above the angle from $t(\vec{e}_1)$ to $t(\vec{e}_2)$ is also counterclockwise,  Jim Hefferon committed Jan 09, 2012 264 265 so it preserves the orientation of the angle. But the second map reverses the orientation.  Jim Hefferon committed Dec 05, 2011 266 A distance-preserving map is  Jim Hefferon committed Jan 09, 2012 267 \definend{direct}\index{direct map}\index{orientation preserving map} if it  Jim Hefferon committed Dec 05, 2011 268 preserves orientations  Jim Hefferon committed Jan 09, 2012 269 and \definend{opposite}\index{opposite map}\index{orientation reversing map}  Jim Hefferon committed Dec 05, 2011 270 271 if it reverses orientation.  Jim Hefferon committed Oct 28, 2013 272 With that, we have characterized the Euclidean study of congruence.  Jim Hefferon committed Jan 09, 2012 273 It considers, for plane figures, the properties that are invariant  Jim Hefferon committed Dec 05, 2011 274 275 276 277 278 under combinations of (i)~a rotation followed by a translation, or (ii)~a reflection followed by a translation (a reflection followed by a non-trivial translation is a \definend{glide reflection}\index{reflection!glide}).  Jim Hefferon committed Oct 28, 2013 279 280 281 Another idea encountered in elementary geometry, besides congruence of figures, is that figures are  Jim Hefferon committed Nov 29, 2013 282 \definend{similar}\index{similar triangles}\index{triangles, similar}  Jim Hefferon committed Dec 05, 2011 283 if they are congruent after a change of scale.  Jim Hefferon committed Oct 28, 2013 284 285 The two triangles below are similar since the second is the same shape as the first but $3/2$-ths the size.  Jim Hefferon committed Dec 05, 2011 286 287 288 \begin{center} \includegraphics{ch3.65} \end{center}  Jim Hefferon committed Jan 09, 2012 289 From the above work we have that figures are similar if there  Jim Hefferon committed Dec 05, 2011 290 is an orthonormal matrix $T$ such that the points $\vec{q}$  Jim Hefferon committed Jan 09, 2012 291 292 on one figure are the images of the points $\vec{p}$ on the other figure by  Jim Hefferon committed Dec 05, 2011 293 294 295 $\vec{q}=(kT)\vec{v}+\vec{p}_0$ for some nonzero real number $k$ and constant vector $\vec{p}_0$.  Jim Hefferon committed Jan 09, 2012 296 297 Although these ideas are from Euclid, mathematics is timeless and they are still in use today.  Jim Hefferon committed Dec 05, 2011 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 One application of the maps studied above is in computer graphics. We can, for example, animate this top view of a cube by putting together film frames of it rotating; that's a rigid motion. \begin{center} \begin{tabular}{ccc} \includegraphics{ch3.66} &\includegraphics{ch3.67} &\includegraphics{ch3.68} \\ Frame 1 &Frame 2 &Frame 3 \end{tabular} \end{center} We could also make the cube appear to be moving away from us by producing film frames of it shrinking, which gives us figures that are similar. \begin{center} \begin{tabular}{ccc} \includegraphics{ch3.69} &\includegraphics{ch3.70} &\includegraphics{ch3.71} \\ Frame 1: &Frame 2: &Frame 3: \end{tabular} \end{center} Computer graphics incorporates techniques from linear algebra in many other ways (see \nearbyexercise{exer:HomoCrds}).  Jim Hefferon committed Jan 09, 2012 322 323 % So the analysis above of distance-preserving maps % is useful as well as interesting.  Jim Hefferon committed Dec 05, 2011 324 A beautiful book that explores some of this area is \cite{Weyl}.  Jim Hefferon committed Jan 09, 2012 325 More on groups, of transformations and otherwise, is in any book  Jim Hefferon committed Dec 05, 2011 326 327 328 329 330 331 332 333 on Modern Algebra, for instance \cite{BirkhoffMaclane}. More on Klein and the Erlanger Program is in \cite{Yaglom}. \begin{exercises} \item Decide if each of these is an orthonormal matrix. \begin{exparts}  Jim Hefferon committed Jan 08, 2012 334  \partsitem $\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 335 336  1/\sqrt{2} &-1/\sqrt{2} \\ -1/\sqrt{2} &-1/\sqrt{2}  Jim Hefferon committed Jan 08, 2012 337 338  \end{mat}$ \partsitem $\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 339 340  1/\sqrt{3} &-1/\sqrt{3} \\ -1/\sqrt{3} &-1/\sqrt{3}  Jim Hefferon committed Jan 08, 2012 341 342  \end{mat}$ \partsitem $\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 343 344  1/\sqrt{3} &-\sqrt{2}/\sqrt{3} \\ -\sqrt{2}/\sqrt{3} &-1/\sqrt{3}  Jim Hefferon committed Jan 08, 2012 345  \end{mat}$  Jim Hefferon committed Dec 05, 2011 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369  \end{exparts} \begin{answer} \begin{exparts} \partsitem Yes. \partsitem No, the columns do not have length one. \partsitem Yes. \end{exparts} \end{answer} \item Write down the formula for each of these distance-preserving maps. \begin{exparts} \partsitem the map that rotates $\pi/6$ radians, and then translates by $\vec{e}_2$ \partsitem the map that reflects about the line $y=2x$ \partsitem the map that reflects about $y=-2x$ and translates over $1$ and up $1$ \end{exparts} \begin{answer} Some of these are nonlinear, because they involve a nontrivial translation. \begin{exparts} \partsitem $\colvec{x \\ y} \mapsto  Jim Hefferon committed Jan 08, 2012 370  \begin{mat}  Jim Hefferon committed Dec 05, 2011 371 372  x\cdot\cos(\pi/6)-y\cdot\sin(\pi/6) \\ x\cdot\sin(\pi/6)+y\cdot\cos(\pi/6)  Jim Hefferon committed Jan 08, 2012 373 374 375  \end{mat} +\colvec[r]{0 \\ 1} =\begin{mat}  Jim Hefferon committed Dec 05, 2011 376 377  x\cdot(\sqrt{3}/2)-y\cdot(1/2)+0 \\ x\cdot(1/2)+y\cdot\cos(\sqrt{3}/2)+1  Jim Hefferon committed Jan 08, 2012 378  \end{mat}$  Jim Hefferon committed Dec 05, 2011 379 380 381 382 383 384  \partsitem 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{equation*} \colvec{x \\ y} \mapsto  Jim Hefferon committed Jan 08, 2012 385  \begin{mat}  Jim Hefferon committed Dec 05, 2011 386 387  x\cdot(1/\sqrt{5})-y\cdot(2/\sqrt{5}) \\ x\cdot(2/\sqrt{5})+y\cdot(1/\sqrt{5})  Jim Hefferon committed Jan 08, 2012 388  \end{mat}  Jim Hefferon committed Dec 05, 2011 389 390 391 392  \end{equation*} \partsitem $\colvec{x \\ y} \mapsto  Jim Hefferon committed Jan 08, 2012 393  \begin{mat}  Jim Hefferon committed Dec 05, 2011 394 395  x\cdot(1/\sqrt{5})-y\cdot(-2/\sqrt{5}) \\ x\cdot(-2/\sqrt{5})+y\cdot(1/\sqrt{5})  Jim Hefferon committed Jan 08, 2012 396 397 398  \end{mat} +\colvec[r]{1 \\ 1} =\begin{mat}  Jim Hefferon committed Dec 05, 2011 399 400  x/\sqrt{5}+2y/\sqrt{5}+1 \\ -2x/\sqrt{5}+y/\sqrt{5}+1  Jim Hefferon committed Jan 08, 2012 401  \end{mat}$  Jim Hefferon committed Dec 05, 2011 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443  \end{exparts} \end{answer} \item \label{exer:IsometryFacts} \begin{exparts} \partsitem 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. \partsitem Prove that congruence is an equivalence relation between plane figures. \end{exparts} \begin{answer} \begin{exparts} \partsitem 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. \partsitem Any plane figure $F$ is congruent to itself via the identity map $\map{\identity}{\Re^2}{\Re^2}$, 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 $\composed{g}{f}$, which is easily checked to be distance-preserving. \end{exparts} \end{answer} \item \label{exer:HomoCrds} 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{equation*}  Jim Hefferon committed Jan 08, 2012 444  \begin{mat}  Jim Hefferon committed Dec 05, 2011 445 446  a &c \\ b &d  Jim Hefferon committed Jan 08, 2012 447  \end{mat}  Jim Hefferon committed Dec 05, 2011 448 449 450  \colvec{x \\ y} +\colvec{e \\ f} \qquad  Jim Hefferon committed Jan 08, 2012 451  \begin{mat}  Jim Hefferon committed Dec 05, 2011 452 453 454  a &c &e \\ b &d &f \\ 0 &0 &1  Jim Hefferon committed Jan 08, 2012 455  \end{mat}  Jim Hefferon committed Dec 05, 2011 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478  \colvec{x \\ y \\ 1} \end{equation*} (These are \definend{homogeneous coordinates}\index{homogeneous coordinates}; see the Topic on Projective Geometry). \begin{answer} The first two components of each are $ax+cy+e$ and $bx+dy+f$. \end{answer} \item \label{exer:GeomInvDistPre} \begin{exparts} \partsitem Verify that the properties described in the second paragraph of this Topic as invariant under distance-preserving maps are indeed so. \partsitem Give two more properties that are of interest in Euclidean geometry from your experience in studying that subject that are also invariant under distance-preserving maps. \partsitem Give a property that is not of interest in Euclidean geometry and is not invariant under distance-preserving maps. \end{exparts} \begin{answer} \begin{exparts} \partsitem The Pythagorean Theorem gives that three points are  Jim Hefferon committed Oct 28, 2013 479  collinear if and only if  Jim Hefferon committed Dec 05, 2011 480 481 482 483 484 485  (for some ordering of them into $P_1$, $P_2$, and $P_3$), $\dist(P_1,P_2)+\dist(P_2,P_3)=\dist(P_1,P_3)$. Of course, where $f$ is distance-preserving, this holds if and only if $\dist(f(P_1),f(P_2))+\dist(f(P_2),f(P_3))=\dist(f(P_1),f(P_3))$, which, again by Pythagoras, is true if and only if  Jim Hefferon committed Oct 28, 2013 486  $f(P_1)$, $f(P_2)$, and $f(P_3)$ are collinear.  Jim Hefferon committed Dec 05, 2011 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519  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 $\dist(P,Q)=r$. Applying the distance-preserving map $f$ gives that the image $f(C)$ is the set of all $f(Q)$ subject to the condition that $\dist(P,Q)=r$. Since $\dist(P,Q)=\dist(f(P),f(Q))$, the set $f(C)$ is also a circle, with center $f(P)$ and radius $r$. \partsitem Here are two that are easy to verify: (i)~the property of being a right triangle, and (ii)~the property of two lines being parallel. \partsitem 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. \end{exparts} \end{answer} \end{exercises} \index{matrix!orthonormal|)} \index{orthonormal matrix|)} \endinput