Commit f4eef155 by Jim Hefferon

### erlang; all of chapter 3 done

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 ... ... @@ -22249,14 +22249,14 @@ octave:6> gplot z \begin{exparts} \partsitem The Pythagorean Theorem gives that three points are colinear if and only if collinear if and only if (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 $f(P_1)$, $f(P_2)$, and $f(P_3)$ are colinear. $f(P_1)$, $f(P_2)$, and $f(P_3)$ are collinear. The argument for betweeness is similar (above, $P_2$ is between $P_1$ and $P_3$).
 % Chapter 3, Topic _Linear Algebra_ Jim Hefferon % http://joshua.smcvt.edu/linalg.html % http://joshua.smcvt.edu/linearalgebra % 2001-Jun-12 \topic{Orthonormal Matrices} \index{orthonormal matrix|(} ... ... @@ -21,17 +21,16 @@ but he uses it often \cite{Casey}.) \begin{center} \includegraphics{ch3.61} \end{center} In modern terminology, picking the plane up~\ldots'' is considering a In modern terms picking the plane up~\ldots'' is taking a map from the plane to itself. Euclid considers only transformations of the plane Euclid considers only transformations that may slide or turn the plane but not bend or stretch it. Accordingly, we define a map $\map{f}{\Re^2}{\Re^2}$ to be Accordingly, define a map $\map{f}{\Re^2}{\Re^2}$ to be \definend{distance-preserving}\index{distance-preserving}% \index{map!distance-preserving} or a \definend{rigid motion}\index{rigid motion} or an \definend{isometry},\index{isometry} \definend{isometry}\index{isometry} 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 $P_2$. ... ... @@ -64,13 +63,13 @@ which proposes the organizing principle that we can describe each kind of geometry\Dash Euclidean, projective, etc.\Dash as the study of the properties that are invariant under some group of transformations. The word group' here means more than just collection', The word group' here means more than just collection' but that lies outside of our scope.) We can use linear algebra to characterize the distance-preserving maps of the plane. We must first observe that To begin, observe that there are distance-preserving transformations of the plane that are not linear. The obvious example is this \emph{translation}.\index{translation} \begin{equation*} ... ... @@ -79,19 +78,19 @@ The obvious example is this \emph{translation}.\index{translation} \colvec{x \\ y}+\colvec[r]{1 \\ 0}=\colvec{x+1 \\ y} \end{equation*} However, this example turns out to be the only example, in the sense that if $f$ is distance-preserving and sends $\zero$ to $\vec{v}_0$ this example turns out to be the only one, in that if $f$ is distance-preserving and sends $\zero$ to $\vec{v}_0$ 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. To prove this equivalent statement, let To prove this equivalent statement, consider the standard basis and suppose that \begin{equation*} t(\vec{e}_1)=\colvec{a \\ b} \qquad t(\vec{e}_2)=\colvec{c \\ d} \end{equation*} for some $a,b,c,d\in\Re$. Then to show that $t$ is linear we can show that To show that $t$ is linear we can show that it can be represented by a matrix, that is, that $t$ acts in this way for all $x,y\in\Re$. \begin{equation*} ... ... @@ -110,7 +109,7 @@ 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). In fact, because $t$ is distance-preserving, we can say more:~for the Because $t$ is distance-preserving we can say more:~for the 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 $\vec{e}_2$, ... ... @@ -141,10 +140,11 @@ and suffices to show that ($*$) describes $t$. Those checks are routine. Thus we can write any distance-preserving $\map{f}{\Re^2}{\Re^2}$ as Thus any distance-preserving $\map{f}{\Re^2}{\Re^2}$ is a linear map plus a translation, $f(\vec{v})=t(\vec{v})+\vec{v}_0$ for some constant vector $\vec{v}_0$ and linear map $t$ that is distance-preserving. So what is left in order to understand distance-preserving maps is to So in order to understand distance-preserving maps what remains is to understand distance-preserving linear maps. Not every linear map is distance-preserving. ... ... @@ -191,12 +191,12 @@ are of length one and are mutually orthogonal. This is an \definend{orthonormal matrix}\index{orthonormal matrix}% \index{matrix!orthonormal} or (or, more informally, \definend{orthogonal matrix}\index{orthogonal matrix}\index{matrix!orthogonal} (people often use the second term to mean not just that the columns are since people often use this term to mean not just that the columns are orthogonal but also that they have length one). We can use this to We can leverage this characterization to understand the geometric actions of distance-preserving maps. Because $\norm{t(\vec{v}\,)}=\norm{\vec{v}\,}$, the map~$t$ ... ... @@ -231,7 +231,7 @@ or to one where it goes a quarter circle counterclockwise. \end{mat}$\end{center} We can geometrically describe these two cases. The geometric description of these two cases is easy. Let$\theta$be the counterclockwise angle between the$x$-axis and the image of$\vec{e}_1$. The first matrix above represents, with respect to the standard bases, ... ... @@ -269,19 +269,20 @@ preserves orientations and \definend{opposite}\index{opposite map}\index{orientation reversing map} if it reverses orientation. So, we have characterized the Euclidean study of congruence. With that, we have characterized the Euclidean study of congruence. It considers, for plane figures, the properties that are invariant 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}). Another idea, besides congruence of figures, encountered in elementary geometry is that figures are \definend{similar}\index{similar triangles}\index{triangles!similar} Another idea encountered in elementary geometry, besides congruence of figures, is that figures are \definend{similar}\index{similar triangles}\index{triangles!similar} if they are congruent after a change of scale. These two triangles are similar since the second is the same shape as the first, but$3/2$-ths the size. The two triangles below are similar since the second is the same shape as the first but$3/2$-ths the size. \begin{center} \includegraphics{ch3.65} \end{center} ... ... @@ -475,14 +476,14 @@ More on Klein and the Erlanger Program is in \cite{Yaglom}. \begin{exparts} \partsitem The Pythagorean Theorem gives that three points are colinear if and only if collinear if and only if (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$f(P_1)$,$f(P_2)$, and$f(P_3)$are colinear.$f(P_1)$,$f(P_2)$, and$f(P_3)$are collinear. The argument for betweeness is similar (above,$P_2$is between$P_1$and$P_3\$). ... ...
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