Commit f4eef155 authored by Jim Hefferon's avatar Jim Hefferon

erlang; all of chapter 3 done

parent e2367cdd
No preview for this file type
......@@ -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 $P_3$).
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment