Commit 314b0607 authored by Jim Hefferon's avatar Jim Hefferon

Typos and small adjustments to slides

parent c0147bc0
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compote\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compote\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compote\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compote\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compote\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compote\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
......@@ -7,10 +7,11 @@
# cd linear-algebra/slides
# ./make_slides.sh 2>&1 | tee make_slides.log
# (you may need to put in "chmod u+x ./make_slides.sh" the first time).
# Warning: running this script takes minutes, not seconds. Perhaps 20 mins.
# Warning: running this script takes minutes, not seconds. Perhaps 10 mins.
# In particular, "asy -f png four_ii_orientation_neg" takes a long time.
#
# 2012-Jun-15 Jim Hefferon
# 2017-Nov-10 Small adjustments so version three will compile from the .git
usage()
{
......@@ -72,6 +73,12 @@ done
# Generate the figures
if [[ -z $REGENERATING ]]
then
echo " Regenerating .. "
echo " This regenerates all the graphics and then runs LaTeX a number of"
echo " times. On my laptop it takes two minutes. Also, a lot of stuff"
echo " flies by on the screen."
echo " "
echo " "
# Make the graphics
cd asy
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer} % add handout option for no pauses
\documentclass[9pt,t]{beamer} % add handout option for no pauses
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......@@ -331,10 +331,11 @@ as required.
We finish by checking that $f$ preserves scalar multiplication.
This is similar to the check for addition.
\begin{align*}
r\cdot f(\,a_0+a_1x+a_2x^2\,)
&=r\cdot\colvec{a_0 \\ a_1 \\ a_2} \\
f(r\cdot(a_0+a_1x+a_2x^2))
&=f(\,(ra_0)+(ra_1)x+(ra_2)x^2\,) \\
&=\colvec{ra_0 \\ ra_1 \\ ra_2} \\
&=f(\,(ra_0)+(ra_1)x+(ra_2)x^2\,)
&=r\cdot\colvec{a_0 \\ a_1 \\ a_2} \\
&=r\cdot f(\,a_0+a_1x+a_2x^2\,)
\end{align*}
\pause
......@@ -495,6 +496,25 @@ Preservation of scalar multiplication is similar.
%..........
\begin{frame}{Preservation is special}
Many functions do not preserve addition and scalar multiplication.
For instance, $\map{f}{\Re^2}{\Re^2}$
\begin{equation*}
f(\colvec{x \\ y})=\colvec{x^2 \\ y^2}
\end{equation*}
does not preserve addition since the sum done one way
\begin{equation*}
f(\colvec{1 \\ 0}+\colvec{2 \\ 0})=f(\colvec{3 \\ 0})=\colvec{9 \\ 0}
\end{equation*}
gives a different result than the sum done the other way.
\begin{equation*}
f(\colvec{1 \\ 0})+f(\colvec{2 \\ 0})=\colvec{1 \\ 0}+\colvec{4 \\ 0}
=\colvec{5 \\ 0}
\end{equation*}
\end{frame}
%..........
\begin{frame}{Special case: Automorphisms}
\df[df:Automorphism]\hspace*{-1em}
......@@ -642,7 +662,7 @@ There is a member of the domain~$L$ that maps to it, namely this one.
To finish, we combine the two structure checks, using the lemma's~(2).
\begin{multline*}
f(\,c_1\cdot\colvec{t_1 \\ 2t_1}+c_2\cdot\colvec{t_2 \\ 2t_2}\,)
=f(\,(c_1t_1++c_2t_2)\cdot\colvec{1 \\ 2}\,) \\
=f(\,(c_1t_1+c_2t_2)\cdot\colvec{1 \\ 2}\,) \\
=c_1t_1+c_2t_2
=
c_1\cdot f(\,\colvec{t_1 \\ 2t_1}\,)+c_2\cdot f(\,\colvec{t_2 \\ 2t_2}\,)
......@@ -686,7 +706,7 @@ via this function.
The inverse $\map{f^{-1}}{\Re}{L}$
given by
\begin{equation*}
f^{-1}(r)=r\cdot\colvec{1 \\ 2}=\colvec{x \\ 2x}
f^{-1}(r)=r\cdot\colvec{1 \\ 2}=\colvec{r \\ 2r}
\end{equation*}
is also an isomorphism.
\end{frame}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......@@ -894,7 +894,7 @@ $h^{-1}(\vec{w}_1)=\set{a_1x^2+1x+c_1\suchthat a_1,c_1\in\Re^2}$.
Members of this set are ``$\vec{w}_1$~vectors.''
\pause
The inverse image of $\vec{w}_2$ is
$h^{-1}(\vec{w}_2)=\set{a_2x^2-1x+c_2\suchthat a_2,c_2\in\Re^2}$;
$h^{-1}(\vec{w}_2)=\set{a_2x^2-1x+c_2\suchthat a_2,c_2\in\Re}$;
these are ``$\vec{w}_2$~vectors.''
The ``$\vec{w}_3$~vectors'' are members of
$h^{-1}(\vec{w}_3)=\set{a_3x^2+0x+c_3\suchthat a_3,c_3\in\Re^2}$.
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......@@ -934,7 +934,7 @@ solve for $x$, $y$, and~$z$.
0 &0 &1 &-a+b \\
0 &0 &-1 &-a+c
\end{amat} \\
&\grstep[-\rho_1+\rho_3]{-\rho_1+\rho_2}
&\grstep{\rho_2+\rho_3}
\begin{amat}{3}
1 &0 &0 &a \\
0 &0 &1 &-a+b \\
......@@ -1069,15 +1069,13 @@ $\rangespace{h}=\Re^2$, and
the rank of~$h$ is as large as it could be,~$2$.
\pause
The similar calculation on the homogeneous system
shows that for all~$\vec{w}$ the associated~$\vec{v}$
is unique.
Doing the homogeneous system by substituting $a=0$, $b=0$ above
shows the null space is trivial.
So~$h$ is one-to-one;
the null space is trivial.
\begin{equation*}
\nullspace{h}=\set{\colvec{0 \\ 0}}
\end{equation*}
A trivial space has an empty basis so the nullity is~$0$.
The trivial null space has an empty basis, so the nullity is~$0$.
\end{frame}
......@@ -1095,7 +1093,7 @@ Suppose that a map~$h$ between real spaces is represented by this.
\end{equation*}
The domain has dimension~$3$ and the codomain has dimension~$2$
so $\map{h}{\Re^3}{\Re^2}$.
\pause
% \pause
\begin{equation*}
\begin{amat}{3}
1 &2 &1 &a \\
......@@ -1114,101 +1112,101 @@ For each $a,b$ there is at least one triple
$x,y,z$ so this map is onto;
$\rank(h)=2$.
\pause
% \pause
For each $a,b$ there is more than one
triple $x,y,z$ so
this map is not one-to-one.
To find the null space set do the reduction on the homogeneous system
by setting $a=b=0$.
The nullspace $\nullspace{h}$ is the set of vectors whose representation
falls into this set.
Said another way, find the null space by substituting $a=0$, $b=0$ above
to solve the homogeneous system.
\begin{equation*}
\set{\colvec{x \\ y \\ z}\suchthat \text{$y+2z=0$ and $x-3z=0$}}
\nullspace{h}
=\set{\colvec{x \\ y \\ z}\suchthat \text{$y+2z=0$ and $x-3z=0$}}
=\set{\colvec{3 \\ -2 \\ 1}z\suchthat z\in\Re}
\end{equation*}
The nullity is~$1$.
\end{frame}
\begin{frame}
\ex Let $h$ be the transformation of~$\Re^2$ represented with respect to the
standard bases $\stdbasis_2,\stdbasis_2$ by this matrix.
\begin{equation*}
H=
\begin{mat}
1 &3 \\
2 &6
\end{mat}
\end{equation*}
The matrix rank is~$1$.
\begin{equation*}
\begin{amat}{2}
1 &3 &a \\
2 &6 &b
\end{amat}
\grstep{-2\rho_1+\rho_2}
\begin{amat}{2}
1 &3 &a \\
0 &0 &-2a+b
\end{amat}
\end{equation*}
\pause
So the map rank is also $1$; the map is not onto.
The range space $\rangespace{h}$ is the set of vectors whose representation
is a member of this set.
\begin{equation*}
\set{\colvec{a \\ b}\suchthat -2a+b=0}
=\set{\colvec{1/2 \\ 1}\cdot b\suchthat b\in\Re}
\end{equation*}
\pause
The map is also not one-to-one; $\nullity(h)=1$.
The null space $\nullspace{h}$ is the set of vectors whose
representation is a member of this set.
\begin{equation*}
\set{\colvec{x \\ y}\suchthat x+3y=0}
=\set{\colvec{-3 \\ 1}\cdot y\suchthat y\in\Re}
\end{equation*}
Note that rank plus nullity equals the dimension of the domain.
\end{frame}
% \begin{frame}
% \ex Let $h$ be the transformation of~$\Re^2$ represented with respect to the
% standard bases $\stdbasis_2,\stdbasis_2$ by this matrix.
% \begin{equation*}
% H=
% \begin{mat}
% 1 &3 \\
% 2 &6
% \end{mat}
% \end{equation*}
% The matrix rank is~$1$.
% \begin{equation*}
% \begin{amat}{2}
% 1 &3 &a \\
% 2 &6 &b
% \end{amat}
% \grstep{-2\rho_1+\rho_2}
% \begin{amat}{2}
% 1 &3 &a \\
% 0 &0 &-2a+b
% \end{amat}
% \end{equation*}
% \pause
% So the map rank is also $1$; the map is not onto.
% The range space $\rangespace{h}$ is the set of vectors whose representation
% is a member of this set.
% \begin{equation*}
% \set{\colvec{a \\ b}\suchthat -2a+b=0}
% =\set{\colvec{1/2 \\ 1}\cdot b\suchthat b\in\Re}
% \end{equation*}
% \pause
% The map is also not one-to-one; $\nullity(h)=1$.
% The null space $\nullspace{h}$ is the set of vectors whose
% representation is a member of this set.
% \begin{equation*}
% \set{\colvec{x \\ y}\suchthat x+3y=0}
% =\set{\colvec{-3 \\ 1}\cdot y\suchthat y\in\Re}
% \end{equation*}
% Note that rank plus nullity equals the dimension of the domain.
% \end{frame}
\begin{frame}
\ex A linear map $\map{h}{\Re^m}{\Re^n}$ represented as here
\begin{equation*}
H=
\begin{mat}
1 &2 \\
1 &3 \\
1 &4
\end{mat}
\end{equation*}
has a dimension~$2$ domain and a dimension~$3$ codomain.
Here is the linear reduction.
\begin{equation*}
\begin{amat}{2}
1 &2 &a \\
1 &3 &b \\
1 &4 &c
\end{amat}
\grstep[-\rho_1+\rho_3]{-\rho_1+\rho_2}
\grstep{-2\rho_2+\rho_3}
\grstep{-2\rho_2+\rho_1}
\begin{amat}{2}
1 &0 &3a-2b \\
0 &1 &-a+b \\
0 &0 &a-2b+c
\end{amat}
\end{equation*}
If we use the standard bases then vectors represent themselves.
\begin{equation*}
\rangespace{h}
=\set{\colvec{a \\ b \\ c}\suchthat a=2b-c}
\qquad
\nullspace{h}
=\set{\colvec{0 \\ 0}}
\end{equation*}
The map~$h$ is not onto, but it is one-to-one.
\end{frame}
% \begin{frame}
% \ex A linear map represented as here
% \begin{equation*}
% H=
% \begin{mat}
% 1 &2 \\
% 1 &3 \\
% 1 &4
% \end{mat}
% \end{equation*}
% has a dimension~$2$ domain and a dimension~$3$ codomain,
% sop we can take $\map{h}{\Re^2}{\Re^3}$.
% Here is the linear reduction.
% \begin{equation*}
% \begin{amat}{2}
% 1 &2 &a \\
% 1 &3 &b \\
% 1 &4 &c
% \end{amat}
% \grstep[-\rho_1+\rho_3]{-\rho_1+\rho_2}
% \grstep{-2\rho_2+\rho_3}
% \grstep{-2\rho_2+\rho_1}
% \begin{amat}{2}
% 1 &0 &3a-2b \\
% 0 &1 &-a+b \\
% 0 &0 &a-2b+c
% \end{amat}
% \end{equation*}
% If we use the standard bases then vectors represent themselves.
% \begin{align*}
% \rangespace{h}
% & =\set{\colvec{a \\ b \\ c}\suchthat a=2b-c}
% =\set{\colvec{2 \\ 1 \\ 0}\cdot b +\colvec{-1 \\ 0 \\ 1}\cdot c\suchthat b,c\in\Re} \\
% \nullspace{h}
% &=\set{\colvec{0 \\ 0}}
% \end{align*}
% The map~$h$ is not onto but it is one-to-one.
% \end{frame}
......@@ -1218,9 +1216,8 @@ So
we can determine things about a map by fixing bases and
calculating with the representation.
The calculations so far involve real spaces and are done with respect to
the standard bases, to reduce the number of factors at play.
We close with an example showing what to do in other cases.
We close with an example showing how to extend those calculations
to the underlying spaces.
% \pause
% Given a domain space~$V$,
......@@ -1244,17 +1241,14 @@ We close with an example showing what to do in other cases.
\begin{frame}
\ex Let the domain and codomain be $\Re^2$ and~$\polyspace_2$,
with these bases.
\begin{equation*}
D=\sequence{\colvec{1 \\ 1 \\ 1},
\colvec{1 \\ 0 \\ 1},
\colvec{0 \\ 2 \\ 0}}
\qquad
B=\sequence{1+x,1-x,x+x^2}
\end{equation*}
As it happens,
\ex Fix these, as well as
the domain and codomain $\Re^2$ and~$\polyspace_2$.
\begin{equation*}
B=\sequence{\colvec{1 \\ 1},
\colvec{1 \\ 0}
},
D=\sequence{1+x,1-x,x+x^2}
\quad
H=\rep{h}{B,D}=
\begin{mat}
1 &2 \\
......@@ -1262,10 +1256,7 @@ As it happens,
1 &4
\end{mat}
\end{equation*}
and the prior example
gives this calculation (the fact that underlying this is
the linear map $\map{h}{\Re^2}{\polyspace_2}$ represented by~$H$
with respect to those bases is not relevant).
Gauss's Method is routine.
\begin{equation*}
\begin{amat}{2}
1 &2 &a \\
......@@ -1281,34 +1272,27 @@ with respect to those bases is not relevant).
0 &0 &a-2b+c
\end{amat}
\end{equation*}
\end{frame}
\begin{frame}
The range contains an output vector if and only if its representation
satisfies that $a-2b+c=0$.
\begin{equation*}
\rangespace{h}
=\set{\vec{p}\in\polyspace_2\suchthat \rep{\vec{p}}{D}=\colvec{2b-c \\ b \\ c}}
\end{equation*}
To get the underlying range elements, the vectors that are represented, expand.
The null space is trivial $\nullspace{h}=\set{\vec{0}}$
so $h$ is one-to-one.
The range contains the output vectors whose representations
satisfy $a-2b+c=0$, so $h$ is not onto.
\pause
To get the underlying elements of the codomain~$\polyspace_2$, expand.
\begin{align*}
\rangespace{h}
&=\set{\vec{p}\in\polyspace_2\suchthat \rep{\vec{p}}{D}=\colvec{2b-c \\ b \\ c}} \\
% \end{equation*}
% \end{frame}
% \begin{frame}
%\begin{align*}
&=\set{(2b-c)\cdot(1+x)+b\cdot(1-x)+c\cdot(x+x^2)\suchthat b,c\in\Re} \\
&=\set{b\cdot(3+x)+c\cdot(-1+x^2)\suchthat b,c\in\Re}
\end{align*}
Inside of the $3$~dimensional codomain~$\polyspace_2$,
the range is a $2$~dimensional subspace with basis $\sequence{3+x,-1+x^2}$.
So ~$h$ is not onto.
The null space is a singleton set, so $h$ is one-to-one.
\begin{equation*}
\nullspace{h}=\set{\colvec{0 \\ 0}}
\end{equation*}
That's the span of $\sequence{3+x,-1+x^2}$, which is linearly independent
so it is a basis.
% \begin{equation*}
% \nullspace{h}=\set{\colvec{0 \\ 0}}
% \end{equation*}
\end{frame}
%...........................
% \begin{frame}
% \ExecuteMetaData[../gr3.tex]{GaussJordanReduction}
% \df[def:RedEchForm]
%
% \end{frame}
\end{document}
This diff is collapsed.
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......
% see: https://groups.google.com/forum/?fromgroups#!topic/comp.text.tex/s6z9Ult_zds
\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass[10pt,t]{beamer}
\documentclass[9pt,t]{beamer}
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\PassOptionsToPackage{pdfpagemode=FullScreen}{hyperref}
......@@ -229,7 +229,7 @@ $c_1\vec{\beta}_1+c_2\vec{\beta}_2+c_3\vec{\beta}_3$ by solving this system.
\end{linsys}
\end{equation*}
By eye we see just what the Theorem says we will see:
there is one, and only one, solution $c_1=3$, $c_2=-1$, and~$c_1=-1$.
there is one, and only one, solution $c_1=3$, $c_2=-1$, and~$c_3=-1$.
\end{frame}
......
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