Commit 285228f1 authored by Jim Hefferon's avatar Jim Hefferon

finish three_ii

parent 14812eef
......@@ -91,7 +91,7 @@ then
done
# pdf's without 3d manipulatable
declare -a pdf_out_files=("two_i_plane" "r3_subspaces" "r3_subspaces_spans" "p2_subspaces" "p2_subspaces_spans" "two_i_a_line" "two_i_a_plane" "three_i_line" "three_i_plane" "three_ii_kernel" "three_ii_rotate_basis" "three_ii_rotate_sum_before" "three_ii_rotate_sum_after" "three_ii_rotate_prod_before" "three_ii_rotate_prod_after" "three_ii_3dproj1" "three_ii_3dproj2" "three_ii_3dproj3" "three_ii_dims" "three_vi_3dprojtoline" "three_vi_3dgramschmidt0" "three_vi_3dgramschmidt1" "four_ii_orientation" "four_ii_orientation_pos" "four_ii_orientation_neg four_ii_negvolbox")
declare -a pdf_out_files=("two_i_plane" "r3_subspaces" "r3_subspaces_spans" "p2_subspaces" "p2_subspaces_spans" "two_i_a_line" "two_i_a_plane" "three_i_line" "three_i_plane" "three_ii_kernel" "three_ii_rotate_basis" "three_ii_rotate_sum_before" "three_ii_rotate_sum_after" "three_ii_rotate_prod_before" "three_ii_rotate_prod_after" "three_ii_3dproj1" "three_ii_3dproj2" "three_ii_3dproj3" "three_ii_dims" "three_ii_inv_img" "three_ii_proj_2d" "three_vi_3dprojtoline" "three_vi_3dgramschmidt0" "three_vi_3dgramschmidt1" "four_ii_orientation" "four_ii_orientation_pos" "four_ii_orientation_neg four_ii_negvolbox")
cmd="asy -f pdf filename" # will substitute for "filename"
for i in ${pdf_out_files[@]}
......
......@@ -701,30 +701,31 @@ we can find a
$\vec{v}\in\Re^3$ that maps to it, specifically any $\vec{v}$ with a
first component~$a$ and second component~$b$.
Thus the rank of $\pi$ is~$2$.
\bigskip\pause
In the book's next section, on computing linear maps,
we will do more examples of determining the range space.
\end{frame}
%..........
\begin{frame}{Homomorphisms organize the domain}
\begin{frame}{Many-to-one}
In moving from isomorphisms to homomorphisms we dropped
the requirement that the maps be onto and one-to-one.
But any homomorphism $\map{h}{V}{W}$ is onto
its range space $\rangespace{h}$,
so dropping the onto condition has, in a way, no effect
on the range.
It doesn't allow any essentially new maps.
\pause
In contrast,
consider dropping the one-to-one condition.
With that, there may be some output vectors $\vec{w}\in W$
for which there are many associated inputs,
$\vec{v}\in V$ such that $h(\vec{v})=\vec{w}$.
Now we can ask:
for any vector in the range $\vec{w}\in\rangespace{h}$
what are the associated domain vectors $\vec{v}\in V$?
consider the effect of dropping the one-to-one condition.
With that, an output vector $\vec{w}\in W$
may have many associated inputs,
many $\vec{v}\in V$ such that $h(\vec{v})=\vec{w}$.
\pause
\smallskip
\ExecuteMetaData[../map2.tex]{InverseImage}
......@@ -776,7 +777,7 @@ the sum $\pi(\vec{u}+\vec{v})=5$.
\pause
That is, a ``$2$ vector'' plus a
``$3$~vector'' is a ``$5$~vector.''
Red plus blue makes purple.
Red plus blue makes magenta.
\centergraphic{asy/three_ii_proj_2d06.pdf}
A similar interpretation holds for preservation of scalar multiplication:
......@@ -817,8 +818,7 @@ The way that the range space vectors add
{\color{red}\colvec{1 \\ 2}}+{\color{blue}\colvec{1.5 \\ 3}}
={\color{magenta}\colvec{2.5 \\ 5}}
\end{equation*}
is reflected in the domain: a red vector plus a blue vector makes a purple
vector.
is reflected in the domain: red plus blue makes magenta.
\begin{center}
\includegraphics{asy/three_ii_inv_img06.pdf}
\end{center}
......@@ -827,7 +827,7 @@ That is, preservation of addition is:
$h({\color{red}\vec{v}_1})+h({\color{blue}\vec{v}_2})
=h({\color{magenta}\vec{v}_1+\vec{v}_2})$.
\end{frame}
\begin{frame}
\begin{frame}{Homomorphisms organize the domain}
So the intuition is that a linear map organizes its domain into inverse
images,
\begin{center}
......@@ -921,20 +921,15 @@ In each of those examples, the homomorphism
$\map{h}{V}{W}$ shows how to view the domain $V$ as organized into the
inverse images $h^{-1}(\vec{w})$.
In the examples these inverse images all look the same so if
we can describe one then we understand how the domain is
In the examples these inverse images are all the same, but shifted.
So if we describe one of them then we understand how the domain is
divided.
% We say ``organized'' because these inverse image sets
% reflect the structure of the range
% in that a ``$\vec{w}_1$~vector'' plus a ``$\vec{w}_2$~vector''
% equals a ``$\vec{w}_1+\vec{w}_2$~vector,'' and likewise for
% scalar multiplication.
Vector spaces have a distinguished element, $\vec{0}$.
So we next consider the inverse image $h^{-1}(\zero)$.
\end{frame}
\begin{frame}
Vector spaces have a distinguished element, $\vec{0}$.
So we next consider the inverse image of that element $h^{-1}(\zero)$.
\lm[le:NullspIsSubSp]\hspace*{-1em}
\ExecuteMetaData[../map2.tex]{lm:NullspIsSubSp}
......@@ -1063,6 +1058,25 @@ A trivial space has an empty basis so $d_{3}$'s nullity is~$0$.
%..........
\begin{frame}{Rank plus nullity}
Recall the example map $\map{h}{\Re^2}{\Re^2}$
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{x+y \\ 2x+2y}
\end{equation*}
whose range space $\rangespace{h}$ is the line $y=2x$
and whose domain is organized into lines,
$\nullspace{h}$ is the line $y=-x$.
There, an entire line's worth of domain vectors collapses to the
single range point.
\begin{center}
\includegraphics{asy/three_ii_inv_img01.pdf}
\quad\raisebox{0.25in}{$\longmapsto$}\quad
\includegraphics{asy/three_ii_inv_img00.pdf}
\end{center}
In moving from domain to range, this maps drops a dimension.
We can account for it by thinking that each output point
absorbs a one-dimensional set.
\end{frame}
\begin{frame}
\th[th:RankPlusNullEqDim]
\ExecuteMetaData[../map2.tex]{th:RankPlusNullEqDim}
......@@ -1070,14 +1084,16 @@ A trivial space has an empty basis so $d_{3}$'s nullity is~$0$.
\pause
\pf
\ExecuteMetaData[../map2.tex]{pf:RankPlusNullEqDim0}
\pause
\ExecuteMetaData[../map2.tex]{pf:RankPlusNullEqDim1}
}{
\medskip
The book contains the proof.
\medskip
\ex Consider this map $\map{h}{\Re^3}{\Re}$.
\begin{equation*}
\colvec{x \\ y \\ z}\mapsunder{h} x/2+y/5+z
......@@ -1099,10 +1115,11 @@ A trivial space has an empty basis so $d_{3}$'s nullity is~$0$.
\begin{frame}
\iftoggle{showallproofs}{
\ExecuteMetaData[../map2.tex]{pf:RankPlusNullEqDim2}
\qed
\ExecuteMetaData[../map2.tex]{pf:RankPlusNullEqDim2}
\qed
}{
\noindent This shows the inverse images $h^{-1}(0)$ and $h^{-1}(1)$
\noindent
This shows the inverse images $h^{-1}(0)$ and $h^{-1}(1)$
lined up on the $z$~axis.
\begin{center}
\includegraphics{asy/three_ii_kernel.pdf}
......@@ -1113,7 +1130,7 @@ A trivial space has an empty basis so $d_{3}$'s nullity is~$0$.
The only difference between these $2$-dimensional subsets
is where they sit in the stack,
shown here as where they intersect $z$~axis.
\pause
That is, $h$ partitions the $3$-dimensional domain
into
$2$-dimensional sets, leaving $1$ dimension of
......@@ -1282,7 +1299,7 @@ The set $\set{1,0,2}\subseteq\Re$ is linearly dependent.
%..........
\begin{frame}{A one-to-one homorphism is an isomorphism}
\begin{frame}{A one-to-one homomorphism is an isomorphism}
\th[th:OOHomoEquivalence]
\ExecuteMetaData[../map2.tex]{th:OOHomoEquivalence}
......
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