Commit 747ebec0 authored by Jim Hefferon's avatar Jim Hefferon

part of map1

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......@@ -10560,7 +10560,7 @@ sage: n(a_by_a)
\chapter{Chapter Three: Maps Between Spaces}
\section{Isomorphisms}
\subsection{Three.I.1: Def{}inition and Examples}
\begin{ans}{Three.I.1.11}
\begin{ans}{Three.I.1.12}
\begin{exparts}
\partsitem Call the map $f$.
\begin{equation*}
......@@ -10643,7 +10643,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.I.1.12}
\begin{ans}{Three.I.1.13}
These are the images.
\begin{exparts*}
\partsitem \( \colvec[r]{5 \\ -2} \)
......@@ -10679,7 +10679,7 @@ sage: n(a_by_a)
\end{align*}
\end{ans}
\begin{ans}{Three.I.1.13}
\begin{ans}{Three.I.1.14}
To verify it is one-to-one, assume that
$f_1(c_1x+c_2y+c_3z)=f_1(d_1x+d_2y+d_3z)$.
Then $c_1+c_2x+c_3x^2=d_1+d_2x+d_3x^2$
......@@ -10715,7 +10715,7 @@ sage: n(a_by_a)
\end{align*}
\end{ans}
\begin{ans}{Three.I.1.14}
\begin{ans}{Three.I.1.15}
\begin{exparts}
\partsitem No; this map is not one-to-one.
In particular, the matrix of all zeroes is mapped to the same
......@@ -10860,14 +10860,14 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.I.1.15}
\begin{ans}{Three.I.1.16}
It is one-to-one and onto, a correspondence,
because it has an inverse (namely, \( f^{-1}(x)=\sqrt[3]{x} \)).
However, it is not an isomorphism.
For instance, \( f(1)+f(1)\neq f(1+1) \).
\end{ans}
\begin{ans}{Three.I.1.16}
\begin{ans}{Three.I.1.17}
Many maps are possible.
Here are two.
\begin{equation*}
......@@ -10878,7 +10878,7 @@ sage: n(a_by_a)
The verifications are straightforward adaptations of the others above.
\end{ans}
\begin{ans}{Three.I.1.17}
\begin{ans}{Three.I.1.18}
Here are two.
\begin{equation*}
a_0+a_1x+a_2x^2 \mapsto \colvec{a_1 \\ a_0 \\ a_2}
......@@ -10893,7 +10893,7 @@ sage: n(a_by_a)
is the image of $(s-t)+tx+ux^2$).
\end{ans}
\begin{ans}{Three.I.1.18}
\begin{ans}{Three.I.1.19}
The space $\Re^2$ is not a subspace of $\Re^3$ because it is not a
subset of $\Re^3$.
The two-tall vectors in $\Re^2$ are not members of $\Re^3$.
......@@ -10934,7 +10934,7 @@ sage: n(a_by_a)
that it preserves combinations of two vectors.
\end{ans}
\begin{ans}{Three.I.1.19}
\begin{ans}{Three.I.1.20}
Here are two:
\begin{equation*}
\colvec{r_1 \\ r_2 \\ \vdots \\ r_{16}}
......@@ -10957,7 +10957,7 @@ sage: n(a_by_a)
Verification that each is an isomorphism is easy.
\end{ans}
\begin{ans}{Three.I.1.20}
\begin{ans}{Three.I.1.21}
When $k$ is the product \( k=mn \), here is an isomorphism.
\begin{equation*}
\begin{mat}
......@@ -10971,7 +10971,7 @@ sage: n(a_by_a)
Checking that this is an isomorphism is easy.
\end{ans}
\begin{ans}{Three.I.1.21}
\begin{ans}{Three.I.1.22}
If \( n\geq 1 \) then \( \polyspace_{n-1}\isomorphicto\Re^n \).
(If we take \( \polyspace_{-1} \) and \( \Re^0 \) to be trivial vector
spaces, then the relationship extends one dimension lower.)
......@@ -10983,7 +10983,7 @@ sage: n(a_by_a)
Checking that it is an isomorphism is straightforward.
\end{ans}
\begin{ans}{Three.I.1.22}
\begin{ans}{Three.I.1.23}
This is the map, expanded.
\begin{align*}
f(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5)
......@@ -11023,25 +11023,25 @@ sage: n(a_by_a)
\end{multline*}
\end{ans}
\begin{ans}{Three.I.1.23}
\begin{ans}{Three.I.1.24}
No vector space has the empty set underlying it.
We can take $\vec{v}$ to be the zero vector.
\end{ans}
\begin{ans}{Three.I.1.24}
\begin{ans}{Three.I.1.25}
Yes; where the two spaces are \( \set{\vec{a}} \) and
\( \set{\vec{b}} \),
the map sending \( \vec{a} \) to \( \vec{b} \) is clearly one-to-one
and onto, and also preserves what little structure there is.
\end{ans}
\begin{ans}{Three.I.1.25}
\begin{ans}{Three.I.1.26}
A linear combination of $n=0$ vectors adds to the zero vector and so
\nearbylemma{le:IsoSendsZeroToZero}
shows that the three statements are equivalent in this case.
\end{ans}
\begin{ans}{Three.I.1.26}
\begin{ans}{Three.I.1.27}
Consider the basis \( \sequence{1} \) for \( \polyspace_0 \)
and let \( f(1)\in\Re \) be \( k \).
For any \( a\in\polyspace_0 \)
......@@ -11053,7 +11053,7 @@ sage: n(a_by_a)
as is easy to check.)
\end{ans}
\begin{ans}{Three.I.1.27}
\begin{ans}{Three.I.1.28}
In each item, following item~(2) of
\nearbylemma{le:PresStructIffPresCombos}, we show that the map preserves
structure by showing that the it preserves linear combinations
......@@ -11095,7 +11095,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.I.1.28}
\begin{ans}{Three.I.1.29}
One direction is easy:~by definition, if \( f \) is one-to-one
then for any
\( \vec{w}\in W \) at most one \( \vec{v}\in V \) has
......@@ -11116,7 +11116,7 @@ sage: n(a_by_a)
$\vec{v}_1=\vec{v}_2$, and so \( f \) is one-to-one.
\end{ans}
\begin{ans}{Three.I.1.29}
\begin{ans}{Three.I.1.30}
We will prove something stronger\Dash not only is the existence of a
dependence preserved by isomorphism, but each instance of a dependence is
preserved, that is,
......@@ -11144,7 +11144,7 @@ sage: n(a_by_a)
to be mapped to the same image by $f$, they must be equal.
\end{ans}
\begin{ans}{Three.I.1.30}
\begin{ans}{Three.I.1.31}
\begin{exparts}
\partsitem
This map is one-to-one because if \( d_s(\vec{v}_1)=d_s(\vec{v}_2) \)
......@@ -11268,7 +11268,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.I.1.31}
\begin{ans}{Three.I.1.32}
First, the map $p(x)\mapsto p(x+k)$ doesn't count
because it is a version of $p(x)\mapsto p(x-k)$.
Here is a correct answer (many others are also correct):
......@@ -11276,7 +11276,7 @@ sage: n(a_by_a)
Verification that this is an isomorphism is straightforward.
\end{ans}
\begin{ans}{Three.I.1.32}
\begin{ans}{Three.I.1.33}
\begin{exparts}
\partsitem For the `only if' half, let \( \map{f}{\Re^1}{\Re^1} \)
to be an isomorphism.
......@@ -11378,7 +11378,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.I.1.33}
\begin{ans}{Three.I.1.34}
There are many answers; two are linear independence and
subspaces.
......@@ -11429,7 +11429,7 @@ sage: n(a_by_a)
a member of $f(U)$.
\end{ans}
\begin{ans}{Three.I.1.34}
\begin{ans}{Three.I.1.35}
\begin{exparts}
\partsitem The association
\begin{equation*}
......@@ -11491,11 +11491,11 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.I.1.35}
\begin{ans}{Three.I.1.36}
See the next subsection.
\end{ans}
\begin{ans}{Three.I.1.36}
\begin{ans}{Three.I.1.37}
\begin{exparts}
\partsitem Most of the conditions in the definition of a
vector space are routine.
......@@ -704,13 +704,15 @@ beginfig(16) % flip over a line map
z0=(3w,2v);
z2=(2w,3v);
pickup pencircle scaled line_width_dark;
draw (-.5w,-.5v)--(3w,3v) dashed withdots scaled .5;
label.ulft(btex {\scriptsize $\ell$} etex,0.5[(0w,0v),(3w,3v)]);
z3=(-.5w,-.5v) shifted codomainshift; z4=(3w,3v) shifted codomainshift;
draw z3--z4 dashed withdots scaled .5;
drawarrow (0w,0v)--z0;
label.rt(btex {\scriptsize $\vec{u}$} etex,z0);
drawarrow ((0w,0v)--z2) shifted codomainshift;
picture pa; % in case the line ell hits this label
pa = thelabel.rt(btex {\scriptsize $f_{\ell}(\vec{u})$} etex,
pa = thelabel.urt(btex {\scriptsize $f_{\ell}(\vec{u})$} etex,
z2 shifted codomainshift);
unfill bbox pa; draw pa;
......
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