Commit b3393d97 authored by Jim Hefferon's avatar Jim Hefferon

map2

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......@@ -12485,7 +12485,7 @@ sage: n(a_by_a)
\end{ans}
\subsection{Three.II.2: Range space and Null space}
\begin{ans}{Three.II.2.22}
\begin{ans}{Three.II.2.21}
First, to answer whether a polynomial is in the null space,
we have to consider it as a member of the domain $\polyspace_3$.
To answer whether it is in the range space, we consider it as a member of
......@@ -12521,7 +12521,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.II.2.23}
\begin{ans}{Three.II.2.22}
\begin{exparts}
\partsitem The null space is
\begin{equation*}
......@@ -12579,7 +12579,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.II.2.24}
\begin{ans}{Three.II.2.23}
For each, use the result
that the rank plus the nullity equals the dimension of the domain.
\begin{exparts*}
......@@ -12590,7 +12590,7 @@ sage: n(a_by_a)
\end{exparts*}
\end{ans}
\begin{ans}{Three.II.2.25}
\begin{ans}{Three.II.2.24}
Because
\begin{equation*}
\frac{d}{dx}\,(a_0+a_1x+\cdots+a_nx^n)
......@@ -12613,11 +12613,11 @@ sage: n(a_by_a)
for \( k\leq n \).
\end{ans}
\begin{ans}{Three.II.2.26}
\begin{ans}{Three.II.2.25}
The shadow of a scalar multiple is the scalar multiple of the shadow.
\end{ans}
\begin{ans}{Three.II.2.27}
\begin{ans}{Three.II.2.26}
\begin{exparts}
\partsitem Setting $a_0+(a_0+a_1)x+(a_2+a_3)x^3=0+0x+0x^2+0x^3$
gives $a_0=0$ and $a_0+a_1=0$ and $a_2+a_3=0$, so the null space is
......@@ -12633,14 +12633,14 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.II.2.28}
\begin{ans}{Three.II.2.27}
All inverse images are lines with slope $-2$.
\begin{center}
\includegraphics{ch3.74}
\end{center}
\end{ans}
\begin{ans}{Three.II.2.29}
\begin{ans}{Three.II.2.28}
These are the inverses.
\begin{exparts}
\partsitem \( a_0+a_1x+a_2x^2+a_3x^3\mapsto
......@@ -12656,7 +12656,7 @@ sage: n(a_by_a)
So this map is actually self-inverse.
\end{ans}
\begin{ans}{Three.II.2.30}
\begin{ans}{Three.II.2.29}
For any vector space $V$, the null space
\begin{equation*}
\set{\vec{v}\in V\suchthat 2\vec{v}=\zero}
......@@ -12671,7 +12671,7 @@ sage: n(a_by_a)
(Thus, this transformation is actually an automorphism.)
\end{ans}
\begin{ans}{Three.II.2.31}
\begin{ans}{Three.II.2.30}
Because the rank plus the nullity equals the dimension of the domain
(here, five), and the rank is at most three,
the possible pairs are:~\( (3,2) \), \( (2,3) \),
......@@ -12680,13 +12680,13 @@ sage: n(a_by_a)
possible is easy.
\end{ans}
\begin{ans}{Three.II.2.32}
\begin{ans}{Three.II.2.31}
No (unless \( \polyspace_n \) is trivial), because the two polynomials
\( f_0(x)=0 \) and \( f_1(x)=1 \) have the same derivative; a map must be
one-to-one to have an inverse.
\end{ans}
\begin{ans}{Three.II.2.33}
\begin{ans}{Three.II.2.32}
The null space is this.
\begin{multline*}
\set{a_0+a_1x+\dots+a_nx^n\suchthat
......@@ -12699,7 +12699,7 @@ sage: n(a_by_a)
Thus the nullity is \( n \).
\end{ans}
\begin{ans}{Three.II.2.34}
\begin{ans}{Three.II.2.33}
\begin{exparts}
\partsitem One direction is obvious:~if the homomorphism
is onto then its range is the codomain and so its
......@@ -12725,7 +12725,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.II.2.35}
\begin{ans}{Three.II.2.34}
We are proving that \( \map{h}{V}{W} \) is one-to-one if and only
if for every linearly independent subset \( S \) of \( V \) the subset
\( h(S)=\set{h(\vec{s})\suchthat \vec{s}\in S} \) of \( W \) is linearly
......@@ -12781,7 +12781,7 @@ sage: n(a_by_a)
and so not all sets have independence preserved.
\end{ans}
\begin{ans}{Three.II.2.36}
\begin{ans}{Three.II.2.35}
(We use the notation from \nearbytheorem{th:HomoDetActOnBasis}.)
Fix a basis $\sequence{\vec{\beta}_1,\dots,\vec{\beta}_n}$ for $V$
and a basis
......@@ -12807,7 +12807,7 @@ sage: n(a_by_a)
Thus the map is onto.
\end{ans}
\begin{ans}{Three.II.2.37}
\begin{ans}{Three.II.2.36}
Yes.
For the transformation of \( \Re^2 \) given by
\begin{equation*}
......@@ -12822,7 +12822,7 @@ sage: n(a_by_a)
We will see more of this in the fifth chapter.
\end{ans}
\begin{ans}{Three.II.2.38}
\begin{ans}{Three.II.2.37}
This is a simple calculation.
\begin{align*}
h(\spanof{S})
......@@ -12836,7 +12836,7 @@ sage: n(a_by_a)
\end{align*}
\end{ans}
\begin{ans}{Three.II.2.39}
\begin{ans}{Three.II.2.38}
\begin{exparts}
\partsitem
We will show that the two sets are equal
......@@ -12901,7 +12901,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.II.2.40}
\begin{ans}{Three.II.2.39}
Because the rank of \( t \) is one, the range space of \( t \)
is a one-dimensional set.
Taking $\sequence{h(\vec{v})}$ as a basis (for some appropriate
......@@ -12924,7 +12924,7 @@ sage: n(a_by_a)
to get the desired conclusion.
\end{ans}
\begin{ans}{Three.II.2.41}
\begin{ans}{Three.II.2.40}
By assumption, \( h \) is not the zero map
and so a vector \( \vec{v}\in V \) exists that is not in the null space.
Note that \( \sequence{h(\vec{v})} \) is a basis for \( \Re \),
......@@ -12948,7 +12948,7 @@ sage: n(a_by_a)
\( h \) to both sides would give a contradiction.
\end{ans}
\begin{ans}{Three.II.2.42}
\begin{ans}{Three.II.2.41}
Fix a basis \( \sequence{\vec{\beta}_1,\dots,\vec{\beta}_n} \)
for \( V \).
We shall prove that this map
......@@ -13005,7 +13005,7 @@ sage: n(a_by_a)
so \( \Phi(r_1h_1+r_2h_2)=r_1\Phi(h_1)+r_2\Phi(h_2) \).
\end{ans}
\begin{ans}{Three.II.2.43}
\begin{ans}{Three.II.2.42}
Let \( \map{h}{V}{W} \) be linear and fix a basis
\( \sequence{\vec{\beta}_1,\dots,\vec{\beta}_n} \) for \( V \).
Consider these \( n \) maps from \( V \) to \( W \)
......@@ -13023,7 +13023,7 @@ sage: n(a_by_a)
\( h_i(r\vec{v}+s\vec{u})=rc_i+sd_i=rh_i(\vec{v})+sh_i(\vec{u}) \).
\end{ans}
\begin{ans}{Three.II.2.44}
\begin{ans}{Three.II.2.43}
Either yes (trivially) or no (nearly trivially).
If we take \( V \) `is homomorphic to' \( W \) to mean there is a
......@@ -13041,7 +13041,7 @@ sage: n(a_by_a)
so the relation is not reflexive.\appendrefs{equivalence relations}
\end{ans}
\begin{ans}{Three.II.2.45}
\begin{ans}{Three.II.2.44}
That they form the chains is obvious.
For the rest, we show here that
\( \rangespace{t^{j+1}}=\rangespace{t^{j}} \)
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