Commit 40939edd authored by Jim Hefferon's avatar Jim Hefferon

map1 edits first pass

parent d53bc95c
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......@@ -10787,7 +10787,7 @@
\end{ans}
\subsection{Subsection Three.I.2: Dimension Characterizes Isomorphism}
\begin{ans}{Three.I.2.8}
\begin{ans}{Three.I.2.9}
Each pair of spaces is isomorphic if and only if the two have the
same dimension.
We can, when there is an isomorphism, state
......@@ -10819,7 +10819,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Three.I.2.9}
\begin{ans}{Three.I.2.10}
\begin{exparts*}
\partsitem \( \rep{3-2x}{B}=\colvec[r]{5 \\ -2} \)
\partsitem \( \colvec[r]{0 \\ 2} \)
......@@ -10827,25 +10827,25 @@
\end{exparts*}
\end{ans}
\begin{ans}{Three.I.2.10}
\begin{ans}{Three.I.2.11}
They have different dimensions.
\end{ans}
\begin{ans}{Three.I.2.11}
\begin{ans}{Three.I.2.12}
Yes, both are \( mn \)-dimensional.
\end{ans}
\begin{ans}{Three.I.2.12}
\begin{ans}{Three.I.2.13}
Yes, any two (nondegenerate) planes are both two-dimensional
vector spaces.
\end{ans}
\begin{ans}{Three.I.2.13}
\begin{ans}{Three.I.2.14}
There are many answers, one is the set of \( \polyspace_k \)
(taking \( \polyspace_{-1} \) to be the trivial vector space).
\end{ans}
\begin{ans}{Three.I.2.14}
\begin{ans}{Three.I.2.15}
False (except when \( n=0 \)).
For instance,
if \( \map{f}{V}{\Re^n} \) is an isomorphism then multiplying by any
......@@ -10854,7 +10854,7 @@
possible is $\zero_V\mapsto 0_W$.)
\end{ans}
\begin{ans}{Three.I.2.15}
\begin{ans}{Three.I.2.16}
No.
A proper subspace has a strictly lower dimension than it's superspace;
if $U$ is a proper subspace of $V$ then any linearly independent subset
......@@ -10862,7 +10862,7 @@
be a basis for $V$, and $U$ wouldn't be proper.
\end{ans}
\begin{ans}{Three.I.2.16}
\begin{ans}{Three.I.2.17}
Where \( B=\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n} \), the
inverse is this.
\begin{equation*}
......@@ -10871,11 +10871,11 @@
\end{equation*}
\end{ans}
\begin{ans}{Three.I.2.17}
\begin{ans}{Three.I.2.18}
All three spaces have dimension equal to the rank of the matrix.
\end{ans}
\begin{ans}{Three.I.2.18}
\begin{ans}{Three.I.2.19}
We must show that if \( \vec{a}=\vec{b} \) then
\( f(\vec{a})=f(\vec{b}) \).
So suppose that
......@@ -10894,11 +10894,11 @@
and so the function is well-defined.
\end{ans}
\begin{ans}{Three.I.2.19}
\begin{ans}{Three.I.2.20}
Yes, because a zero-dimensional space is a trivial space.
\end{ans}
\begin{ans}{Three.I.2.20}
\begin{ans}{Three.I.2.21}
\begin{exparts}
\partsitem No, this collection has no spaces of odd dimension.
\partsitem Yes, because $\polyspace_{k}\isomorphicto\Re^{k+1}$.
......@@ -10907,7 +10907,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Three.I.2.21}
\begin{ans}{Three.I.2.22}
One direction is easy:~if the two are isomorphic via \( f \)
then for any basis \( B\subseteq V \),
the set \( D=f(B) \) is also a basis (this is shown in
......@@ -10935,7 +10935,7 @@
and the scalar multiplication calculation is similar.
\end{ans}
\begin{ans}{Three.I.2.22}
\begin{ans}{Three.I.2.23}
\begin{exparts}
\partsitem Pulling the definition back from
\( \Re^4 \) to \( \polyspace_3 \)
......@@ -10950,7 +10950,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Three.I.2.23}
\begin{ans}{Three.I.2.24}
Yes.
Assume that \( V \) is a vector space with basis
......@@ -10990,7 +10990,7 @@
Preservation of scalar multiplication is similar.
\end{ans}
\begin{ans}{Three.I.2.24}
\begin{ans}{Three.I.2.25}
Because \( V_1\intersection V_2=\set{\zero_V} \) and \( f \) is
one-to-one we have that \( f(V_1)\intersection f(V_2)=\set{\zero_U} \).
To finish, count the dimensions:
......@@ -10998,7 +10998,7 @@
as required.
\end{ans}
\begin{ans}{Three.I.2.25}
\begin{ans}{Three.I.2.26}
Rational numbers have many representations, e.g.,
\( 1/2=3/6 \), and the numerators can vary among
representations.
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