Commit 6b7148f8 authored by Jim Hefferon's avatar Jim Hefferon

draft of map2 edits

parent 21c1e531
......@@ -13,6 +13,7 @@ TODO list for Linear Algebra http://joshua.smcvt.edu/linearalgebra
** subsectionoptional results in pdf errors in the log file about math shift
* New material
** topic on Google's algorithm?
......@@ -25,6 +26,10 @@ TODO list for Linear Algebra http://joshua.smcvt.edu/linearalgebra
** Hestenes?
** Topic on magic squares?
http://mathoverflow.net/questions/33911/why-linear-algebra-is-funor
** Discuss writegood mode in 1st read me. Mention that C-cG turns it off.
* Refactor some code
......@@ -58,6 +63,8 @@ one file to read from another? That is, cannot put files by chapter?
See http://tex.stackexchange.com/questions/37167/sorting-logs-into-different-folders
** Use the numprint package?
http://www.ctan.org/pkg/numprint
......
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......@@ -11054,8 +11054,8 @@
\end{ans}
\begin{ans}{Three.II.1.18}
For each, we must either check that linear combinations
are preserved, or give an example of a linear combination that is
For each, we must either check that the map preserves linear combinations
or give an example of a linear combination that is
not.
\begin{exparts*}
\partsitem Yes.
......@@ -11219,7 +11219,7 @@
0 &1
\end{mat}
\end{equation*}
are mapped to the same member of the codomain, $1\in\Re$.
map to the same member of the codomain, $1\in\Re$.
\end{ans}
\begin{ans}{Three.II.1.22}
......@@ -11394,8 +11394,7 @@
(This argument has already appeared, as part of the proof that
isomorphism is an equivalence.)
Let $\map{f}{U}{V}$ and $\map{g}{V}{W}$ be linear.
For any $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$ combinations
are preserved.
The composition preserves linear combinations
\begin{multline*}
\composed{g}{f}(c_1\vec{u}_1+c_2\vec{u}_2)
=g(\,f(c_1\vec{u}_1+c_2\vec{u}_2)\,)
......@@ -11404,6 +11403,7 @@
=c_1\cdot \composed{g}{f}(\vec{u}_1)
+c_2\cdot \composed{g}{f}(\vec{u}_2)
\end{multline*}
where $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$
\end{ans}
\begin{ans}{Three.II.1.33}
......@@ -11528,7 +11528,7 @@
Because the endpoints of $\ell$ are in the image of $C$, there are
members of $C$ that map to them, say $h(\vec{c}_1)=\vec{d}_1$
and $h(\vec{c}_2)=\vec{d}_2$.
Now, where $\hat{t}$ is the scalar that is fixed in the first
Now, where $\hat{t}$ is the scalar that we fixed in the first
sentence of this paragraph, observe that
$h(\hat{t}\cdot\vec{c}_1+(1-\hat{t})\cdot\vec{c}_2)
=\hat{t}\cdot h(\vec{c}_1)+(1-\hat{t})\cdot h(\vec{c}_2)
......@@ -11642,7 +11642,7 @@
If $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$ doesn't span the space
then the map needn't be unique.
For instance, if we try to define a map from $\Re^2$ to itself by
specifying only that $\vec{e}_1$ is sent to itself, then
specifying only that $\vec{e}_1$ maps to itself, then
there is more than
one homomorphism possible; both the identity map and the projection map
onto the first component fit this condition.
......@@ -11688,7 +11688,7 @@
\end{equation*}
They are linear because they are the composition of linear functions,
and the fact that the compoistion of linear functions is linear
was shown as part of the proof that isomorphism is an equivalence
was part of the proof that isomorphism is an equivalence
relation (alternatively, the check that they are linear is
straightforward).
\partsitem In general, a map from a vector space \( V \) to an
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