Commit 14812eef authored by Jim Hefferon's avatar Jim Hefferon

graphics for slid three_ii

parent ad8ef3da
......@@ -720,13 +720,17 @@ 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 $h(\vec{v})=\vec{w}$.
This is a real difference.
Now, we can ask:
Now we can ask:
for any vector in the range $\vec{w}\in\rangespace{h}$
what are the associated domain vectors $\vec{v}\in V$?
\pause
\smallskip
\ExecuteMetaData[../map2.tex]{InverseImage}
\smallskip
The structure of the inverse image sets
will give us insight into the definition of homomorphism.
\end{frame}
......@@ -787,8 +791,9 @@ This function $\map{h}{\Re^2}{\Re^2}$ is linear.
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{x+y \\ 2x+2y}
\end{equation*}
Here are elements of $h^{-1}(\colvec{1 \\ 2})$ going to $\colvec{1 \\ 2}$.
(Only one inverse image element is shown as a vector, most are shown as dots.)
Here are elements of $h^{-1}(\colvec{1 \\ 2})$.
(Only one inverse image element is shown as a vector, most are indicated
with dots.)
\begin{center}
\includegraphics{asy/three_ii_inv_img01.pdf}
\quad\raisebox{0.25in}{$\longmapsto$}\quad
......@@ -821,10 +826,14 @@ vector.
That is, preservation of addition is:
$h({\color{red}\vec{v}_1})+h({\color{blue}\vec{v}_2})
=h({\color{magenta}\vec{v}_1+\vec{v}_2})$.
\pause
\end{frame}
\begin{frame}
So the intuition is that a linear map organizes its domain into inverse
images, such that those sets reflect the structure of the range.
images,
\begin{center}
\includegraphics{../ch3.5} % bean to bean; many to one
\end{center}
such that those sets reflect the structure of the range.
\end{frame}
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment