Commit 89399dac authored by Jim Hefferon's avatar Jim Hefferon

three_v

parent bb5636b0
......@@ -93,9 +93,15 @@ This is the arrow diagram.
\begin{frame}
\lm[le:ChBasisMatDoesChBases]
\ExecuteMetaData[../map5.tex]{lm:ChBasisMatDoesChBases}
\pf
\ExecuteMetaData[../map5.tex]{pf:ChBasisMatDoesChBases}
\qed
\iftoggle{showallproofs}{
\pf
\ExecuteMetaData[../map5.tex]{pf:ChBasisMatDoesChBases}
\qed
}{
\medskip
The book has the proof.
}
\end{frame}
......@@ -115,7 +121,7 @@ and so represent those elements with respect to~$D$.
\quad
\rep{1+x+x^2}{D}=\colvec{0 \\ 1 \\ 1}
\end{equation*}
To get the change of basis matrix just concatenate.
The change of basis matrix is the concatenation of those.
\begin{equation*}
\rep{id}{B,D}=
\begin{mat}
......@@ -153,13 +159,14 @@ For example, we can translate the representation of $\vec{v}=2-x+3x^2$.
}{
\medskip
The book contains the proof; we give an example that illustrates.
The book contains the proof; we will do an example.
\medskip
\ex
To show that any nonsingular matrix $M$ changes bases recall that
it decomposes into a product of elementary
reduction matrices $M=R_1\cdots R_r$; we will be finished
if we show that each $R_i$ changes bases.
Recall that any nonsingular matrix $M$
decomposes into a product of elementary
reduction matrices $M=R_1\cdots R_r$.
We can show that $M$
changes bases by showing that each $R_i$ changes bases.
Consider this $\nbyn{3}$ case.
\begin{equation*}
C_{1,3}(-4)\cdot\rep{\vec{v}}{B}
......@@ -271,6 +278,7 @@ For example, we can translate the representation of $\vec{v}=2-x+3x^2$.
The natural next step is to see how to convert $\rep{h}{B,D}$
to $\rep{h}{\hat{B},\hat{D}}$.
Here is the arrow diagram.
\ExecuteMetaData[../map5.tex]{ChangeRepresentationOfMapArrowDiagram}
\end{frame}
......@@ -363,16 +371,16 @@ using the change of basis matrices.
Here is the diagram, specialized for this case.
\begin{equation*}
\begin{CD}
\Re^2_{\wrt{B}} @>t_{\pi/6}>H> \Re^2_{\wrt{D}} \\
\Re^2_{\wrt{\stdbasis_2}} @>t_{\pi/6}>\rep{t_{\pi/6}}{\stdbasis_2,\stdbasis_2}> \Re^2_{\wrt{\stdbasis_2}} \\
@V{\text{\scriptsize$\identity$}} VV @V{\text{\scriptsize$\identity$}} VV \\
\Re^2_{\wrt{\stdbasis_2}} @>t_{\pi/6}>\rep{t_{\pi/6}}{\stdbasis_2,\stdbasis_2}> \Re^2_{\wrt{\stdbasis_2}}
\Re^2_{\wrt{B}} @>t_{\pi/6}>\hat{H}> \Re^2_{\wrt{D}}
\end{CD}
\end{equation*}
To get $H$ we move down from the upper left, across, and then back up.
\end{frame}
\begin{frame}
With respect to the standard basis real vectors represent themselves,
so the matrix representing moving down is easy.
so the matrix representing moving up is easy.
\begin{equation*}
\rep{id}{B,\stdbasis_2}
\begin{mat}
......@@ -380,7 +388,7 @@ so the matrix representing moving down is easy.
1 &-1
\end{mat}
\end{equation*}
The matrix for moving back up is the inverse.
The matrix for moving down is the inverse of the prior one.
\begin{equation*}
\rep{id}{\stdbasis_2,D}
=(\rep{id}{D,\stdbasis_2})^{-1}
......
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