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

three_v

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