diff --git a/slides/three_v.tex b/slides/three_v.tex index e5307d36fecffb1ee80ca865b6304ad3999f0137..8ab0c7882bf97d6d5523f0281aa1e8a8ac7e94a1 100644 --- a/slides/three_v.tex +++ b/slides/three_v.tex @@ -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}