Commit 89399dac by 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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