Commit 0e85430c by Jim Hefferon

### more on three_ii including some asy graphics

parent 918e407b
 ... ... @@ -171,7 +171,8 @@ % functions \newcommand{\map}[3]{\mbox{$#1\colon #2\to #3$}} \newcommand{\mapsunder}[1]{\stackrel{#1}{\longmapsto}} \newcommand{\xmapsunder}[1]{\xmapsto{#1}} % doesn't seem to be in mathtools: \newcommand{\mapsunder}[1]{\xrightmapsto{#1}} \newcommand{\xmapsunder}[1]{\xrightmapsto{#1}} \newcommand{\composed}[2]{#1\mathbin{\circ} #2} \newcommand{\identity}{\mbox{id}} \newcommand{\restrictionmap}[2]{{#1}\mathpunct\upharpoonright\hbox{}_{#2}} ... ...
 ... ... @@ -1020,8 +1020,9 @@ is more fruitful and more central to further progress. \end{answer} \item Generalize \nearbyexample{ex:MatTransMapLinear} by proving that the matrix transpose map is linear. What is the domain and codomain? by proving that for every appropriate domain and codomain the matrix transpose map is linear. What are the appropriate domains and codomains? \begin{answer} Recall that the entry in row~$i$ and column~$j$ of the transpose of $M$ is the entry $m_{j,i}$ from row~$j$ and column~$i$ of $M$. ... ... @@ -1452,12 +1453,14 @@ many members of the domain. \begin{center} \includegraphics{ch3.5} % bean to bean; many to one \end{center} %<*InverseImage> Recall that for any function $\map{h}{V}{W}$, the set of elements of $V$ that map to $$\vec{w}\in W$$ is the \definend{inverse image\/}\index{inverse image}% \index{function! inverse image} $h^{-1}(\vec{w})=\set{\vec{v}\in V\suchthat h(\vec{v})=\vec{w}}$. Above, the left bean shows three inverse image sets. % \begin{example} Consider the projection\index{projection} ... ...
slides/asy/jh.asy 0 → 100644
 // jh.asy Asmyptote common definitions import fontsize; defaultpen(fontsize(9.24994pt)); import texcolors; pen FILLCOLOR=rgb("fff0ca"); pen MAINPEN=linecap(0) +linewidth(0.4pt); pen VECTORPEN=linecap(0) +linewidth(0.8pt); real VECTORHEADSIZE=5; pen THINPEN=linecap(0) +linewidth(0.25pt); pen DASHPEN=linecap(0) +linewidth(0.4pt) +linetype(new real[] {8,8}); pen FCNPEN=linecap(0) +gray(0.3) +linewidth(1.5pt) +opacity(.5,"Normal"); pen AXISPEN=linecap(0) +gray(0.3) +linewidth(0.4pt) +opacity(.5,"Normal"); pen DXPEN=linecap(0) +red +linewidth(1pt); texpreamble("\usepackage{conc}");
 // three_ii_proj.asy import jh; real height; height=2.25cm; size(0,height); import graph; pen inverse_image_pen=linecap(1) +linetype(new real[] {0,2}) +linewidth(1.5pt); pair domain_vec_tip; path domain_vec; for(int i=0; i < 6; ++i) { domain_vec_tip=(2,2+1.5*i/3.0); domain_vec=(0,0)--domain_vec_tip; // this gives an endpoint to the vector that is 0.15 ps points from end draw((0,0)--arcpoint(reverse(domain_vec),.15),VECTORPEN,arrow=Arrow(DefaultHead,VECTORHEADSIZE)); dot(domain_vec_tip,inverse_image_pen+blue); } dot((2,0),inverse_image_pen+red); real[] xticks={4,-3,-2,-1,1,2,3,4}; real[] yticks={-1,1,2,3,4}; xaxis("",-4.1,4.1,AXISPEN,RightTicks("%",Size=2,xticks)); yaxis("",-1.1,4.1,AXISPEN,LeftTicks("%",Size=2,yticks)); // path xaxis=(-4.1,0)--(4.1,0); // path yaxis=(0,-1.1)--(0,4.1); // draw(xaxis,AXISPEN); // draw(yaxis,AXISPEN); // draw(unitcircle,THINPEN); // draw(e1,VECTORPEN,arrow=Arrow(DefaultHead,VECTORHEADSIZE)); // draw(e2,VECTORPEN,arrow=Arrow(DefaultHead,VECTORHEADSIZE)); // draw(e1_rotated,VECTORPEN+blue,arrow=Arrow(DefaultHead,VECTORHEADSIZE)); // draw(e2_rotated,VECTORPEN+blue,arrow=Arrow(DefaultHead,VECTORHEADSIZE));
 ... ... @@ -220,16 +220,16 @@ and $d/dx\,(x^2+1)=2x$. \pause This is a homomorphism. \begin{gather*} \begin{multline*} d/dx\,\big(\,r_1(a_1x^2+b_1x+c_1)+r_2(a_2x^2+b_2x+c_2)\,\big) \hspace*{5em} \\ \begin{align*} \begin{aligned} &=d/dx\,\big(\,(r_1a_1+r_2a_2)x^2+(r_1b_1+r_2b_2)x+(r_1c_1+r_2c_2)\,\big) \\ &=2(r_1a_1+r_2a_2)x+(r_1b_1+r_2b_2) \\ &=(2r_1a_1x+r_1b_1)+(2r_2a_2x+r_2b_2) \\ &=r_1\cdot d/dx\,(a_1x^2+b_1x+c_1) +r_2\cdot d/dx\,(a_2x^2+b_2x+c_2) \end{align*} \end{gather*} \end{aligned} \end{multline*} \end{frame} ... ... @@ -307,10 +307,95 @@ This map is linear. %.......... \begin{frame} \ex One basis of the space of quadratic polynomials $\polyspace_2$ is $B=\sequence{x^2,x,1}$. We can define a map $\map{\text{eval}_3}{\polyspace_2}{\Re}$ by specifying its action on that basis \begin{equation*} x^2\mapsunder{\text{eval}_3}9 \quad x\mapsunder{\text{eval}_3}3 \quad 1\mapsunder{\text{eval}_3}1 \end{equation*} and then extending linearly. \begin{equation*} \text{eval}_3(ax^2+bx+c)=a\cdot\text{eval}_3(x^2) +b\cdot\text{eval}_3(x) +c\cdot\text{eval}_3(1) =9a+3b+c \end{equation*} \pause The action of this map on the basis elements is to plug the value~$3$ in for $x$. That remains true when we extend linearly, so $\text{eval}_3(\,p(x)\,)=p(3)$. \end{frame} %.......... \begin{frame} \ex Consider the standard basis $\stdbasis_2$ for the vector space $\Re^2$. We can specify a rotation of the two basis vectors as here. \begin{equation*} \colvec{1 \\ 0}\mapsto\colvec[r]{\cos\theta \\ \sin\theta} \quad \colvec{0 \\ 1}\mapsto\colvec[r]{-\sin\theta \\ \cos\theta} \qquad \vcenteredhbox{\includegraphics{asy/three_ii_rotate.pdf}} \end{equation*} \pause We get a homomorphism $\map{t_{\theta}}{\Re^2}{\Re^2}$ by extending that linearly. \begin{align*} t_{\theta}(\colvec{x \\ y}) &= t_{\theta}(x\cdot\colvec{1 \\ 0}+y\cdot\colvec{0 \\ 1}) \\ &= x\cdot t_{\theta}(\colvec{1 \\ 0})+y\cdot t_{\theta}(\colvec{0 \\ 1}) \\ &=x\cdot\colvec[r]{\cos\theta \\ \sin\theta} +y\cdot\colvec[r]{-\sin\theta \\ \cos\theta} \\ &=\colvec{x\cos\theta-y\sin\theta \\ x\sin\theta+y\cos\theta} \end{align*} \pause This homomorphism is both one-to-one and onto, so it is an automorphism. \end{frame} %.......... \begin{frame} \df[df:LinearTransformation] \ExecuteMetaData[../map2.tex]{df:LinearTransformation} \pause \ex For any vector space $V$ the \definend{identity} map $\map{\textrm{id}}{V}{V}$ given by $\vec{v}\mapsto\vec{v}$ is a linear transformation. The check is easy. \pause \ex In $\Re^3$ the function $t$ that acts in this way \begin{equation*} \colvec{x \\ y \\ z}\mapsunder{t}\colvec{-x \\ y \\ z} \end{equation*} is a transformation. We have this. \begin{multline*} t(r_1\colvec{x_1 \\ y_1 \\ z_1}+r_2\colvec{x_2 \\ y_2 \\ z_2}) =t(\colvec{r_1x_1+r_2x_2 \\ r_1y_1+r_2y_2 \\ r_1z_1+r_2z_2}) =\colvec{-(r_1x_1+r_2x_2) \\ r_1y_1+r_2y_2 \\ r_1z_1+r_2z_2} \\ =r_1\colvec{-x_1 \\ y_1 \\ z_1}+r_2\colvec{-x_2 \\ y_2 \\ z_2} =r_1t(\colvec{x_1 \\ y_1 \\ z_1})+r_2t(\colvec{x_2 \\ y_2 \\ z_2}) \end{multline*} \end{frame} ... ... @@ -351,6 +436,82 @@ This map is linear. \begin{frame}{Range space} \df[df:RangeSpace] \ExecuteMetaData[../map2.tex]{df:RangeSpace} \pause \ex Projection $\map{\pi}{\Re^3}{\Re^2}$ onto the $xy$-plane \begin{equation*} \colvec{x \\ y \\ z}\mapsto\colvec{x \\ y} \end{equation*} is a linear map (the check is easy). The range space \begin{equation*} \rangespace{\pi} =\set{\colvec{x \\ y \\ 0}\suchthat x,y\in\Re} \end{equation*} is two-dimensional so the rank of $\pi$ is $2$. \end{frame} \begin{frame} \ex The derivative map $\map{d/dx}{\Re^4}{\Re^4}$ is linear. Its range is $\rangespace{d/dx}=\set{a_0+a_1x+a_2x^2\suchthat a_i\in\Re}$. (Verifying that every member of that space is the derivative of a fourth degree polynomial is easy.) Thus the rank of the derivative is $3$. \pause \ex This map is linear; the check is routine. \begin{equation*} \begin{mat} a &b \\ c &d \end{mat} \mapsunder{h} \colvec{a+b \\ 2a+2b} \end{equation*} The rangespace is this line through the origin \begin{equation*} \set{\colvec{t \\ 2t} \suchthat t\in\Re} \end{equation*} (every member of that set is the image \begin{equation*} \colvec{t \\ 2t} = h( \begin{mat} t &0 \\ 0 &0 \end{mat}\ ) \end{equation*} of some matrix). The rank of this map is $1$. \end{frame} %.......... \begin{frame}{Inverse image} \centergraphic{../ch3.5} \ExecuteMetaData[../map2.tex]{InverseImage} \ex Consider the projection map $\map{\pi}{\Re^2}{\Re}$ defined by \begin{equation*} \pi(\colvec{x \\ y}) =x \end{equation*} that just forgets'' the second component. The check that $\pi$ is a homomorphism is routine. \end{frame} \begin{frame} This function is many-to-one; for example many vectors are mapped to the member $2$ of the codomain. \centergraphic{asy/three_ii_proj.pdf} \end{frame} ... ...