Commit b49411da by Jim Hefferon

### more three_ii slides

parent 97fd5d08
 ... ... @@ -237,7 +237,7 @@ there exists a homomorphism from $$V$$ to $$W$$ sending each \begin{proof} %<*pf:HomoDetActOnBasis0> We will define the map by associating $\vec{\beta}_i$ with $\vec{w}_i$ associating each $\vec{\beta}_i$ with $\vec{w}_i$ and then extending linearly to all of the domain. That is, given the input $\vec{v}$, we find its coordinates with respect to the basis ... ... @@ -251,7 +251,7 @@ the representation of each domain vector $$\vec{v}$$ is unique. %<*pf:HomoDetActOnBasis1> This map is a homomorphism since it preserves linear combinations; where $$\vec{v_1}=c_1\vec{\beta}_1+\cdots+c_n\vec{\beta}_n$$ and $$\vec{v_2}=d_1\vec{\beta}_1+\cdots+d_n\vec{\beta}_n$$, $$\vec{v_2}=d_1\vec{\beta}_1+\cdots+d_n\vec{\beta}_n$$ then we have this. \begin{align*} h(r_1\vec{v}_1+r_2\vec{v}_2) ... ... @@ -262,9 +262,9 @@ we have this. % %<*pf:HomoDetActOnBasis2> And, this map is unique since if $$\map{\hat{h}}{V}{W}$$ This map is unique since if $$\map{\hat{h}}{V}{W}$$ is another homomorphism satisfying that $$\hat{h}(\vec{\beta}_i)=\vec{w}_i$$ for each $$i$$ for each $$i$$, then $$h$$ and $$\hat{h}$$ agree on all of the vectors in the domain. \begin{multline*} \hat{h}(\vec{v}) ... ... @@ -395,7 +395,7 @@ from $$V$$ to $$W$$. %<*pf:SpLinFcns> This set is non-empty because it contains the zero homomorphism. So to show that it is a subspace we need only check that it is closed under linear combinations. closed under the operations. Let $$\map{f,g}{V}{W}$$ be linear. Then the sum of the two is linear \begin{align*} ... ... @@ -1456,10 +1456,10 @@ many members of the domain. %<*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}% is the \definend{inverse image set\/}\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. Above, the left side shows three inverse image sets. % \begin{example} ... ... @@ -1673,12 +1673,20 @@ is a subspace of the domain. % \end{lemma} \begin{note} This result is about inverse images of sets $h^{-1}(S)=\set{\vec{v}\in V\suchthat h(\vec{v})\in S}$ whereas the examples above consider inverse images of single vectors. We are using the same term in both cases by identifying the inverse image of the element $h^{-1}(\vec{w})$ with the inverse image of the one-element set $h^{-1}(\set{\vec{w}})$. \end{note} \begin{proof} %<*pf:NullspIsSubSp> Let $\map{h}{V}{W}$ be a homomorphism and let $S$ be a subspace of the range space of $h$. Consider the inverse image $h^{-1}(S)=\set{\vec{v}\in V\suchthat h(\vec{v})\in S}$. Consider the inverse image of $S$. It is nonempty because it contains $\zero_V$, since $$h(\zero_V)=\zero_W$$ and $$\zero_W$$ is an element $S$, as $S$ is a subspace. ... ...
 ... ... @@ -170,14 +170,14 @@ The \definend{inclusion map} $\map{\iota}{\Re^2}{\Re^3}$ =\colvec{x \\ y \\ z} \end{equation*} is a homomorphism. Here is the check. Here is the verification. \begin{align*} \iota(c_1\colvec{x_1 \\ y_1}+c_2\colvec{x_2 \\ y_2}) \iota(c_1\cdot\colvec{x_1 \\ y_1}+c_2\cdot\colvec{x_2 \\ y_2}) &=\iota(\colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2}) \\ &=\colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ 0} \\ &=\colvec{c_1x_1 \\ c_1y_1 \\ 0} +\colvec{c_2x_2 \\ c_2y_2 \\ 0} \\ &=c_1\iota(\colvec{x_1 \\ y_1})+c_2\iota(\colvec{x_2 \\ y_2}) &=c_1\cdot\iota(\colvec{x_1 \\ y_1})+c_2\cdot\iota(\colvec{x_2 \\ y_2}) \end{align*} \end{frame} ... ... @@ -197,14 +197,15 @@ and $h(3-x)=-1-x$. \pause This function is linear. The verification is straightforward. \begin{align*} h(\,c_1(a_1+b_1x)+c_2(a_2+b_2x)\,) &=h(\,(c_1a_1+c_2a_2)+(c_1b_1+c_2b_2)x\,) \\ &=(c_1b_1+c_2b_2)+(c_1b_1+c_2b_2)x \\ &=(c_1b_1+c_1b_1x)+(c_2b_2+c_2b_2x) \\ &=c_1h(a_1+b_1x)+c_2h(a_2+b_2x) \end{align*} \begin{multline*} h(\,c_1\cdot (a_1+b_1x)+c_2\cdot(a_2+b_2x)\,) \\ \begin{aligned} &=h(\,(c_1a_1+c_2a_2)+(c_1b_1+c_2b_2)x\,) \\ &=(c_1b_1+c_2b_2)+(c_1b_1+c_2b_2)x \\ &=(c_1b_1+c_1b_1x)+(c_2b_2+c_2b_2x) \\ &=c_1\cdot h(a_1+b_1x)+c_2\cdot h(a_2+b_2x) \end{aligned} \end{multline*} \end{frame} ... ... @@ -217,11 +218,10 @@ The derivative map $\map{d/dx}{\polyspace_2}{\polyspace_1}$ is given by $d/dx\,(ax^2+bx+c)=2ax+b$. For instance, $d/dx\,(3x^2-2x+4)=6x-2$ and $d/dx\,(x^2+1)=2x$. \pause This is a homomorphism. It is a homomorphism. \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} \\ 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{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) \\ ... ... @@ -237,7 +237,7 @@ This is a homomorphism. %.......... \begin{frame} The \definend{trace} of a square matrix is the sum down the uppper-left to lower-right diagonal is the sum down the uppper-left to lower-right diagonal. Thus $\map{\trace}{\matspace_{\nbyn{2}}}{\Re}$ is this. ... ... @@ -248,41 +248,43 @@ is this. \end{mat} =a+b \end{equation*} This map is linear. \begin{align*} \trace( r_1\begin{mat} It is linear. \begin{multline*} \trace(\, r_1\cdot\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} + r_2\begin{mat} r_2\cdot\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat} ) &=\trace( \begin{mat} r_1a_1+r_2a_2 &r_1b_1+r_2b_2 \\ r_1c_1+r_2c_2 &r_1d_1+r_2d_2 \end{mat} ) \\ &=(r_1a_1+r_2a_2)+(r_1d_1+r_2d_2) \\ &=r_1(a_1+d_1)+r_2(a_2+d_2) \\ &=r_1\trace( \begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} ) + r_2\trace( \begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat} ) \end{align*} \,) \\ \begin{aligned} &=\trace( \begin{mat} r_1a_1+r_2a_2 &r_1b_1+r_2b_2 \\ r_1c_1+r_2c_2 &r_1d_1+r_2d_2 \end{mat} ) \\ &=(r_1a_1+r_2a_2)+(r_1d_1+r_2d_2) \\ &=r_1(a_1+d_1)+r_2(a_2+d_2) \\ &=r_1\cdot\trace( \begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} ) + r_2\cdot\trace( \begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat} ) \end{aligned} \end{multline*} \end{frame} ... ... @@ -384,18 +386,18 @@ The check is easy. \pause \ex In $\Re^3$ the function $t$ that acts in this way In $\Re^3$ the function $f_{yz}$ \begin{equation*} \colvec{x \\ y \\ z}\mapsunder{t}\colvec{-x \\ y \\ z} \end{equation*} is a transformation. We have this. that reflects vectors over the $yz$-plane is a linear transformation. \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}) f_{yz}(r_1\colvec{x_1 \\ y_1 \\ z_1}+r_2\colvec{x_2 \\ y_2 \\ z_2}) =f_{yz}(\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}) =r_1f_{yz}(\colvec{x_1 \\ y_1 \\ z_1})+r_2f_{yz}(\colvec{x_2 \\ y_2 \\ z_2}) \end{multline*} \end{frame} ... ... @@ -440,11 +442,11 @@ We have this. \pause \ex Projection $\map{\pi}{\Re^3}{\Re^2}$ onto the $xy$-plane Projection $\map{\pi}{\Re^3}{\Re^2}$ \begin{equation*} \colvec{x \\ y \\ z}\mapsto\colvec{x \\ y} \end{equation*} is a linear map (the check is easy). is a linear map; the check is routine. The range space \begin{equation*} \rangespace{\pi} ... ... @@ -457,7 +459,7 @@ is two-dimensional so the rank of $\pi$ is $2$. The derivative map $\map{d/dx}{\Re^4}{\Re^4}$ is linear. Its range is $\rangespace{d/dx}=\set{a_0+a_1x+a_2x^2\suchthat a_i\in\Re}$. Its range is $\rangespace{d/dx}=\set{a_0+a_1x+a_2x^2+a_3x^3\suchthat a_i\in\Re}$. (Verifying that every member of that space is the derivative of a fourth degree polynomial is easy.) Thus the rank of the derivative is $3$. ... ... @@ -488,7 +490,7 @@ The rangespace is this line through the origin \end{mat}\ ) \end{equation*} of some matrix). of a $\nbyn{2}$ matrix). The rank of this map is $1$. \end{frame} ... ... @@ -526,6 +528,45 @@ so $\vec{u}+\vec{v}$ is a $5$-vector.'' %.......... \begin{frame} \ex Consider $\map{h}{\polyspace_2}{\Re^2}$ \begin{equation*} ax^2+bx+c\mapsto\colvec{b \\ b} \end{equation*} and consider these three members of the range. \begin{equation*} \vec{w}_1=\colvec{1 \\ 1},\; \vec{w}_2=\colvec{-1 \\ -1},\; \vec{w}_3=\colvec{0 \\ 0} \end{equation*} \pause The inverse image of $\vec{w}_1$ is $h^{-1}(\vec{w}_1)=\set{a_1x^2+x+c_1\suchthat a_1,c_1\in\Re^2}$. Think of these as $\vec{w}_1$~vectors.'' Some examples are $3x^2+x+1$, $3x^2+x-4$, and $-2x^2+x$. \pause The inverse image of $\vec{w}_2$ is $h^{-1}(\vec{w}_2)=\set{a_2x^2-x+c_2\suchthat a_2,c_2\in\Re^2}$; these are $\vec{w}_2$~vectors.'' The vectors from the domain associated with $\vec{w}_3$ are members of the set $h^{-1}(\vec{w}_3)=\set{a_3x^2+c_3\suchthat a_3,c_3\in\Re^2}$. \pause As in the prior example, any $\vec{v}_1\in h^{-1}(\vec{w}_1)$ plus any $\vec{v}_2\in h^{-1}(\vec{w}_2)$ totals to a $\vec{v}_3\in h^{-1}(\vec{w}_3)$. This is because $h$ is a homomorphism, so $h(\vec{v}_1+\vec{v}_2)=h(\vec{v}_1)+h(\vec{v}_2)=\vec{w}_1+\vec{w}_2=\vec{w}_3$. That is, the sum of a $\vec{w}_1$~vector'' with a $\vec{w}_2$~vector'' is mapped by $h$ to $\vec{w}_3$. \end{frame} %.......... \begin{frame} \lm[le:NullspIsSubSp] ... ... @@ -544,6 +585,15 @@ so $\vec{u}+\vec{v}$ is a $5$-vector.'' \begin{frame}{Null space} \df[df:NullSpace] \ExecuteMetaData[../map2.tex]{df:NullSpace} \pause \medskip \no Strictly, the nullspace of the codomain is not $\zero_{W}$, it is $\set{\zero_{W}}$. Thus the nullspace should properly be given as $h^{-1}(\set{\zero_{W}})$. But authors often state it as above. \end{frame} ... ...
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