Commit 3a171de3 by Jim Hefferon

### map1 and map2 exercise adjustments

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 ... ... @@ -715,7 +715,7 @@ partitions the collection of vector spaces. \end{split} \end{multline*} \end{answer} \item Verify that this map is an isomorphism: \recommended \item Verify that this map is an isomorphism: $\map{h}{\Re^4}{\matspace_{\nbyn{2}}}$ given by \begin{equation*} \colvec{a \\ b \\ c \\ d} ... ... @@ -803,9 +803,9 @@ partitions the collection of vector spaces. \end{align*} \end{answer} \recommended \item Decide whether each map is an isomorphism (if it is an isomorphism then prove it and if it isn't then state a condition that it fails to satisfy). Decide whether each map is an isomorphism. If it is an isomorphism then prove it and if it isn't then state a condition that it fails to satisfy. \begin{exparts} \partsitem $$\map{f}{\matspace_{\nbyn{2}}}{\Re}$$ given by \begin{equation*} ... ... @@ -1210,7 +1210,7 @@ partitions the collection of vector spaces. (Incidentally, any such map $a\mapsto ka$ is an isomorphism, as is easy to check.) \end{answer} \recommended \item \item These prove that isomorphism is an equivalence relation. \begin{exparts} \partsitem Show that the identity map $\map{\mbox{id}}{V}{V}$ is an ... ... @@ -2218,7 +2218,7 @@ See \nearbyexercise{exer:FcnWellDef}. \begin{corollary} \label{co:FiniteDimensionalIsoToReN} %<*co:FiniteDimensionalIsoToReN> A finite-dimensional vector space is isomorphic to one and only one Each finite-dimensional vector space is isomorphic to one and only one of the $\Re^n$. % \end{corollary} ... ... @@ -2425,6 +2425,26 @@ Any finite-dimensional vector space is actually \partsitem Yes, both have dimension $$2k$$. \end{exparts} \end{answer} \item Which of these spaces are isomorphic to each other? \begin{exparts*} \partsitem $\Re^3$ \partsitem $\matspace_{\nbyn{2}}$ \partsitem $\polyspace_3$ \partsitem $\Re^4$ \partsitem $\polyspace_2$ \end{exparts*} \begin{answer} Just find the dimension of each space, for instance by finding a basis, and then spaces with the same dimension are isomorphic. This lists the dimension of each space. \begin{exparts*} \partsitem $3$ \partsitem $4$ \partsitem $4$ \partsitem $4$ \partsitem $3$ \end{exparts*} \end{answer} \recommended \item Consider the isomorphism $$\map{\rep{\cdot}{B}}{\polyspace_1}{\Re^2}$$ ... ... @@ -2442,6 +2462,58 @@ Any finite-dimensional vector space is actually \partsitem $$\colvec[r]{-1 \\ 1}$$ \end{exparts*} \end{answer} \item For which $n$ is the space isomorphic to~$\Re^n$? \begin{exparts} \partsitem $\polyspace_4$ \partsitem $\polyspace_1$ \partsitem $\matspace_{\nbym{2}{3}}$ \partsitem the plane $2x-y+z=0$ subset of~$\Re^3$ \partsitem the vector space of linear combinations of three letters $\set{ax+by+cz\suchthat a,b,c\in\Re}$ \end{exparts} \begin{answer} For each, the simplest thing is to find the dimension of the space by finding a basis. For each basis given below, We will omit the verification that it is a basis. \begin{exparts} \partsitem It is isomorphic to $\Re^5$. One basis for $\polyspace_4$ is $\set{x^4,x^3,x^2,x,1}$ so the space has dimension~$5$. \partsitem It is isomorphic to $\Re^2$ since one basis for the space $\polyspace_1=\set{a+bx\suchthat a,b\in\Re}$ is $\set{1,x}$. \partsitem It is isomorphic to $\Re^6$. One basis has these six matrices. \begin{equation*} \begin{mat} 1 &0 &0 \\ 0 &0 &0 \end{mat},\; \begin{mat} 0 &1 &0 \\ 0 &0 &0 \end{mat},\; \cdots\; \begin{mat} 0 &0 &0 \\ 0 &0 &1 \end{mat} \end{equation*} \partsitem It is a plane so it is isomorphic to~$\Re^2$. For a more extensive answer, parametrizing the plane gives this vector description \begin{equation*} \set{\colvec{x \\ y \\ z}=\colvec{1/2 \\ 1 \\ 0}y +\colvec{-1/2 \\ 0 \\ 1}z \suchthat y,z\in\Re} \end{equation*} and so it has a basis consisting of those two vectors. \partsitem It is isomorphic to $\Re^3$. One basis is the set of linear combinations $\set{x,y,z}$, that is, $\set{x+0y+0z, 0x+y+0z,0x+0y+z}$. \end{exparts} \end{answer} \recommended \item Show that if $$m\neq n$$ then $$\Re^m\not\isomorphicto\Re^n$$. \begin{answer} ... ...
 ... ... @@ -829,7 +829,7 @@ is more fruitful and more central to progress. =r\vec{v} \). \end{exparts} \end{answer} \recommended \item \item Consider the vector space $$\Re^+$$ where vector addition and scalar multiplication are not the ones inherited from $\Re$ but rather are these: ... ... @@ -849,8 +849,8 @@ is more fruitful and more central to progress. the exponential map, so it is a correspondence, and therefore it is an isomorphism. \end{answer} \recommended \item Consider this transformation of $$\Re^2$$. \item Consider this transformation of the plane $$\Re^2$$. \begin{equation*} \colvec{x \\ y} \mapsto \colvec{x/2 \\ y/3} \end{equation*} ... ... @@ -1022,7 +1022,7 @@ is more fruitful and more central to progress. \end{multline*} where $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$ \end{answer} \recommended \item \item Where $$\map{f}{V}{W}$$ is linear, suppose that $$f(\vec{v}_1)=\vec{w}_1$$, \ldots, $$f(\vec{v}_n)=\vec{w}_n$$ for some vectors $$\vec{w}_1$$, \ldots, $$\vec{w}_n$$ from $$W$$. ... ... @@ -2313,6 +2313,51 @@ point and the map is an isomorphism between the domain and the range. \end{equation*} for $$k\leq n$$. \end{answer} \item For the map $\map{h}{\Re^3}{\Re^2}$ given by \begin{equation*} \colvec{x \\ y \\ z} \mapsto \colvec{x+y \\ x+z} \end{equation*} find the range space, rank, null space, and nullity. \begin{answer} To see that the range space is all of~$\Re^2$ note that for any $u,v\in\Re$ this system has a solution. \begin{equation*} \begin{linsys}{3} x &+ &y & & &= &u \\ x & & &+ &z &= &v \end{linsys} \grstep{-\rho_1+\rho_2} \begin{linsys}{3} x &+ &y & & &= &u \\ & &-y &+ &z &= &-u+v \end{linsys} \end{equation*} (In fact, because there is a free variable, $z$, it has infinitely many solutions.) Thus the rank, the dimension of the range space, is~$2$. Since the rank plus the nullity equals the dimension of the domain, we know that the nullity is~$1$. Finding the null space verifies that: \begin{equation*} \begin{linsys}{3} x &+ &y & & &= &0 \\ x & & &+ &z &= &0 \end{linsys} \grstep{-\rho_1+\rho_2} \begin{linsys}{3} x &+ &y & & &= &0 \\ & &-y &+ &z &= &0 \end{linsys} \end{equation*} the nullspace is this set \begin{equation*} \set{\colvec{-1 \\ 1 \\ 1}\cdot z\suchthat z \in \Re} \end{equation*} which is one-dimensional. \end{answer} \item \label{exer:CondTwoProjMap} \nearbyexample{ex:RThreeHomoRTwo} restates the first condition in the definition of homomorphism as `the shadow of a sum is the sum of the ... ... @@ -2604,7 +2649,7 @@ point and the map is an isomorphism between the domain and the range. &=\spanof{h(S)} \end{align*} \end{answer} \recommended \item \label{exer:Cosets} \item \label{exer:Cosets} \begin{exparts} \partsitem Prove that for any linear map $$\map{h}{V}{W}$$ and any $$\vec{w}\in W$$, the set $$h^{-1}(\vec{w})$$ has the form ... ...
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