Commit 21c1e531 by Jim Hefferon

### first run thru edits of map2

 ... ... @@ -1680,7 +1680,7 @@ so its nullity is $1$. \begin{example} The map from \nearbyexample{ex:MatToPolyRnge} has this nullspace. has this nullspace an nullity $2$. \begin{equation*} \nullspace{h}=\set{\begin{mat} a &b \\ ... ... @@ -1690,22 +1690,24 @@ has this nullspace. \end{example} Now for the second insight from the above pictures. In \nearbyexample{ex:RThreeHomoRTwo}, each of the vertical lines is squashed down to a single point\Dash $\pi$, in passing from the domain to the range, takes all of these one-dimensional vertical lines and zeroes them out'', In \nearbyexample{ex:RThreeHomoRTwo} each of the vertical lines is squashed down to a single point\Dash in passing from the domain to the range $\pi$ takes all of these one-dimensional vertical lines and squashes them to a point, leaving the range one dimension smaller than the domain. Similarly, in \nearbyexample{ex:RTwoHomoRHardOne}, the Similarly, in \nearbyexample{ex:RTwoHomoRHardOne} the two-dimensional domain is mapped to a one-dimensional range by breaking the domain into lines (here, they are diagonal lines), and compressing each of those lines to a single member of the range. Finally, in \nearbyexample{ex:PicRThreeToRTwo}, the domain into the diagonal lines and compressing each of those to a single member of the range. Finally, in \nearbyexample{ex:PicRThreeToRTwo} the domain breaks into planes which get zeroed out'', and so the map starts with a three-dimensional domain squashed to a point and so the map starts with a three-dimensional domain but ends with a one-dimensional range\Dash this map subtracts'' two from the dimension. (Notice that, in this third example, the codomain is one-dimensional range. (Notice that in this third example the codomain is two-dimensional but the range of the map is only one-dimensional, and it is the dimension of the range that is of interest.) the dimension of the range that we are studying.) \begin{theorem} \label{th:RankPlusNullEqDim} ... ... @@ -1717,30 +1719,31 @@ A linear map's rank plus its nullity equals the dimension of its domain. Let $$\map{h}{V}{W}$$ be linear and let $$B_N=\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_k}$$ be a basis for the nullspace. Extend that to a basis Expand that to a basis $$B_V=\sequence{\vec{\beta}_1,\dots,\vec{\beta}_k, \vec{\beta}_{k+1},\dots,\vec{\beta}_n}$$ for the entire domain. for the entire domain, using Corollary~Two.III.\ref{cor:LIExpBas}. We shall show that $$B_R=\sequence{ h(\vec{\beta}_{k+1}),\dots,h(\vec{\beta}_n)}$$ is a basis for the rangespace. Then counting the size of these bases gives the result. With that, counting the size of these bases gives the result. To see that $$B_R$$ is linearly independent, consider the equation $$c_{k+1}h(\vec{\beta}_{k+1})+\dots+c_nh(\vec{\beta}_n)=\zero_W$$. This gives that $$h(c_{k+1}\vec{\beta}_{k+1}+\dots+c_n\vec{\beta}_n)=\zero_W$$ and so $$c_{k+1}\vec{\beta}_{k+1}+\dots+c_n\vec{\beta}_n$$ consider $$\zero_W=c_{k+1}h(\vec{\beta}_{k+1})+\dots+c_nh(\vec{\beta}_n)$$. The function is linear so we have $$\vec{0_W}=h(c_{k+1}\vec{\beta}_{k+1}+\dots+c_n\vec{\beta}_n)$$ and therefore $$c_{k+1}\vec{\beta}_{k+1}+\dots+c_n\vec{\beta}_n$$ is in the nullspace of $h$. As $$B_N$$ is a basis for this nullspace, there are scalars $$c_1,\dots,c_k\in\Re$$ satisfying this relationship. As $$B_N$$ is a basis for the nullspace there are scalars $$c_1,\dots,c_k$$ satisfying this relationship. \begin{equation*} c_1\vec{\beta}_1+\dots+c_k\vec{\beta}_k = c_{k+1}\vec{\beta}_{k+1}+\dots+c_n\vec{\beta}_n \end{equation*} But $$B_V$$ is a basis for $$V$$ so each scalar equals zero. But this is an equation among the members of $$B_V$$, which is a basis for $$V$$, so each $c_i$ equals $0$. Therefore $$B_R$$ is linearly independent. To show that $$B_R$$ spans the rangespace, ... ... @@ -1776,43 +1779,42 @@ the rangespace and nullspace are \nullspace{h}= \set{\colvec{0 \\ 0 \\ z}\suchthat z\in\Re } \end{equation*} and so the rank of $h$ is two while the nullity is one. and so the rank of $h$ is $2$ while the nullity is $1$. \end{example} \begin{example} If $$\map{t}{\Re}{\Re}$$ is the linear transformation $$x\mapsto -4x,$$ then the range is $$\rangespace{t}=\Re^1$$, and so the rank of $t$ is one and the nullity is zero. the rank of $t$ is $1$ and the nullity is $0$. \end{example} \begin{corollary} \label{cor:RankDecreases} The rank of a linear map is less than or equal to the dimension of the domain. Equality holds if and only if the nullity of the map is zero. Equality holds if and only if the nullity of the map is $0$. \end{corollary} We know that an isomorphism exists between two spaces if and only if their dimensions are equal. Here we see that for a homomorphism to exist, if and only if the dimension of the range equals the dimension of the domain. We have now seen that for a homomorphism to exist, the dimension of the range must be less than or equal to the dimension of the domain. For instance, there is no homomorphism from $$\Re^2$$ onto $$\Re^3$$. There are many homomorphisms from $$\Re^2$$ into $$\Re^3$$, but none is onto all of three-space. from $$\Re^2$$ into $$\Re^3$$, but none onto. The rangespace of a linear map can be of dimension strictly less than the dimension of the domain (\nearbyexample{ex:DerivMapRnge}'s derivative transformation on $\polyspace_3$ has a domain of dimension four but a range of dimension three). Thus, under a homomorphism, and so linearly independent sets in the domain may map to linearly dependent sets in the range (for instance, the derivative sends may map to linearly dependent sets in the range. (\nearbyexample{ex:DerivMapRnge}'s derivative transformation on $\polyspace_3$ has a domain of dimension~$4$ but a range of dimension~$3$ and the derivative sends $\set{1,x,x^2,x^3}$ to $\set{0,1,2x,3x^2}$). That is, under a homomorphism, independence may be lost. That is, under a homomorphism independence may be lost. In contrast, dependence stays. \begin{lemma} ... ... @@ -1831,7 +1833,7 @@ $c_i$. \end{proof} When is independence not lost? One obvious sufficient condition is when the homomorphism is an isomorphism. The obvious sufficient condition is when the homomorphism is an isomorphism. This condition is also necessary; see \nearbyexercise{exer:NonSingIffPreservLI}. We will finish this subsection comparing homomorphisms with isomorphisms ... ... @@ -1856,14 +1858,16 @@ This one-to-one homomorphism $$\map{\iota}{\Re^2}{\Re^3}$$ \colvec{x \\ y \\ 0} \end{equation*} gives a correspondence between $$\Re^2$$ and the $$xy$$-plane inside of $$\Re^3$$. subset of $$\Re^3$$. \end{example} The prior observation allows us to adapt some results about isomorphisms. % The prior observation allows us to adapt some results about isomorphisms. \begin{theorem} \label{th:OOHomoEquivalence} In an $$n$$-dimensional vector space $$V$$, these are equivalent statements about a linear map $$\map{h}{V}{W}$$. \begin{tfae} \item $$h$$ is one-to-one \item $$h$$ has an inverse, from its range to its domain, that is linear ... ... @@ -1874,8 +1878,6 @@ In an $$n$$-dimensional vector space $$V$$, these $$\sequence{h(\vec{\beta}_1),\dots,h(\vec{\beta}_n)}$$ is a basis for $$\rangespace{h}$$ \end{tfae} are equivalent statements about a linear map $$\map{h}{V}{W}$$. \end{theorem} \begin{proof} ... ... @@ -1887,8 +1889,8 @@ We will then show that \). For $$\text{(1)} \Longrightarrow \text{(2)}$$, suppose that the linear map $h$ is one-to-one, and so has an inverse. The domain of that inverse is the range of $h$ and so a linear combination suppose that the linear map $h$ is one-to-one and so has an inverse. The domain of that inverse is the range of $h$ and thus a linear combination of two members of that domain has the form $c_1h(\vec{v}_1)+c_2h(\vec{v}_2)$. On that combination, the inverse $$h^{-1}$$ gives this. \begin{align*} ... ... @@ -1896,11 +1898,12 @@ On that combination, the inverse $$h^{-1}$$ gives this. &=h^{-1}(h(c_1\vec{v}_1+c_2\vec{v}_2)) \\ &=\composed{h^{-1}}{h}\,(c_1\vec{v}_1+c_2\vec{v}_2) \\ &=c_1\vec{v}_1+c_2\vec{v}_2 \\ &=c_1\composed{h^{-1}}{h}\,(\vec{v}_1) +c_2\composed{h^{-1}}{h}\,(\vec{v}_2) \\ % &=c_1\composed{h^{-1}}{h}\,(\vec{v}_1) % +c_2\composed{h^{-1}}{h}\,(\vec{v}_2) \\ &=c_1\cdot h^{-1}(h(\vec{v}_1))+c_2\cdot h^{-1}(h(\vec{v}_2)) \end{align*} Thus the inverse of a one-to-one linear map is automatically linear. Thus if a linear map is one-to-one, that is, if it has an inverse, then the inverse must be linear. But this also gives the $$\text{(2)} \Longrightarrow \text{(1)}$$ implication, because the inverse itself must be one-to-one. ... ... @@ -1930,7 +1933,7 @@ is a basis for $$V$$ so that $$\sequence{h(\vec{\beta}_1),\dots,h(\vec{\beta}_n)}$$ is a basis for $$\rangespace{h}$$. Then every $$\vec{w}\in\rangespace{h}$$ a the unique representation $$\vec{w}\in\rangespace{h}$$ has the unique representation $$\vec{w}=c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n)$$. Define a map from $$\rangespace{h}$$ to $V$ by \begin{equation*} ... ... @@ -1941,8 +1944,8 @@ Checking that it is linear and that it is the inverse of $h$ are easy. \end{proof} We've now seen that a linear map shows how the structure of the domain is like that of the range. We have now seen that a linear map expresses how the structure of the domain is like that of the range. Such a map can be thought to organize the domain space into inverse images of points in the range. In the special case that the map is one-to-one, each inverse image is a single ... ...
 ... ... @@ -78,7 +78,7 @@ is that the third chapter, on linear maps, does not begin with the definition of homomorphism. Rather, we start with the definition of isomorphism, which is natural: students themselves observe that some spaces are just like'' others. observe that some spaces are the same'' as others. After that, the next section takes the reasonable step of isolating the operation-preservation idea ... ... @@ -99,8 +99,8 @@ taken from various journals, competitions, or problems collections. These are marked with a `\puzzlemark' and as part of the fun the original wording has been retained as much as possible. as part of the fun I have retained the original wording as much as possible. That is, as with the rest of the book, the exercises are aimed to both build an ability at, ... ...