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 ... ... @@ -10560,7 +10560,7 @@ sage: n(a_by_a) \chapter{Chapter Three: Maps Between Spaces} \section{Isomorphisms} \subsection{Three.I.1: Def{}inition and Examples} \begin{ans}{Three.I.1.11} \begin{ans}{Three.I.1.12} \begin{exparts} \partsitem Call the map $f$. \begin{equation*} ... ... @@ -10643,7 +10643,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.I.1.12} \begin{ans}{Three.I.1.13} These are the images. \begin{exparts*} \partsitem $$\colvec[r]{5 \\ -2}$$ ... ... @@ -10679,7 +10679,7 @@ sage: n(a_by_a) \end{align*} \end{ans} \begin{ans}{Three.I.1.13} \begin{ans}{Three.I.1.14} To verify it is one-to-one, assume that $f_1(c_1x+c_2y+c_3z)=f_1(d_1x+d_2y+d_3z)$. Then $c_1+c_2x+c_3x^2=d_1+d_2x+d_3x^2$ ... ... @@ -10715,7 +10715,7 @@ sage: n(a_by_a) \end{align*} \end{ans} \begin{ans}{Three.I.1.14} \begin{ans}{Three.I.1.15} \begin{exparts} \partsitem No; this map is not one-to-one. In particular, the matrix of all zeroes is mapped to the same ... ... @@ -10860,14 +10860,14 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.I.1.15} \begin{ans}{Three.I.1.16} It is one-to-one and onto, a correspondence, because it has an inverse (namely, $$f^{-1}(x)=\sqrt{x}$$). However, it is not an isomorphism. For instance, $$f(1)+f(1)\neq f(1+1)$$. \end{ans} \begin{ans}{Three.I.1.16} \begin{ans}{Three.I.1.17} Many maps are possible. Here are two. \begin{equation*} ... ... @@ -10878,7 +10878,7 @@ sage: n(a_by_a) The verifications are straightforward adaptations of the others above. \end{ans} \begin{ans}{Three.I.1.17} \begin{ans}{Three.I.1.18} Here are two. \begin{equation*} a_0+a_1x+a_2x^2 \mapsto \colvec{a_1 \\ a_0 \\ a_2} ... ... @@ -10893,7 +10893,7 @@ sage: n(a_by_a) is the image of $(s-t)+tx+ux^2$). \end{ans} \begin{ans}{Three.I.1.18} \begin{ans}{Three.I.1.19} The space $\Re^2$ is not a subspace of $\Re^3$ because it is not a subset of $\Re^3$. The two-tall vectors in $\Re^2$ are not members of $\Re^3$. ... ... @@ -10934,7 +10934,7 @@ sage: n(a_by_a) that it preserves combinations of two vectors. \end{ans} \begin{ans}{Three.I.1.19} \begin{ans}{Three.I.1.20} Here are two: \begin{equation*} \colvec{r_1 \\ r_2 \\ \vdots \\ r_{16}} ... ... @@ -10957,7 +10957,7 @@ sage: n(a_by_a) Verification that each is an isomorphism is easy. \end{ans} \begin{ans}{Three.I.1.20} \begin{ans}{Three.I.1.21} When $k$ is the product $$k=mn$$, here is an isomorphism. \begin{equation*} \begin{mat} ... ... @@ -10971,7 +10971,7 @@ sage: n(a_by_a) Checking that this is an isomorphism is easy. \end{ans} \begin{ans}{Three.I.1.21} \begin{ans}{Three.I.1.22} If $$n\geq 1$$ then $$\polyspace_{n-1}\isomorphicto\Re^n$$. (If we take $$\polyspace_{-1}$$ and $$\Re^0$$ to be trivial vector spaces, then the relationship extends one dimension lower.) ... ... @@ -10983,7 +10983,7 @@ sage: n(a_by_a) Checking that it is an isomorphism is straightforward. \end{ans} \begin{ans}{Three.I.1.22} \begin{ans}{Three.I.1.23} This is the map, expanded. \begin{align*} f(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5) ... ... @@ -11023,25 +11023,25 @@ sage: n(a_by_a) \end{multline*} \end{ans} \begin{ans}{Three.I.1.23} \begin{ans}{Three.I.1.24} No vector space has the empty set underlying it. We can take $\vec{v}$ to be the zero vector. \end{ans} \begin{ans}{Three.I.1.24} \begin{ans}{Three.I.1.25} Yes; where the two spaces are $$\set{\vec{a}}$$ and $$\set{\vec{b}}$$, the map sending $$\vec{a}$$ to $$\vec{b}$$ is clearly one-to-one and onto, and also preserves what little structure there is. \end{ans} \begin{ans}{Three.I.1.25} \begin{ans}{Three.I.1.26} A linear combination of $n=0$ vectors adds to the zero vector and so \nearbylemma{le:IsoSendsZeroToZero} shows that the three statements are equivalent in this case. \end{ans} \begin{ans}{Three.I.1.26} \begin{ans}{Three.I.1.27} Consider the basis $$\sequence{1}$$ for $$\polyspace_0$$ and let $$f(1)\in\Re$$ be $$k$$. For any $$a\in\polyspace_0$$ ... ... @@ -11053,7 +11053,7 @@ sage: n(a_by_a) as is easy to check.) \end{ans} \begin{ans}{Three.I.1.27} \begin{ans}{Three.I.1.28} In each item, following item~(2) of \nearbylemma{le:PresStructIffPresCombos}, we show that the map preserves structure by showing that the it preserves linear combinations ... ... @@ -11095,7 +11095,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.I.1.28} \begin{ans}{Three.I.1.29} One direction is easy:~by definition, if $$f$$ is one-to-one then for any $$\vec{w}\in W$$ at most one $$\vec{v}\in V$$ has ... ... @@ -11116,7 +11116,7 @@ sage: n(a_by_a) $\vec{v}_1=\vec{v}_2$, and so $$f$$ is one-to-one. \end{ans} \begin{ans}{Three.I.1.29} \begin{ans}{Three.I.1.30} We will prove something stronger\Dash not only is the existence of a dependence preserved by isomorphism, but each instance of a dependence is preserved, that is, ... ... @@ -11144,7 +11144,7 @@ sage: n(a_by_a) to be mapped to the same image by $f$, they must be equal. \end{ans} \begin{ans}{Three.I.1.30} \begin{ans}{Three.I.1.31} \begin{exparts} \partsitem This map is one-to-one because if $$d_s(\vec{v}_1)=d_s(\vec{v}_2)$$ ... ... @@ -11268,7 +11268,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.I.1.31} \begin{ans}{Three.I.1.32} First, the map $p(x)\mapsto p(x+k)$ doesn't count because it is a version of $p(x)\mapsto p(x-k)$. Here is a correct answer (many others are also correct): ... ... @@ -11276,7 +11276,7 @@ sage: n(a_by_a) Verification that this is an isomorphism is straightforward. \end{ans} \begin{ans}{Three.I.1.32} \begin{ans}{Three.I.1.33} \begin{exparts} \partsitem For the `only if' half, let $$\map{f}{\Re^1}{\Re^1}$$ to be an isomorphism. ... ... @@ -11378,7 +11378,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.I.1.33} \begin{ans}{Three.I.1.34} There are many answers; two are linear independence and subspaces. ... ... @@ -11429,7 +11429,7 @@ sage: n(a_by_a) a member of $f(U)$. \end{ans} \begin{ans}{Three.I.1.34} \begin{ans}{Three.I.1.35} \begin{exparts} \partsitem The association \begin{equation*} ... ... @@ -11491,11 +11491,11 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.I.1.35} \begin{ans}{Three.I.1.36} See the next subsection. \end{ans} \begin{ans}{Three.I.1.36} \begin{ans}{Three.I.1.37} \begin{exparts} \partsitem Most of the conditions in the definition of a vector space are routine.
 ... ... @@ -704,13 +704,15 @@ beginfig(16) % flip over a line map z0=(3w,2v); z2=(2w,3v); pickup pencircle scaled line_width_dark; draw (-.5w,-.5v)--(3w,3v) dashed withdots scaled .5; label.ulft(btex {\scriptsize $\ell$} etex,0.5[(0w,0v),(3w,3v)]); z3=(-.5w,-.5v) shifted codomainshift; z4=(3w,3v) shifted codomainshift; draw z3--z4 dashed withdots scaled .5; drawarrow (0w,0v)--z0; label.rt(btex {\scriptsize $\vec{u}$} etex,z0); drawarrow ((0w,0v)--z2) shifted codomainshift; picture pa; % in case the line ell hits this label pa = thelabel.rt(btex {\scriptsize $f_{\ell}(\vec{u})$} etex, pa = thelabel.urt(btex {\scriptsize $f_{\ell}(\vec{u})$} etex, z2 shifted codomainshift); unfill bbox pa; draw pa; ... ...
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