Commit b3393d97 by Jim Hefferon

### map2

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 ... ... @@ -12485,7 +12485,7 @@ sage: n(a_by_a) \end{ans} \subsection{Three.II.2: Range space and Null space} \begin{ans}{Three.II.2.22} \begin{ans}{Three.II.2.21} First, to answer whether a polynomial is in the null space, we have to consider it as a member of the domain $\polyspace_3$. To answer whether it is in the range space, we consider it as a member of ... ... @@ -12521,7 +12521,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.II.2.23} \begin{ans}{Three.II.2.22} \begin{exparts} \partsitem The null space is \begin{equation*} ... ... @@ -12579,7 +12579,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.II.2.24} \begin{ans}{Three.II.2.23} For each, use the result that the rank plus the nullity equals the dimension of the domain. \begin{exparts*} ... ... @@ -12590,7 +12590,7 @@ sage: n(a_by_a) \end{exparts*} \end{ans} \begin{ans}{Three.II.2.25} \begin{ans}{Three.II.2.24} Because \begin{equation*} \frac{d}{dx}\,(a_0+a_1x+\cdots+a_nx^n) ... ... @@ -12613,11 +12613,11 @@ sage: n(a_by_a) for $$k\leq n$$. \end{ans} \begin{ans}{Three.II.2.26} \begin{ans}{Three.II.2.25} The shadow of a scalar multiple is the scalar multiple of the shadow. \end{ans} \begin{ans}{Three.II.2.27} \begin{ans}{Three.II.2.26} \begin{exparts} \partsitem Setting $a_0+(a_0+a_1)x+(a_2+a_3)x^3=0+0x+0x^2+0x^3$ gives $a_0=0$ and $a_0+a_1=0$ and $a_2+a_3=0$, so the null space is ... ... @@ -12633,14 +12633,14 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.II.2.28} \begin{ans}{Three.II.2.27} All inverse images are lines with slope $-2$. \begin{center} \includegraphics{ch3.74} \end{center} \end{ans} \begin{ans}{Three.II.2.29} \begin{ans}{Three.II.2.28} These are the inverses. \begin{exparts} \partsitem $$a_0+a_1x+a_2x^2+a_3x^3\mapsto ... ... @@ -12656,7 +12656,7 @@ sage: n(a_by_a) So this map is actually self-inverse. \end{ans} \begin{ans}{Three.II.2.30} \begin{ans}{Three.II.2.29} For any vector space V, the null space \begin{equation*} \set{\vec{v}\in V\suchthat 2\vec{v}=\zero} ... ... @@ -12671,7 +12671,7 @@ sage: n(a_by_a) (Thus, this transformation is actually an automorphism.) \end{ans} \begin{ans}{Three.II.2.31} \begin{ans}{Three.II.2.30} Because the rank plus the nullity equals the dimension of the domain (here, five), and the rank is at most three, the possible pairs are:~\( (3,2)$$, $$(2,3)$$, ... ... @@ -12680,13 +12680,13 @@ sage: n(a_by_a) possible is easy. \end{ans} \begin{ans}{Three.II.2.32} \begin{ans}{Three.II.2.31} No (unless $$\polyspace_n$$ is trivial), because the two polynomials $$f_0(x)=0$$ and $$f_1(x)=1$$ have the same derivative; a map must be one-to-one to have an inverse. \end{ans} \begin{ans}{Three.II.2.33} \begin{ans}{Three.II.2.32} The null space is this. \begin{multline*} \set{a_0+a_1x+\dots+a_nx^n\suchthat ... ... @@ -12699,7 +12699,7 @@ sage: n(a_by_a) Thus the nullity is $$n$$. \end{ans} \begin{ans}{Three.II.2.34} \begin{ans}{Three.II.2.33} \begin{exparts} \partsitem One direction is obvious:~if the homomorphism is onto then its range is the codomain and so its ... ... @@ -12725,7 +12725,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.II.2.35} \begin{ans}{Three.II.2.34} We are proving that $$\map{h}{V}{W}$$ is one-to-one if and only if for every linearly independent subset $$S$$ of $$V$$ the subset $$h(S)=\set{h(\vec{s})\suchthat \vec{s}\in S}$$ of $$W$$ is linearly ... ... @@ -12781,7 +12781,7 @@ sage: n(a_by_a) and so not all sets have independence preserved. \end{ans} \begin{ans}{Three.II.2.36} \begin{ans}{Three.II.2.35} (We use the notation from \nearbytheorem{th:HomoDetActOnBasis}.) Fix a basis $\sequence{\vec{\beta}_1,\dots,\vec{\beta}_n}$ for $V$ and a basis ... ... @@ -12807,7 +12807,7 @@ sage: n(a_by_a) Thus the map is onto. \end{ans} \begin{ans}{Three.II.2.37} \begin{ans}{Three.II.2.36} Yes. For the transformation of $$\Re^2$$ given by \begin{equation*} ... ... @@ -12822,7 +12822,7 @@ sage: n(a_by_a) We will see more of this in the fifth chapter. \end{ans} \begin{ans}{Three.II.2.38} \begin{ans}{Three.II.2.37} This is a simple calculation. \begin{align*} h(\spanof{S}) ... ... @@ -12836,7 +12836,7 @@ sage: n(a_by_a) \end{align*} \end{ans} \begin{ans}{Three.II.2.39} \begin{ans}{Three.II.2.38} \begin{exparts} \partsitem We will show that the two sets are equal ... ... @@ -12901,7 +12901,7 @@ sage: n(a_by_a) \end{exparts} \end{ans} \begin{ans}{Three.II.2.40} \begin{ans}{Three.II.2.39} Because the rank of $$t$$ is one, the range space of $$t$$ is a one-dimensional set. Taking $\sequence{h(\vec{v})}$ as a basis (for some appropriate ... ... @@ -12924,7 +12924,7 @@ sage: n(a_by_a) to get the desired conclusion. \end{ans} \begin{ans}{Three.II.2.41} \begin{ans}{Three.II.2.40} By assumption, $$h$$ is not the zero map and so a vector $$\vec{v}\in V$$ exists that is not in the null space. Note that $$\sequence{h(\vec{v})}$$ is a basis for $$\Re$$, ... ... @@ -12948,7 +12948,7 @@ sage: n(a_by_a) $$h$$ to both sides would give a contradiction. \end{ans} \begin{ans}{Three.II.2.42} \begin{ans}{Three.II.2.41} Fix a basis $$\sequence{\vec{\beta}_1,\dots,\vec{\beta}_n}$$ for $$V$$. We shall prove that this map ... ... @@ -13005,7 +13005,7 @@ sage: n(a_by_a) so $$\Phi(r_1h_1+r_2h_2)=r_1\Phi(h_1)+r_2\Phi(h_2)$$. \end{ans} \begin{ans}{Three.II.2.43} \begin{ans}{Three.II.2.42} Let $$\map{h}{V}{W}$$ be linear and fix a basis $$\sequence{\vec{\beta}_1,\dots,\vec{\beta}_n}$$ for $$V$$. Consider these $$n$$ maps from $$V$$ to $$W$$ ... ... @@ -13023,7 +13023,7 @@ sage: n(a_by_a) $$h_i(r\vec{v}+s\vec{u})=rc_i+sd_i=rh_i(\vec{v})+sh_i(\vec{u})$$. \end{ans} \begin{ans}{Three.II.2.44} \begin{ans}{Three.II.2.43} Either yes (trivially) or no (nearly trivially). If we take $$V$$ `is homomorphic to' $$W$$ to mean there is a ... ... @@ -13041,7 +13041,7 @@ sage: n(a_by_a) so the relation is not reflexive.\appendrefs{equivalence relations} \end{ans} \begin{ans}{Three.II.2.45} \begin{ans}{Three.II.2.44} That they form the chains is obvious. For the rest, we show here that $$\rangespace{t^{j+1}}=\rangespace{t^{j}}$$
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