Commit 633557a5 by Jim Hefferon

parent c0ca25a5
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 ... ... @@ -11994,9 +11994,10 @@ sage: solve(system, x,y) \partsitem Yes. This is the check that it preserves combinations of two members of the domain. \begin{align*} \begin{multline*} h(r_1\cdot\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} +r_2\cdot\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat}) +r_2\cdot\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat}) \\ \begin{aligned} &=h(\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 ... ... @@ -12007,9 +12008,11 @@ sage: solve(system, x,y) +r_2(2a_2+3b_2+c_2-d_2) \\ &=r_1\cdot h(\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} +r_2\cdot h(\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat}) \end{align*} \end{aligned} \end{multline*} \partsitem No. An example of a combination that is not preserved is this. An example of a combination that is not preserved is that doing it one way gives this \begin{equation*} h(\begin{mat}[r] 1 &0 \\ ... ... @@ -12024,7 +12027,9 @@ sage: solve(system, x,y) 0 &0 \end{mat}) =4 \quad\text{while}\quad \end{equation*} while the other way gives this. \begin{equation*} h(\begin{mat}[r] 1 &0 \\ 0 &0 ... ... @@ -12249,9 +12254,10 @@ sage: solve(system, x,y) \begin{exparts} \partsitem Where $$c$$ and $$d$$ are scalars, we have this. \begin{align*} \begin{multline*} h(c\cdot \colvec{x_1 \\ \vdots \\ x_n} +d\cdot \colvec{y_1 \\ \vdots \\ y_n}) +d\cdot \colvec{y_1 \\ \vdots \\ y_n}) \\ \begin{aligned} &=h(\colvec{cx_1+dy_1 \\ \vdots \\ cx_n+dy_n}) \\ &=\colvec{a_{1,1}(cx_1+dy_1)+\dots+a_{1,n}(cx_n+dy_n) \\ \vdots \\ ... ... @@ -12264,13 +12270,14 @@ sage: solve(system, x,y) a_{m,1}y_1+\dots+a_{m,n}y_n} \\ &=c\cdot h(\colvec{x_1 \\ \vdots \\ x_n}) +d\cdot h(\colvec{y_1 \\ \vdots \\ y_n}) \end{align*} \end{aligned} \end{multline*} \partsitem Each power $i$ of the derivative operator is linear because of these rules familiar from calculus. \begin{equation*} \frac{d^i}{dx^i}(\,f(x)+g(x)\,)=\frac{d^i}{dx^i}f(x) +\frac{d^i}{dx^i}g(x) \quad\text{and}\quad \qquad \frac{d^i}{dx^i}\,r\cdot f(x)=r\cdot\frac{d^i}{dx^i}f(x) \end{equation*} Thus the given map is a linear transformation of $$\polyspace_n$$ ... ... @@ -12344,7 +12351,7 @@ sage: solve(system, x,y) Recall that the entry in row~$i$ and column~$j$ of the transpose of $M$ is the entry $m_{j,i}$ from row~$j$ and column~$i$ of $M$. Now, the check is routine. \begin{align*} \begin{multline*} \trans{[r\cdot\begin{mat} \ &\vdots \\ \cdots &a_{i,j} &\cdots \\ ... ... @@ -12354,7 +12361,8 @@ sage: solve(system, x,y) \ &\vdots \\ \cdots &b_{i,j} &\cdots \\ &\vdots \end{mat}]} \end{mat}]} \\ \begin{aligned} &=\trans{\begin{mat} \ &\vdots \\ \cdots &ra_{i,j}+sb_{i,j} &\cdots \\ ... ... @@ -12385,7 +12393,8 @@ sage: solve(system, x,y) \cdots &b_{j,i} &\cdots \\ &\vdots \end{mat} } \end{align*} \end{aligned} \end{multline*} The domain is $$\matspace_{\nbym{m}{n}}$$ while the codomain is $$\matspace_{\nbym{n}{m}}$$. ... ... @@ -12438,8 +12447,8 @@ sage: solve(system, x,y) If $$h(\vec{v}_1)$$ is the zero vector then this line is degenerate. \partsitem A $$k$$-dimensional linear surface in $$\Re^n$$ maps to a (possibly degenerate) $$k$$-dimensional linear surface in $$\Re^m$$. a $$k$$-dimensional linear surface in $$\Re^m$$ (possibly it is degenerate). The proof is just like that the one for the line. \end{exparts} ... ... @@ -12505,7 +12514,7 @@ sage: solve(system, x,y) subspace of is a subspace. Suppose that $$X$$ is a subspace of $$W$$. Note that $$\zero_W\in X$$ so the set Note that $$\zero_W\in X$$ so that the set $$\set{\vec{v}\in V \suchthat h(\vec{v})\in X}$$ is not empty. To show that this set is closed under combinations, let $$\vec{v}_1,\dots,\vec{v}_n$$ be elements of $$V$$ ... ... @@ -12550,14 +12559,14 @@ sage: solve(system, x,y) \begin{ans}{Three.II.1.42} \begin{exparts} \partsitem Briefly, the check of linearity is this. \begin{equation*} \begin{multline*} F(r_1\cdot \vec{v}_1+r_2\cdot \vec{v}_2) =\colvec{f_1(r_1\vec{v}_1+r_2\vec{v}_2) \\ f_2(r_1\vec{v}_1+r_2\vec{v}_2)} f_2(r_1\vec{v}_1+r_2\vec{v}_2)} \\ =r_1\colvec{f_1(\vec{v}_1) \\ f_2(\vec{v}_1)} +r_2\colvec{f_1(\vec{v}_2) \\ f_2(\vec{v}_2)} =r_1\cdot F(\vec{v}_1)+r_2\cdot F(\vec{v}_2) \end{equation*} \end{multline*} \partsitem Yes. Let $$\map{\pi_1}{\Re^2}{\Re^1}$$ and $$\map{\pi_2}{\Re^2}{\Re^1}$$ be the projections ... ... @@ -12941,7 +12950,7 @@ sage: solve(system, x,y) \begin{ans}{Three.II.2.38} \begin{exparts} \partsitem We will show that the two sets are equal We will show the sets are equal $h^{-1}(\vec{w}) =\set{\vec{v}+\vec{n}\suchthat \vec{n}\in\nullspace{h} }$ by mutual inclusion. ... ... @@ -13094,8 +13103,9 @@ sage: solve(system, x,y) c_1h_2(\vec{\beta}_1)+\dots+c_nh_2(\vec{\beta}_n) \end{align*} we have \begin{align*} (r_1h_1+r_2h_2)(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) \begin{multline*} (r_1h_1+r_2h_2)(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) \\ \begin{aligned} &= c_1(r_1h_1(\vec{\beta}_1)+r_2h_2(\vec{\beta}_1)) +\dots ... ... @@ -13103,7 +13113,8 @@ sage: solve(system, x,y) &= r_1(c_1h_1(\vec{\beta}_1)+\dots+c_nh_1(\vec{\beta}_n)) + r_2(c_1h_2(\vec{\beta}_1)+\dots+c_nh_2(\vec{\beta}_n)) \end{align*} \end{aligned} \end{multline*} so $$\Phi(r_1h_1+r_2h_2)=r_1\Phi(h_1)+r_2\Phi(h_2)$$. \end{ans}
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 ... ... @@ -562,9 +562,10 @@ is more fruitful and more central to progress. \partsitem Yes. This is the check that it preserves combinations of two members of the domain. \begin{align*} \begin{multline*} h(r_1\cdot\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} +r_2\cdot\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat}) +r_2\cdot\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat}) \\ \begin{aligned} &=h(\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 ... ... @@ -575,9 +576,11 @@ is more fruitful and more central to progress. +r_2(2a_2+3b_2+c_2-d_2) \\ &=r_1\cdot h(\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} +r_2\cdot h(\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat}) \end{align*} \end{aligned} \end{multline*} \partsitem No. An example of a combination that is not preserved is this. An example of a combination that is not preserved is that doing it one way gives this \begin{equation*} h(\begin{mat}[r] 1 &0 \\ ... ... @@ -592,7 +595,9 @@ is more fruitful and more central to progress. 0 &0 \end{mat}) =4 \quad\text{while}\quad \end{equation*} while the other way gives this. \begin{equation*} h(\begin{mat}[r] 1 &0 \\ 0 &0 ... ... @@ -922,9 +927,10 @@ is more fruitful and more central to progress. \begin{exparts} \partsitem Where $$c$$ and $$d$$ are scalars, we have this. \begin{align*} \begin{multline*} h(c\cdot \colvec{x_1 \\ \vdots \\ x_n} +d\cdot \colvec{y_1 \\ \vdots \\ y_n}) +d\cdot \colvec{y_1 \\ \vdots \\ y_n}) \\ \begin{aligned} &=h(\colvec{cx_1+dy_1 \\ \vdots \\ cx_n+dy_n}) \\ &=\colvec{a_{1,1}(cx_1+dy_1)+\dots+a_{1,n}(cx_n+dy_n) \\ \vdots \\ ... ... @@ -937,13 +943,14 @@ is more fruitful and more central to progress. a_{m,1}y_1+\dots+a_{m,n}y_n} \\ &=c\cdot h(\colvec{x_1 \\ \vdots \\ x_n}) +d\cdot h(\colvec{y_1 \\ \vdots \\ y_n}) \end{align*} \end{aligned} \end{multline*} \partsitem Each power $i$ of the derivative operator is linear because of these rules familiar from calculus. \begin{equation*} \frac{d^i}{dx^i}(\,f(x)+g(x)\,)=\frac{d^i}{dx^i}f(x) +\frac{d^i}{dx^i}g(x) \quad\text{and}\quad \qquad \frac{d^i}{dx^i}\,r\cdot f(x)=r\cdot\frac{d^i}{dx^i}f(x) \end{equation*} Thus the given map is a linear transformation of $$\polyspace_n$$ ... ... @@ -1038,7 +1045,7 @@ is more fruitful and more central to progress. Recall that the entry in row~$i$ and column~$j$ of the transpose of $M$ is the entry $m_{j,i}$ from row~$j$ and column~$i$ of $M$. Now, the check is routine. \begin{align*} \begin{multline*} \trans{[r\cdot\begin{mat} \ &\vdots \\ \cdots &a_{i,j} &\cdots \\ ... ... @@ -1048,7 +1055,8 @@ is more fruitful and more central to progress. \ &\vdots \\ \cdots &b_{i,j} &\cdots \\ &\vdots \end{mat}]} \end{mat}]} \\ \begin{aligned} &=\trans{\begin{mat} \ &\vdots \\ \cdots &ra_{i,j}+sb_{i,j} &\cdots \\ ... ... @@ -1079,7 +1087,8 @@ is more fruitful and more central to progress. \cdots &b_{j,i} &\cdots \\ &\vdots \end{mat} } \end{align*} \end{aligned} \end{multline*} The domain is $$\matspace_{\nbym{m}{n}}$$ while the codomain is $$\matspace_{\nbym{n}{m}}$$. \end{answer} ... ... @@ -1153,8 +1162,8 @@ is more fruitful and more central to progress. If $$h(\vec{v}_1)$$ is the zero vector then this line is degenerate. \partsitem A $$k$$-dimensional linear surface in $$\Re^n$$ maps to a (possibly degenerate) $$k$$-dimensional linear surface in $$\Re^m$$. a $$k$$-dimensional linear surface in $$\Re^m$$ (possibly it is degenerate). The proof is just like that the one for the line. \end{exparts} \end{answer} ... ... @@ -1235,7 +1244,7 @@ is more fruitful and more central to progress. subspace of is a subspace. Suppose that $$X$$ is a subspace of $$W$$. Note that $$\zero_W\in X$$ so the set Note that $$\zero_W\in X$$ so that the set $$\set{\vec{v}\in V \suchthat h(\vec{v})\in X}$$ is not empty. To show that this set is closed under combinations, let $$\vec{v}_1,\dots,\vec{v}_n$$ be elements of $$V$$ ... ... @@ -1303,14 +1312,14 @@ is more fruitful and more central to progress. \begin{answer} \begin{exparts} \partsitem Briefly, the check of linearity is this. \begin{equation*} \begin{multline*} F(r_1\cdot \vec{v}_1+r_2\cdot \vec{v}_2) =\colvec{f_1(r_1\vec{v}_1+r_2\vec{v}_2) \\ f_2(r_1\vec{v}_1+r_2\vec{v}_2)} f_2(r_1\vec{v}_1+r_2\vec{v}_2)} \\ =r_1\colvec{f_1(\vec{v}_1) \\ f_2(\vec{v}_1)} +r_2\colvec{f_1(\vec{v}_2) \\ f_2(\vec{v}_2)} =r_1\cdot F(\vec{v}_1)+r_2\cdot F(\vec{v}_2) \end{equation*} \end{multline*} \partsitem Yes. Let $$\map{\pi_1}{\Re^2}{\Re^1}$$ and $$\map{\pi_2}{\Re^2}{\Re^1}$$ be the projections ... ... @@ -2256,7 +2265,7 @@ point and the map is an isomorphism between the domain and the range. \end{answer} \recommended \item Each of these transformations of $$\polyspace_3$$ is one-to-one. Find the inverse of each. For each, find the inverse. \begin{exparts} \partsitem $$a_0+a_1x+a_2x^2+a_3x^3\mapsto a_0+a_1x+2a_2x^2+3a_3x^3$$ \partsitem $$a_0+a_1x+a_2x^2+a_3x^3\mapsto a_0+a_2x+a_1x^2+a_3x^3$$ ... ... @@ -2551,7 +2560,7 @@ point and the map is an isomorphism between the domain and the range. transformation of $$\polyspace_n$$ for each $$k$$. Prove that this map is a linear transformation of that space is a linear transformation of the space \begin{equation*} f\mapsto \frac{d^k}{dx^k}f+c_{k-1}\frac{d^{k-1}}{dx^{k-1}}f ... ... @@ -2564,7 +2573,7 @@ point and the map is an isomorphism between the domain and the range. \begin{answer} \begin{exparts} \partsitem We will show that the two sets are equal We will show the sets are equal $h^{-1}(\vec{w}) =\set{\vec{v}+\vec{n}\suchthat \vec{n}\in\nullspace{h} }$ by mutual inclusion. ... ... @@ -2735,8 +2744,9 @@ point and the map is an isomorphism between the domain and the range. c_1h_2(\vec{\beta}_1)+\dots+c_nh_2(\vec{\beta}_n) \end{align*} we have \begin{align*} (r_1h_1+r_2h_2)(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) \begin{multline*} (r_1h_1+r_2h_2)(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) \\ \begin{aligned} &= c_1(r_1h_1(\vec{\beta}_1)+r_2h_2(\vec{\beta}_1)) +\dots ... ... @@ -2744,7 +2754,8 @@ point and the map is an isomorphism between the domain and the range. &= r_1(c_1h_1(\vec{\beta}_1)+\dots+c_nh_1(\vec{\beta}_n)) + r_2(c_1h_2(\vec{\beta}_1)+\dots+c_nh_2(\vec{\beta}_n)) \end{align*} \end{aligned} \end{multline*} so $$\Phi(r_1h_1+r_2h_2)=r_1\Phi(h_1)+r_2\Phi(h_2)$$. \end{answer} \item ... ...
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