Commit 633557a5 authored by Jim Hefferon's avatar Jim Hefferon

map2 answers

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}
No preview for this file type
......@@ -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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