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Line breaks, Chapter LT

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# File src/section-ILT.xml

 <archetypepart acro="R" part="ltdefn" /></equation>
 </p>

-<p>To establish that $R$ is injective we must begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$ and somehow arrive from this at the conclusion that $\vect{x}=\vect{y}$.  Here we go,
-<!--was a gather-->
+<p>To establish that $R$ is injective we must begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$ and somehow arrive at the conclusion that $\vect{x}=\vect{y}$.  Here we go,
 <alignmath>
-\lt{T}{\vect{x}}=\lt{T}{\vect{y}}\\
-\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}=\lt{T}{\colvector{y_1\\y_2\\y_3\\y_4\\y_5}}\\
+\colvector{0\\0\\0\\0\\0}
+<![CDATA[&=]]>\lt{T}{\vect{x}}-\lt{T}{\vect{y}}\\
+<![CDATA[&=]]>\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}-\lt{T}{\colvector{y_1\\y_2\\y_3\\y_4\\y_5}}\\
+<![CDATA[&=]]>
 \colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\
 36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\
 -44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\
 34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\
-12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5}
-=
+12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5}\\
+<![CDATA[&\quad\quad-]]>
 \colvector{-65 y_1 + 128 y_2 + 10 y_3 - 262 y_4 + 40 y_5\\
 36 y_1 - 73 y_2 - y_3 + 151 y_4 - 16 y_5\\
 -44 y_1 + 88 y_2 + 5 y_3 - 180 y_4 + 24 y_5\\
 34 y_1 - 68 y_2 - 3 y_3 + 140 y_4 - 18 y_5\\
 12 y_1 - 24 y_2 - y_3 + 49 y_4 - 5 y_5}\\
-\colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\
-36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\
--44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\
-34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\
-12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5}
--
-\colvector{-65 y_1 + 128 y_2 + 10 y_3 - 262 y_4 + 40 y_5\\
-36 y_1 - 73 y_2 - y_3 + 151 y_4 - 16 y_5\\
--44 y_1 + 88 y_2 + 5 y_3 - 180 y_4 + 24 y_5\\
-34 y_1 - 68 y_2 - 3 y_3 + 140 y_4 - 18 y_5\\
-12 y_1 - 24 y_2 - y_3 + 49 y_4 - 5 y_5}
-=
-\colvector{0\\0\\0\\0\\0}\\
+<![CDATA[&=]]>
 \colvector{-65 (x_1-y_1) + 128 (x_2-y_2) + 10 (x_3-y_3) - 262 (x_4-y_4) + 40 (x_5-y_5)\\
 36 (x_1-y_1) - 73 (x_2-y_2) - (x_3-y_3) + 151 (x_4-y_4) - 16 (x_5-y_5)\\
 -44 (x_1-y_1) + 88 (x_2-y_2) + 5 (x_3-y_3) - 180 (x_4-y_4) + 24 (x_5-y_5)\\
 34 (x_1-y_1) - 68 (x_2-y_2) - 3 (x_3-y_3) + 140 (x_4-y_4) - 18 (x_5-y_5)\\
-12 (x_1-y_1) - 24 (x_2-y_2) - (x_3-y_3) + 49 (x_4-y_4) - 5 (x_5-y_5)}
-=
-\colvector{0\\0\\0\\0\\0}\\
+12 (x_1-y_1) - 24 (x_2-y_2) - (x_3-y_3) + 49 (x_4-y_4) - 5 (x_5-y_5)}\\
+<![CDATA[&=]]>
 \begin{bmatrix}
 <![CDATA[-65&128&10&-262&40\\]]>
 <![CDATA[36&-73&-1&151&-16\\]]>
 <![CDATA[12&-24&-1&49&-5]]>
 \end{bmatrix}
 \colvector{x_1-y_1\\x_2-y_2\\x_3-y_3\\x_4-y_4\\x_5-y_5}
-=
-\colvector{0\\0\\0\\0\\0}
 </alignmath>
 </p>

 <theorem acro="ILTLI" index="linear independence!injective linear transformation">
 <title>Injective Linear Transformations and Linear Independence</title>
 <statement>
-<p>Suppose that $\ltdefn{T}{U}{V}$ is an injective linear transformation and $S=\set{\vectorlist{u}{t}}$ is a linearly independent subset of $U$.  Then $R=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}}$ is a linearly independent subset of $V$.</p>
-
+<p>Suppose that $\ltdefn{T}{U}{V}$ is an injective linear transformation and
+<alignmath>
+<![CDATA[S&=\set{\vectorlist{u}{t}}]]>
+</alignmath>
+is a linearly independent subset of $U$.  Then
+<alignmath>
+<![CDATA[R&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}}]]>
+</alignmath>
+is a linearly independent subset of $V$.</p>
 </statement>

 <proof>
 <theorem acro="ILTB" index="injective linear transformation!bases">
 <title>Injective Linear Transformations and Bases</title>
 <statement>
-<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and $B=\set{\vectorlist{u}{m}}$ is a basis of $U$.  Then $T$ is injective if and only if $C=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}$ is a linearly independent subset of $V$.</p>
+<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and
+<alignmath>
+<![CDATA[B&=\set{\vectorlist{u}{m}}]]>
+</alignmath>
+is a basis of $U$.  Then $T$ is injective if and only if
+<alignmath>
+<![CDATA[C&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}]]>
+</alignmath>
+is a linearly independent subset of $V$.</p>

 </statement>


# File src/section-IVLT.xml


 <p>Then
 <alignmath>
-<![CDATA[\lt{\left(\compose{T}{S}\right)}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}&=]]>
-<![CDATA[\lt{T}{\lt{S}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}}\\]]>
-<![CDATA[&=\lt{T}{(a - c - d)+ (c + d)x +\frac{1}{2}(a - b - c - d)x^2+cx^3}\\]]>
-<![CDATA[&=\begin{bmatrix}]]>
+<![CDATA[&\lt{\left(\compose{T}{S}\right)}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}\\]]>
+<![CDATA[&\quad\quad=\lt{T}{\lt{S}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}}\\]]>
+<![CDATA[&\quad\quad=\lt{T}{(a - c - d)+ (c + d)x +\frac{1}{2}(a - b - c - d)x^2+cx^3}\\]]>
+<![CDATA[&\quad\quad=\begin{bmatrix}]]>
 <![CDATA[(a - c - d)+ (c + d)&(a - c - d)-2(\frac{1}{2}(a - b - c - d))\\c&(c + d)-c]]>
 \end{bmatrix}\\
-<![CDATA[&=\begin{bmatrix}a&b\\c&d\end{bmatrix}\\]]>
-<![CDATA[&=\lt{I_{M_{22}}}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}]]>
+<![CDATA[&\quad\quad=\begin{bmatrix}a&b\\c&d\end{bmatrix}\\]]>
+<![CDATA[&\quad\quad=\lt{I_{M_{22}}}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}]]>
 <intertext>and</intertext>
-<![CDATA[\lt{\left(\compose{S}{T}\right)}{a+bx+cx^2+dx^3}&=]]>
-\lt{S}{\lt{T}{a+bx+cx^2+dx^3}}\\
-<![CDATA[&=\lt{S}{\begin{bmatrix}]]>
+<![CDATA[&\lt{\left(\compose{S}{T}\right)}{a+bx+cx^2+dx^3}\\]]>
+<![CDATA[&\quad\quad=\lt{S}{\lt{T}{a+bx+cx^2+dx^3}}]]>\\
+<![CDATA[&\quad\quad=\lt{S}{\begin{bmatrix}]]>
 <![CDATA[a+b&a-2c\\d&b-d]]>
 \end{bmatrix}}\\
-<![CDATA[&=((a+b)-d-(b-d))+]]>
+<![CDATA[&\quad\quad=((a+b)-d-(b-d))+]]>
 (d+(b-d))x\\
 <![CDATA[&\quad\quad+\left(\frac{1}{2}((a+b)-(a-2c)-d-(b-d))\right)x^2+]]>
 (d)x^3\\
-<![CDATA[&=a+bx+cx^2+dx^3\\]]>
-<![CDATA[&=\lt{I_{P_3}}{a+bx+cx^2+dx^3}]]>
+<![CDATA[&\quad\quad=a+bx+cx^2+dx^3\\]]>
+<![CDATA[&\quad\quad=\lt{I_{P_3}}{a+bx+cx^2+dx^3}]]>
 </alignmath>
 </p>

 </alignmath>
 </p>

-<p>This says that for any vector, $\vect{u}$, from $U$, there exist scalars ($\scalarlist{d}{s},\,\scalarlist{c}{r}$) that form $\vect{u}$ as a linear combination of the vectors in the set $B$.  In other words, $B$ spans $U$ (<acroref type="definition" acro="SS" />).</p>
+<p>This says that for any vector, $\vect{u}$, from $U$, there exist scalars ($\scalarlist{d}{s}$, $\scalarlist{c}{r}$) that form $\vect{u}$ as a linear combination of the vectors in the set $B$.  In other words, $B$ spans $U$ (<acroref type="definition" acro="SS" />).</p>

 <p>So $B$ is a basis (<acroref type="definition" acro="B" />) of $U$ with $s+r$ vectors, and thus
 <equation>

# File src/section-LT.xml


 <p>Then
 <alignmath>
-<![CDATA[\lt{S}{p(x)+q(x)}&=(x-2)(p(x)+q(x))=(x-2)p(x)+(x-2)q(x)=\lt{S}{p(x)}+\lt{S}{q(x)}\\]]>
+<![CDATA[\lt{S}{p(x)+q(x)}&=(x-2)(p(x)+q(x))\\]]>
+<![CDATA[&=(x-2)p(x)+(x-2)q(x)=\lt{S}{p(x)}+\lt{S}{q(x)}\\]]>
 <![CDATA[\lt{S}{\alpha p(x)}&=(x-2)(\alpha p(x))=(x-2)\alpha p(x)=\alpha(x-2)p(x)=\alpha\lt{S}{p(x)}]]>
 </alignmath>
 </p>
 </theorem>

 <p>Return to <acroref type="example" acro="NLT" /> and compute $\lt{S}{\colvector{0\\0\\0}}=\colvector{0\\0\\-2}$ to quickly see again that $S$ is not a linear transformation, while in <acroref type="example" acro="LTPM" />  compute
-<![CDATA[$\lt{S}{0+0x+0x^2+0x^3}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$]]>
+<alignmath>
+<![CDATA[\lt{S}{0+0x+0x^2+0x^3}&=\begin{bmatrix}0&0\\0&0\end{bmatrix}]]>
+</alignmath>
 as an example of <acroref type="theorem" acro="LTTZZ" /> at work.</p>

 <sageadvice acro="LTS" index="linear transformation!symbolic">
 <theorem acro="LTDB" index="linear transformation!defined on a basis">
 <title>Linear Transformation Defined on a Basis</title>
 <statement>
-<p>Suppose $B=\set{\vectorlist{u}{n}}$ is a basis for the vector space $U$ and $\vectorlist{v}{n}$ is a list of vectors from the vector space $V$ (which are not necessarily distinct).   Then there is a unique linear transformation, $\ltdefn{T}{U}{V}$, such that $\lt{T}{\vect{u}_i}=\vect{v}_i$, $1\leq i\leq n$.</p>
+<p>Suppose $U$ is a vector space with basis $B=\set{\vectorlist{u}{n}}$ and the vector space $V$ contains the vectors $\vectorlist{v}{n}$ (which may not be distinct).   Then there is a unique linear transformation, $\ltdefn{T}{U}{V}$, such that $\lt{T}{\vect{u}_i}=\vect{v}_i$, $1\leq i\leq n$.</p>

 </statement>

 <![CDATA[\lt{T}{\vect{w}+\vect{x}}&=]]>
 \lt{T}{
 a_1\vect{u}_1+
-a_2\vect{u}_2+
 \cdots+
 a_n\vect{u}_n+
 b_1\vect{u}_1+
-b_2\vect{u}_2+
 \cdots+
 b_n\vect{u}_n
 }\\
 <![CDATA[&=]]>
 \lt{T}{
 \left(a_1+b_1\right)\vect{u}_1+
-\left(a_2+b_2\right)\vect{u}_2+
 \cdots+
 \left(a_n+b_n\right)\vect{u}_n
 }
 <![CDATA[&&]]>\text{<acroref type="definition" acro="VS" />}\\
 <![CDATA[&=]]>
 \left(a_1+b_1\right)\vect{v}_1+
-\left(a_2+b_2\right)\vect{v}_2+
 \cdots+
 \left(a_n+b_n\right)\vect{v}_n
 <![CDATA[&&\text{Definition of $T$}\\]]>
 <![CDATA[&=]]>
 a_1\vect{v}_1+
-a_2\vect{v}_2+
 \cdots+
 a_n\vect{v}_n+
 b_1\vect{v}_1+
-b_2\vect{v}_2+
 \cdots+
 b_n\vect{v}_n
 <![CDATA[&&]]>\text{<acroref type="definition" acro="VS" />}\\
 </p>

 <p>Then by <acroref type="definition" acro="LTA" />, we have
-<equation>
-\lt{(T+S)}{\colvector{x_1\\x_2}}
-=
-\lt{T}{\colvector{x_1\\x_2}}+\lt{S}{\colvector{x_1\\x_2}}
-=
-\colvector{x_1+2x_2\\3x_1-4x_2\\5x_1+2x_2}+
-\colvector{4x_1-x_2\\x_1+3x_2\\-7x_1+5x_2}
-=
-\colvector{5x_1+x_2\\4x_1-x_2\\-2x_1+7x_2}
-</equation>
+<alignmath>
+<![CDATA[\lt{(T+S)}{\colvector{x_1\\x_2}}&=\lt{T}{\colvector{x_1\\x_2}}+\lt{S}{\colvector{x_1\\x_2}}\\]]>
+<![CDATA[&=\colvector{x_1+2x_2\\3x_1-4x_2\\5x_1+2x_2}+\colvector{4x_1-x_2\\x_1+3x_2\\-7x_1+5x_2}]]>
+=\colvector{5x_1+x_2\\4x_1-x_2\\-2x_1+7x_2}
+</alignmath>
 and by <acroref type="theorem" acro="SLTLT" /> we know $T+S$ is also a linear transformation from $\complex{2}$ to $\complex{3}$.
 </p>

 </p>

 <p>For the sake of an example, choose $\alpha=2$, so by <acroref type="definition" acro="LTSM" />, we have
-<equation>
-\lt{\alpha T}{\colvector{x_1\\x_2\\x_3\\x_4}}
-=
-2\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4}}
-=
-2\colvector{x_1+2x_2-x_3+2x_4\\x_1+5x_2-3x_3+x_4\\-2x_1+3x_2-4x_3+2x_4}
-=
-\colvector{2x_1+4x_2-2x_3+4x_4\\2x_1+10x_2-6x_3+2x_4\\-4x_1+6x_2-8x_3+4x_4}
-</equation>
+<alignmath>
+<![CDATA[\lt{\alpha T}{\colvector{x_1\\x_2\\x_3\\x_4}}]]>
+<![CDATA[&=2\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4}}]]>
+<![CDATA[ =2\colvector{x_1+2x_2-x_3+2x_4\\x_1+5x_2-3x_3+x_4\\-2x_1+3x_2-4x_3+2x_4}\\]]>
+<![CDATA[&=\colvector{2x_1+4x_2-2x_3+4x_4\\2x_1+10x_2-6x_3+2x_4\\-4x_1+6x_2-8x_3+4x_4}]]>
+</alignmath>
 and by <acroref type="theorem" acro="MLTLT" /> we know $2T$ is also a linear transformation from $\complex{4}$ to $\complex{3}$.</p>

 </example>
 <p>Then by <acroref type="definition" acro="LTC" />
 <alignmath>
 <![CDATA[\lt{(\compose{S}{T})}{\colvector{x_1\\x_2}}&=]]>
-\lt{S}{\lt{T}{\colvector{x_1\\x_2}}}\\
-<![CDATA[&=\lt{S}{\colvector{x_1+2x_2\\3x_1-4x_2\\5x_1+2x_2\\6x_1-3x_2}}\\]]>
+\lt{S}{\lt{T}{\colvector{x_1\\x_2}}}
+<![CDATA[=\lt{S}{\colvector{x_1+2x_2\\3x_1-4x_2\\5x_1+2x_2\\6x_1-3x_2}}\\]]>
 <![CDATA[&=\colvector{]]>
 2(x_1+2x_2)-(3x_1-4x_2)+(5x_1+2x_2)-(6x_1-3x_2)\\
 5(x_1+2x_2)-3(3x_1-4x_2)+8(5x_1+2x_2)-2(6x_1-3x_2)\\
 </exercise>

 <exercise type="C" number="16" rough="find matrix rep R3 --> R4, given formula">
-<problem contributor="chrisblack">Find the matrix representation of $\ltdefn{T}{\complex{3}}{\complex{4}}$ given by
+<problem contributor="chrisblack">Find the matrix representation of $\ltdefn{T}{\complex{3}}{\complex{4}}$,
 $\lt{T}{\colvector{x\\y\\z}} = \colvector{3x + 2y + z\\ x + y + z \\ x - 3y \\2x + 3y + z }$.
 </problem>
 <solution contributor="chrisblack"><![CDATA[Answer: $A_T = \begin{bmatrix} 3 & 2 & 1\\ 1 & 1 & 1\\ 1 & -3 & 0\\ 2 & 3 & 1\end{bmatrix}$.]]>

# File src/section-SLT.xml

 \rng{T}=\spn{<archetypepart acro="Q" part="ltrangebasis" />}</equation>
 </p>

-<p>Suppose an element of the range $\vect{v^*}$ has its first 4 components equal to $-1, 2, 3, -1$, in that order.  Then to be an element of $\rng{T}$, we would have
+<p>Suppose an element of the range $\vect{v^*}$ has its first 4 components equal to $-1$, $2$, $3$, $-1$, in that order.  Then to be an element of $\rng{T}$, we would have
 <equation>
 \vect{v^*}=(-1)\colvector{1\\0\\0\\0\\1}+(2)\colvector{0\\1\\0\\0\\-1}+(3)
 \colvector{0\\0\\1\\0\\-1}+(-1)\colvector{0\\0\\0\\1\\2}
 <theorem acro="SSRLT" index="linear transformation!spanning range">
 <title>Spanning Set for Range of a Linear Transformation</title>
 <statement>
-<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and $S=\set{\vectorlist{u}{t}}$ spans $U$.  Then
+<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and
 <alignmath>
-R=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}}
+<![CDATA[S&=\set{\vectorlist{u}{t}}]]>
+</alignmath>
+spans $U$.  Then
+<alignmath>
+<![CDATA[R&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}}]]>
 </alignmath>
 spans $\rng{T}$.</p>

 <theorem acro="SLTB" index="surjective linear transformation!bases">
 <title>Surjective Linear Transformations and Bases</title>
 <statement>
-<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and $B=\set{\vectorlist{u}{m}}$ is a basis of $U$.  Then $T$ is surjective if and only if $C=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}$ is a spanning set for $V$.</p>
+<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and
+<alignmath>
+<![CDATA[B&=\set{\vectorlist{u}{m}}]]>
+</alignmath>
+is a basis of $U$.  Then $T$ is surjective if and only if
+<alignmath>
+<![CDATA[C&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}]]>
+</alignmath>
+is a spanning set for $V$.</p>
 </statement>

 <proof>