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Line breaks, Chapter R

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File src/section-CB.xml

 
 <p>The vector space $P_4$ (<acroref type="example" acro="VSP" />) has two nice bases (<acroref type="example" acro="BP" />),
 <alignmath>
-<![CDATA[B&=\set{1,x,x^2,x^3,x^4}]]>
-<![CDATA[&]]>
+<![CDATA[B&=\set{1,x,x^2,x^3,x^4}\\]]>
 <![CDATA[C&=\set{1,1+x,1+x+x^2,1+x+x^2+x^3,1+x+x^2+x^3+x^4}]]>
 </alignmath>
 </p>
 <p>The change-of-basis matrix from $C$ to $B$ is actually easier to build.  Grab each vector in the basis $C$ and form its representation relative to $B$
 <alignmath>
 \vectrep{B}{1}
-=\vectrep{B}{(1)1}
-<![CDATA[&=\colvector{1\\0\\0\\0\\0}\\]]>
+<![CDATA[&=\vectrep{B}{(1)1}]]>
+=\colvector{1\\0\\0\\0\\0}\\
 \vectrep{B}{1+x}
-=\vectrep{B}{(1)1+(1)x}
-<![CDATA[&=\colvector{1\\1\\0\\0\\0}\\]]>
+<![CDATA[&=\vectrep{B}{(1)1+(1)x}]]>
+=\colvector{1\\1\\0\\0\\0}\\
 \vectrep{B}{1+x+x^2}
-=\vectrep{B}{(1)1+(1)x+(1)x^2}
-<![CDATA[&=\colvector{1\\1\\1\\0\\0}\\]]>
+<![CDATA[&=\vectrep{B}{(1)1+(1)x+(1)x^2}]]>
+=\colvector{1\\1\\1\\0\\0}\\
 \vectrep{B}{1+x+x^2+x^3}
-=\vectrep{B}{(1)1+(1)x+(1)x^2+(1)x^3}
-<![CDATA[&=\colvector{1\\1\\1\\1\\0}\\]]>
+<![CDATA[&=\vectrep{B}{(1)1+(1)x+(1)x^2+(1)x^3}]]>
+=\colvector{1\\1\\1\\1\\0}\\
 \vectrep{B}{1+x+x^2+x^3+x^4}
-=\vectrep{B}{(1)1+(1)x+(1)x^2+(1)x^3+(1)x^4}
-<![CDATA[&=\colvector{1\\1\\1\\1\\1}\\]]>
+<![CDATA[&=\vectrep{B}{(1)1+(1)x+(1)x^2+(1)x^3+(1)x^4}]]>
+=\colvector{1\\1\\1\\1\\1}\\
 </alignmath>
 </p>
 
 -1-2x^2+3x^3,\,
 -3-x+x^3,\,
 -x^2+x^3
-}
-<![CDATA[&]]>
+}\\
 <![CDATA[E&=\set{1,\,x,\,x^2,\,x^3}]]>
 </alignmath>
 </p>
 
 <p>Now for the matrix representation relative to $C$ first compute,
 <alignmath>
-\vectrep{C}{\lt{T}{\vect{y}_1}}
-<![CDATA[&=\vectrep{C}{\begin{bmatrix}-17&-57\\-14&-41\end{bmatrix}}\\]]>
+<![CDATA[&\vectrep{C}{\lt{T}{\vect{y}_1}}]]>
+<![CDATA[=\vectrep{C}{\begin{bmatrix}-17&-57\\-14&-41\end{bmatrix}}\\]]>
 <![CDATA[&=\vectrep{C}{]]>
 <![CDATA[(-17)\begin{bmatrix}1&0\\0&0\end{bmatrix}+]]>
 <![CDATA[(-57)\begin{bmatrix}0&1\\0&0\end{bmatrix}+]]>
 <![CDATA[(-41)\begin{bmatrix}0&0\\0&1\end{bmatrix}]]>
 }
 =\colvector{-17\\-57\\-14\\-41}\\
-\vectrep{C}{\lt{T}{\vect{y}_2}}
-<![CDATA[&=\vectrep{C}{\begin{bmatrix}11&35\\10&25\end{bmatrix}}\\]]>
+<![CDATA[&\vectrep{C}{\lt{T}{\vect{y}_2}}]]>
+<![CDATA[=\vectrep{C}{\begin{bmatrix}11&35\\10&25\end{bmatrix}}\\]]>
 <![CDATA[&=\vectrep{C}{]]>
 <![CDATA[11\begin{bmatrix}1&0\\0&0\end{bmatrix}+]]>
 <![CDATA[35\begin{bmatrix}0&1\\0&0\end{bmatrix}+]]>
 <![CDATA[25\begin{bmatrix}0&0\\0&1\end{bmatrix}]]>
 }
 =\colvector{11\\35\\10\\25}\\
-\vectrep{C}{\lt{T}{\vect{y}_3}}
-<![CDATA[&=\vectrep{C}{\begin{bmatrix}8&24\\6&16\end{bmatrix}}\\]]>
+<![CDATA[&\vectrep{C}{\lt{T}{\vect{y}_3}}]]>
+<![CDATA[=\vectrep{C}{\begin{bmatrix}8&24\\6&16\end{bmatrix}}\\]]>
 <![CDATA[&=\vectrep{C}{]]>
 <![CDATA[8\begin{bmatrix}1&0\\0&0\end{bmatrix}+]]>
 <![CDATA[24\begin{bmatrix}0&1\\0&0\end{bmatrix}+]]>
 <![CDATA[16\begin{bmatrix}0&0\\0&1\end{bmatrix}]]>
 }
 =\colvector{8\\24\\6\\16}\\
-\vectrep{C}{\lt{T}{\vect{y}_4}}
-<![CDATA[&=\vectrep{C}{\begin{bmatrix}-11&-33\\-10&-23\end{bmatrix}}\\]]>
+<![CDATA[&\vectrep{C}{\lt{T}{\vect{y}_4}}]]>
+<![CDATA[=\vectrep{C}{\begin{bmatrix}-11&-33\\-10&-23\end{bmatrix}}\\]]>
 <![CDATA[&=\vectrep{C}{]]>
 <![CDATA[(-11)\begin{bmatrix}1&0\\0&0\end{bmatrix}+]]>
 <![CDATA[(-33)\begin{bmatrix}0&1\\0&0\end{bmatrix}+]]>
 <p>Then to build the matrix representation of $S$ relative to $B$ compute,
 <alignmath>
 <![CDATA[\vectrep{B}{\lt{S}{\vect{x}_1}}&=]]>
-<![CDATA[\vectrep{B}{\begin{bmatrix}0 & -14 \\ 18 & -6\end{bmatrix}}=]]>
-\vectrep{B}{0\vect{x}_1+(-14)\vect{x}_2+18\vect{x}_3+(-6)\vect{x}_4}=
+<![CDATA[\vectrep{B}{\begin{bmatrix}0 & -14 \\ 18 & -6\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{B}{0\vect{x}_1+(-14)\vect{x}_2+18\vect{x}_3+(-6)\vect{x}_4}=]]>
 \colvector{0\\-14\\18\\-6}\\
 <![CDATA[\vectrep{B}{\lt{S}{\vect{x}_2}}&=]]>
-<![CDATA[\vectrep{B}{\begin{bmatrix}-1 & -15\\21 & -7\end{bmatrix}}=]]>
-\vectrep{B}{(-1)\vect{x}_1+(-15)\vect{x}_2+21\vect{x}_3+(-7)\vect{x}_4}=
+<![CDATA[\vectrep{B}{\begin{bmatrix}-1 & -15\\21 & -7\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{B}{(-1)\vect{x}_1+(-15)\vect{x}_2+21\vect{x}_3+(-7)\vect{x}_4}=]]>
 \colvector{-1\\-15\\21\\-7}\\
 <![CDATA[\vectrep{B}{\lt{S}{\vect{x}_3}}&=]]>
-<![CDATA[\vectrep{B}{\begin{bmatrix}-1 & -13\\19 & -7\end{bmatrix}}=]]>
-\vectrep{B}{(-1)\vect{x}_1+(-13)\vect{x}_2+19\vect{x}_3+(-7)\vect{x}_4}=
+<![CDATA[\vectrep{B}{\begin{bmatrix}-1 & -13\\19 & -7\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{B}{(-1)\vect{x}_1+(-13)\vect{x}_2+19\vect{x}_3+(-7)\vect{x}_4}=]]>
 \colvector{-1\\-13\\19\\-7}\\
 <![CDATA[\vectrep{B}{\lt{S}{\vect{x}_4}}&=]]>
-<![CDATA[\vectrep{B}{\begin{bmatrix}-3 & 1\\3 & -3\end{bmatrix}}=]]>
-\vectrep{B}{(-3)\vect{x}_1+1\vect{x}_2+3\vect{x}_3+(-3)\vect{x}_4}=
+<![CDATA[\vectrep{B}{\begin{bmatrix}-3 & 1\\3 & -3\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{B}{(-3)\vect{x}_1+1\vect{x}_2+3\vect{x}_3+(-3)\vect{x}_4}=]]>
 \colvector{-3\\1\\3\\-3}
 </alignmath>
 </p>
 <p>Then to build the matrix representation of $S$ relative to $D$ compute,
 <alignmath>
 <![CDATA[\vectrep{D}{\lt{S}{\vect{y}_1}}&=]]>
-<![CDATA[\vectrep{D}{\begin{bmatrix}0 & -14\\18 & -6\end{bmatrix}}=]]>
-\vectrep{D}{14\vect{y}_1+(-32)\vect{y}_2+24\vect{y}_3+(-6)\vect{y}_4}=
+<![CDATA[\vectrep{D}{\begin{bmatrix}0 & -14\\18 & -6\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{D}{14\vect{y}_1+(-32)\vect{y}_2+24\vect{y}_3+(-6)\vect{y}_4}=]]>
 \colvector{14\\-32\\24\\-6}\\
 <![CDATA[\vectrep{D}{\lt{S}{\vect{y}_2}}&=]]>
-<![CDATA[\vectrep{D}{\begin{bmatrix}-1 & -29 \\ 39 & -13\end{bmatrix}}=]]>
-\vectrep{D}{28\vect{y}_1+(-68)\vect{y}_2+52\vect{y}_3+(-13)\vect{y}_4}=
+<![CDATA[\vectrep{D}{\begin{bmatrix}-1 & -29 \\ 39 & -13\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{D}{28\vect{y}_1+(-68)\vect{y}_2+52\vect{y}_3+(-13)\vect{y}_4}=]]>
 \colvector{28\\-68\\52\\-13}\\
 <![CDATA[\vectrep{D}{\lt{S}{\vect{y}_3}}&=]]>
-<![CDATA[\vectrep{D}{\begin{bmatrix}-2 & -42 \\ 58 & -20\end{bmatrix}}=]]>
-\vectrep{D}{40\vect{y}_1+(-100)\vect{y}_2+78\vect{y}_3+(-20)\vect{y}_4}=
+<![CDATA[\vectrep{D}{\begin{bmatrix}-2 & -42 \\ 58 & -20\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{D}{40\vect{y}_1+(-100)\vect{y}_2+78\vect{y}_3+(-20)\vect{y}_4}=]]>
 \colvector{40\\-100\\78\\-20}\\
 <![CDATA[\vectrep{D}{\lt{S}{\vect{y}_4}}&=]]>
-<![CDATA[\vectrep{D}{\begin{bmatrix}-5 & -41 \\ 61 & -23\end{bmatrix}}=]]>
-\vectrep{D}{36\vect{y}_1+(-102)\vect{y}_2+84\vect{y}_3+(-23)\vect{y}_4}=
+<![CDATA[\vectrep{D}{\begin{bmatrix}-5 & -41 \\ 61 & -23\end{bmatrix}}\\]]>
+<![CDATA[&=\vectrep{D}{36\vect{y}_1+(-102)\vect{y}_2+84\vect{y}_3+(-23)\vect{y}_4}=]]>
 \colvector{36\\-102\\84\\-23}\\
 </alignmath>
 </p>
 <![CDATA[ -69 & -29 & 18-2 i & 0 & -7 \\]]>
 <![CDATA[ 159 & 73 & -44 & -16-2 i & 2 \\]]>
 <![CDATA[ -195 & -87 & 55 & 10 & -16-2 i]]>
-\end{bmatrix}
-\rref
+\end{bmatrix}\\
+<![CDATA[&\quad\quad\rref]]>
 \begin{bmatrix}
 <![CDATA[ 1 & 0 & 0 & 0 &  -\frac{3}{4}+\frac{i}{4} \\]]>
 <![CDATA[ 0 & 1 & 0 & 0 &  \frac{7}{4}-\frac{i}{4} \\]]>
 <![CDATA[ -69 & -29 & 18+2 i & 0 & -7 \\]]>
 <![CDATA[ 159 & 73 & -44 & -16+2 i & 2 \\]]>
 <![CDATA[ -195 & -87 & 55 & 10 & -16+2 i]]>
-\end{bmatrix}
-\rref
+\end{bmatrix}\\
+<![CDATA[&\quad\quad\rref]]>
 \begin{bmatrix}
 <![CDATA[ 1 & 0 & 0 & 0 &  -\frac{3}{4}-\frac{i}{4} \\]]>
 <![CDATA[ 0 & 1 & 0 & 0 &  \frac{7}{4}+\frac{i}{4} \\]]>

File src/section-MR.xml

 <example acro="OLTTR" index="matrix representations">
 <title>One linear transformation, three representations</title>
 
-<p>Consider the linear transformation
-<equation>
-\ltdefn{S}{P_3}{M_{22}},\quad \lt{S}{a+bx+cx^2+dx^3}=
+<p>Consider the linear transformation, $\ltdefn{S}{P_3}{M_{22}}$, given by
+<alignmath>
+\lt{S}{a+bx+cx^2+dx^3}=
 \begin{bmatrix}
 <![CDATA[3a+7b-2c-5d & 8a+14b-2c-11d\\]]>
 <![CDATA[-4a-8b+2c+6d & 12a+22b-4c-17d]]>
 \end{bmatrix}
-</equation>
+</alignmath>
 </p>
 
 <p>First, we build a representation relative to the bases,
 <alignmath>
 <![CDATA[&\vectrep{C}{\lt{S}{1+2x+x^2-x^3}}]]>
 <![CDATA[=\vectrep{C}{\begin{bmatrix}20 & 45 \\ -24 & 69\end{bmatrix}}\\]]>
-<![CDATA[&\quad\quad=\vectrep{C}{]]>
+<![CDATA[&\quad=\vectrep{C}{]]>
 <![CDATA[(-90)\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[37\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+]]>
 <![CDATA[(-40)\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+]]>
 =\colvector{-90\\37\\-40\\4}\\
 <![CDATA[&\vectrep{C}{\lt{S}{1+3x+x^2+x^3}}]]>
 <![CDATA[=\vectrep{C}{\begin{bmatrix}17 & 37 \\ -20 & 57\end{bmatrix}}\\]]>
-<![CDATA[&\quad\quad=\vectrep{C}{]]>
+<![CDATA[&\quad=\vectrep{C}{]]>
 <![CDATA[(-72)\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[29\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+]]>
 <![CDATA[(-34)\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+]]>
 =\colvector{-72\\29\\-34\\3}\\
 <![CDATA[&\vectrep{C}{\lt{S}{-1-2x+2x^3}}]]>
 <![CDATA[=\vectrep{C}{\begin{bmatrix}-27 & -58 \\ 32 & -90\end{bmatrix}}\\]]>
-<![CDATA[&\quad\quad=\vectrep{C}{]]>
+<![CDATA[&\quad=\vectrep{C}{]]>
 <![CDATA[114\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[(-46)\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+]]>
 <![CDATA[54\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+]]>
 =\colvector{114\\-46\\54\\-5}\\
 <![CDATA[&\vectrep{C}{\lt{S}{2+3x+2x^2-5x^3}}]]>
 <![CDATA[=\vectrep{C}{\begin{bmatrix}48 & 109 \\ -58 & 167\end{bmatrix}}\\]]>
-<![CDATA[&\quad\quad=\vectrep{C}{]]>
-<![CDATA[(-220)\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+]]>
+<![CDATA[&\quad=\vectrep{C}{]]>
+<![CDATA[-220\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[91\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+]]>
-<![CDATA[-96\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+]]>
+<![CDATA[(-96)\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+]]>
 <![CDATA[10\begin{bmatrix}-1 & -4 \\ -2 & -4\end{bmatrix}]]>
 }
 =\colvector{-220\\91\\-96\\10}\\
 <![CDATA[(-11)\begin{bmatrix}0&1\\0&0\end{bmatrix}+]]>
 <![CDATA[6\begin{bmatrix}0&0\\1&0\end{bmatrix}+]]>
 <![CDATA[(-17)\begin{bmatrix}0&0\\0&1\end{bmatrix}]]>
-}
-=\colvector{-5 \\ -11 \\ 6 \\ -17}\\
+}\\
+<![CDATA[&=\colvector{-5 \\ -11 \\ 6 \\ -17}]]>
 </alignmath>
 </p>
 
 \lt{S}{p(x)}
 <![CDATA[&=\vectrepinv{G}{\matrixrep{S}{F}{G}\vectrep{F}{p(x)}}\\]]>
 <![CDATA[&=\vectrepinv{G}{\matrixrep{S}{F}{G}\vectrep{F}{3-x+2x^2-5x^3}}\\]]>
-<![CDATA[&=\vectrepinv{G}{\matrixrep{S}{F}{G}\vectrep{F}{]]>
+<![CDATA[&=\vectrepinvname{G}\left(\matrixrep{S}{F}{G}\vectrepname{F}\left(]]>
 32(1+x-x^2+2x^3)
--7(-1+2x+2x^3)
--17(2+x-2x^2+3x^3)
+-7(-1+2x+2x^3)\right.\right.\\
+<![CDATA[&\left.\left.\quad\quad\quad\quad\quad\quad\quad -17(2+x-2x^2+3x^3)]]>
 -2(1+x+2x^3)
-}}\\
+\right)\right)\\
 <![CDATA[&=\vectrepinv{G}{\matrixrep{S}{F}{G}\colvector{32\\-7\\-17\\-2}}\\]]>
 <![CDATA[&=\vectrepinv{G}{]]>
 \begin{bmatrix}
 
 <p>Begin by computing the new linear transformation that is the composition of $T$ and $S$ (<acroref type="definition" acro="LTC" />, <acroref type="theorem" acro="CLTLT" />),  $\ltdefn{\left(\compose{S}{T}\right)}{\complex{2}}{M_{22}}$,
 <alignmath>
-\lt{\left(\compose{S}{T}\right)}{\colvector{a\\b}}
-<![CDATA[&=\lt{S}{\lt{T}{\colvector{a\\b}}}\\]]>
-<![CDATA[&=\lt{S}{(-a + 3b)+(2a + 4b)x+(a - 2b)x^2}\\]]>
-<![CDATA[&=]]>
+<![CDATA[&\lt{\left(\compose{S}{T}\right)}{\colvector{a\\b}}=\lt{S}{\lt{T}{\colvector{a\\b}}}\\]]>
+<![CDATA[&\quad\quad=\lt{S}{(-a + 3b)+(2a + 4b)x+(a - 2b)x^2}\\]]>
+<![CDATA[&\quad\quad=]]>
 \begin{bmatrix}
 <![CDATA[2(-a + 3b) + (2a + 4b) + 2(a - 2b) & (-a + 3b) + 4(2a + 4b) - (a - 2b)\\]]>
 <![CDATA[-(-a + 3b) + 3(a - 2b) & 3(-a + 3b) + (2a + 4b) + 2(a - 2b)]]>
 \end{bmatrix}\\
-<![CDATA[&=]]>
+<![CDATA[&\quad\quad=]]>
 \begin{bmatrix}
 <![CDATA[2a + 6b & 6a + 21b\\]]>
 <![CDATA[4a - 9b &a + 9b]]>
 <p>Now compute the matrix representations (<acroref type="definition" acro="MR" />) for each of these three linear transformations ($T$, $S$, $\compose{S}{T}$), relative to the appropriate bases.  First for $T$,
 <alignmath>
 <![CDATA[\vectrep{C}{\lt{T}{\colvector{3\\1}}}&=\vectrep{C}{10x+x^2}\\]]>
-<![CDATA[&=\vectrep{C}{28(1-2x+x^2)+28(-1+3x)+(-9)(2x+3x^2)}=\colvector{28\\28\\-9}\\]]>
+<![CDATA[&=\vectrep{C}{28(1-2x+x^2)+28(-1+3x)+(-9)(2x+3x^2)}\\]]>
+<![CDATA[&=\colvector{28\\28\\-9}\\]]>
 <![CDATA[\vectrep{C}{\lt{T}{\colvector{2\\1}}}&=\vectrep{C}{1+8x}\\]]>
-<![CDATA[&=\vectrep{C}{33(1-2x+x^2)+32(-1+3x)+(-11)(2x+3x^2)}=\colvector{33\\32\\-11}\\]]>
+<![CDATA[&=\vectrep{C}{33(1-2x+x^2)+32(-1+3x)+(-11)(2x+3x^2)}\\]]>
+<![CDATA[&=\colvector{33\\32\\-11}\\]]>
 </alignmath>
 </p>
 
 
 <p>Now, a representation of $S$,
 <alignmath>
-<![CDATA[\vectrep{D}{\lt{S}{1-2x+x^2}}&=\vectrep{D}{\begin{bmatrix}2 & -8 \\ 2 & 3 \end{bmatrix}}\\]]>
-<![CDATA[&=\vectrep{D}{]]>
+<![CDATA[&\vectrep{D}{\lt{S}{1-2x+x^2}}=\vectrep{D}{\begin{bmatrix}2 & -8 \\ 2 & 3 \end{bmatrix}}\\]]>
+<![CDATA[&\quad\quad=\vectrep{D}{]]>
 <![CDATA[(-11)\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+]]>
 <![CDATA[(-21)\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[0\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+]]>
 <![CDATA[(17)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix}]]>
 }\\
-<![CDATA[&=\colvector{-11\\-21\\0\\17}\\]]>
-<![CDATA[\vectrep{D}{\lt{S}{-1+3x}}&=\vectrep{D}{\begin{bmatrix}1 & 11 \\ 1 & 0\end{bmatrix}}\\]]>
-<![CDATA[&=\vectrep{D}{]]>
+<![CDATA[&\quad\quad=\colvector{-11\\-21\\0\\17}\\]]>
+<![CDATA[&\vectrep{D}{\lt{S}{-1+3x}}=\vectrep{D}{\begin{bmatrix}1 & 11 \\ 1 & 0\end{bmatrix}}\\]]>
+<![CDATA[&\quad\quad=\vectrep{D}{]]>
 <![CDATA[26\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+]]>
 <![CDATA[51\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[0\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+]]>
 <![CDATA[(-38)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix}]]>
 }\\
-<![CDATA[&=\colvector{26\\51\\0\\-38}\\]]>
-<![CDATA[\vectrep{D}{\lt{S}{2x+3x^2}}&=\vectrep{D}{\begin{bmatrix}8 & 5 \\ 9 & 8\end{bmatrix}}\\]]>
-<![CDATA[&=\vectrep{D}{]]>
+<![CDATA[&\quad\quad=\colvector{26\\51\\0\\-38}\\]]>
+<![CDATA[&\vectrep{D}{\lt{S}{2x+3x^2}}=\vectrep{D}{\begin{bmatrix}8 & 5 \\ 9 & 8\end{bmatrix}}\\]]>
+<![CDATA[&\quad\quad=\vectrep{D}{]]>
 <![CDATA[34\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+]]>
 <![CDATA[67\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[1\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+]]>
 <![CDATA[(-46)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix}]]>
 }\\
-<![CDATA[&=\colvector{34\\67\\1\\-46}\\]]>
+<![CDATA[&\quad\quad=\colvector{34\\67\\1\\-46}\\]]>
 </alignmath>
 </p>
 
 
 <p>Finally, a representation of $\compose{S}{T}$,
 <alignmath>
-\vectrep{D}{\lt{\left(\compose{S}{T}\right)}{\colvector{3\\1}}}
-<![CDATA[&=\vectrep{D}{\begin{bmatrix}12 & 39\\3 &12\end{bmatrix}}\\]]>
-<![CDATA[&=\vectrep{D}{]]>
+<![CDATA[&\vectrep{D}{\lt{\left(\compose{S}{T}\right)}{\colvector{3\\1}}}=\vectrep{D}{\begin{bmatrix}12 & 39\\3 &12\end{bmatrix}}\\]]>
+<![CDATA[&\quad\quad=\vectrep{D}{]]>
 <![CDATA[114\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+]]>
 <![CDATA[237\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[(-9)\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+]]>
 <![CDATA[(-174)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix}]]>
 }\\
-<![CDATA[&=\colvector{114\\237\\-9\\-174}\\]]>
-\vectrep{D}{\lt{\left(\compose{S}{T}\right)}{\colvector{2\\1}}}
-<![CDATA[&=\vectrep{D}{\begin{bmatrix}10 & 33\\-1 & 11\end{bmatrix}}\\]]>
-<![CDATA[&=\vectrep{D}{]]>
+<![CDATA[&\quad\quad=\colvector{114\\237\\-9\\-174}\\]]>
+<![CDATA[&\vectrep{D}{\lt{\left(\compose{S}{T}\right)}{\colvector{2\\1}}}=\vectrep{D}{\begin{bmatrix}10 & 33\\-1 & 11\end{bmatrix}}\\]]>
+<![CDATA[&\quad\quad=\vectrep{D}{]]>
 <![CDATA[95\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+]]>
 <![CDATA[202\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+]]>
 <![CDATA[(-11)\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+]]>
 <![CDATA[(-149)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix}]]>
 }\\
-<![CDATA[&=\colvector{95\\202\\-11\\-149}\\]]>
+<![CDATA[&\quad\quad=\colvector{95\\202\\-11\\-149}\\]]>
 </alignmath>
 </p>
 
 <example acro="KVMR" index="kernel!via matrix representation">
 <title>Kernel via matrix representation</title>
 
-<p>Consider the kernel of the linear transformation
+<p>Consider the kernel of the linear transformation, $\ltdefn{T}{M_{22}}{P_2}$, given by
 <equation>
-<![CDATA[\ltdefn{T}{M_{22}}{P_2},\quad \lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}=]]>
+<![CDATA[\lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}=]]>
 (2a-b+c-5d)+(a+4b+5c+2d)x+(3a-2b+c-8d)x^2
 </equation>
 </p>
 
 <example acro="RVMR" index="range!via matrix representation">
 <title>Range via matrix representation</title>
-
-<p>In this example, we will recycle the linear transformation $T$ and the bases $B$ and $C$ of <acroref type="example" acro="KVMR" /> but now we will compute the range of $T$,
+<p>In this example, we will recycle the linear transformation $T$ and the bases $B$ and $C$ of <acroref type="example" acro="KVMR" /> but now we will compute the range of $\ltdefn{T}{M_{22}}{P_2}$, given by
 <equation>
-<![CDATA[\ltdefn{T}{M_{22}}{P_2},\quad \lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}=]]>
+<![CDATA[\lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}=]]>
 (2a-b+c-5d)+(a+4b+5b+2d)x+(3a-2b+c-8d)x^2
 </equation>
 </p>

File src/section-VR.xml

 </statement>
 
 <proof>
-<p>We will appeal to <acroref type="theorem" acro="KILT" />.  Suppose $U$ is a vector space of dimension $n$, so vector representation is of the form $\ltdefn{\vectrepname{B}}{U}{\complex{n}}$.  Let $B=\set{\vectorlist{u}{n}}$ be the basis of $U$ used in the definition of $\vectrepname{B}$.  Suppose $\vect{u}\in\krn{\vectrepname{B}}$.  We write $\vect{u}$ as a linear combination of the vectors in the basis $B$ where the scalars are the components of the vector representation, \lt{\vectrepname{B}}{\vect{u}}.
+<p>We will appeal to <acroref type="theorem" acro="KILT" />.  Suppose $U$ is a vector space of dimension $n$, so vector representation is $\ltdefn{\vectrepname{B}}{U}{\complex{n}}$.  Let $B=\set{\vectorlist{u}{n}}$ be the basis of $U$ used in the definition of $\vectrepname{B}$.  Suppose $\vect{u}\in\krn{\vectrepname{B}}$.  We write $\vect{u}$ as a linear combination of the vectors in the basis $B$ where the scalars are the components of the vector representation, \lt{\vectrepname{B}}{\vect{u}}.
 <alignmath>
 \vect{u}
 <![CDATA[&=]]>
 \vectorentry{\lt{\vectrepname{B}}{\vect{u}}}{1}\vect{u}_1+
 \vectorentry{\lt{\vectrepname{B}}{\vect{u}}}{2}\vect{u}_2+
-\vectorentry{\lt{\vectrepname{B}}{\vect{u}}}{3}\vect{u}_3+
 \cdots+
 \vectorentry{\lt{\vectrepname{B}}{\vect{u}}}{n}\vect{u}_n
 <![CDATA[&&]]>\text{<acroref type="definition" acro="VR" />}\\
 <![CDATA[&=]]>
 \vectorentry{\zerovector}{1}\vect{u}_1+
 \vectorentry{\zerovector}{2}\vect{u}_2+
-\vectorentry{\zerovector}{3}\vect{u}_3+
 \cdots+
 \vectorentry{\zerovector}{n}\vect{u}_n
 <![CDATA[&&]]>\text{<acroref type="definition" acro="KLT" />}\\
-<![CDATA[&= 0\vect{u}_1+ 0\vect{u}_2+ 0\vect{u}_3+ \cdots+ 0\vect{u}_n]]>
+<![CDATA[&= 0\vect{u}_1+ 0\vect{u}_2+ \cdots+ 0\vect{u}_n]]>
 <![CDATA[&&]]>\text{<acroref type="definition" acro="ZCV" />}\\
-<![CDATA[&=\zerovector+\zerovector+\zerovector+\cdots+\zerovector]]>
+<![CDATA[&=\zerovector+\zerovector+\cdots+\zerovector]]>
 <![CDATA[&&]]>\text{<acroref type="theorem" acro="ZSSM" />}\\
 <![CDATA[&=\zerovector]]>
 <![CDATA[&&]]>\text{<acroref type="property" acro="Z" />}
 </statement>
 
 <proof>
-<p>We will appeal to <acroref type="theorem" acro="RSLT" />.  Suppose $U$ is a vector space of dimension $n$, so vector representation is of the form $\ltdefn{\vectrepname{B}}{U}{\complex{n}}$.  Let $B=\set{\vectorlist{u}{n}}$ be the basis of $U$ used in the definition of $\vectrepname{B}$.  Suppose $\vect{v}\in\complex{n}$.
+<p>We will appeal to <acroref type="theorem" acro="RSLT" />.  Suppose $U$ is a vector space of dimension $n$, so vector representation is $\ltdefn{\vectrepname{B}}{U}{\complex{n}}$.  Let $B=\set{\vectorlist{u}{n}}$ be the basis of $U$ used in the definition of $\vectrepname{B}$.  Suppose $\vect{v}\in\complex{n}$.
 Define the vector $\vect{u}$ by
 <equation>
 \vect{u}
 </equation>
 </p>
 
-<p>Then for $1\leq i\leq n$
+<p>Then for $1\leq i\leq n$, by <acroref type="definition" acro="VR" />,
 <alignmath>
 \vectorentry{\vectrep{B}{\vect{u}}}{i}
 <![CDATA[&=\vectorentry{\vectrep{B}{]]>
 \vectorentry{\vect{v}}{3}\vect{u}_3+
 \cdots+
 \vectorentry{\vect{v}}{n}\vect{u}_n
-}}{i}\\
-<![CDATA[&=\vectorentry{\vect{v}}{i}&&]]>\text{<acroref type="definition" acro="VR" />}
+}}{i}
+<![CDATA[=\vectorentry{\vect{v}}{i}&&]]>\text{}
 </alignmath>
 so the entries of vectors $\vectrep{B}{\vect{u}}$ and $\vect{v}$ are equal and <acroref type="definition" acro="CVE" /> yields the vector equality $\vectrep{B}{\vect{u}}=\vect{v}$.  This demonstrates that $\vect{v}\in\rng{\vectrepname{B}}$, so $\complex{n}\subseteq\rng{\vectrepname{B}}$.  Since $\rng{\vectrepname{B}}\subseteq\complex{n}$ by <acroref type="definition" acro="RLT" />, we have $\rng{\vectrepname{B}}=\complex{n}$ and <acroref type="theorem" acro="RSLT" /> says $\vectrepname{B}$ is surjective.</p>
 
 <theorem acro="CLI" index="coordinatization!linear independence">
 <title>Coordinatization and Linear Independence</title>
 <statement>
-<p>Suppose that $U$ is a vector space with a basis $B$ of size $n$.  Then $S=\set{\vectorlist{u}{k}}$ is a linearly independent subset of $U$ if and only if $R=\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}$ is a linearly independent subset of $\complex{n}$.</p>
-
+<p>Suppose that $U$ is a vector space with a basis $B$ of size $n$.  Then
+<alignmath>
+S=\set{\vectorlist{u}{k}}
+</alignmath>
+is a linearly independent subset of $U$ if and only if
+<alignmath>
+R=\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}
+</alignmath>
+is a linearly independent subset of $\complex{n}$.</p>
 </statement>
 
 <proof>
 <theorem acro="CSS" index="coordinatization!spanning sets">
 <title>Coordinatization and Spanning Sets</title>
 <statement>
-<p>Suppose that $U$ is a vector space with a basis $B$ of size $n$.  Then $\vect{u}\in\spn{\set{\vectorlist{u}{k}}}$  if and only if $\vectrep{B}{\vect{u}}\in\spn{\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}}$.</p>
-
+<p>Suppose that $U$ is a vector space with a basis $B$ of size $n$.  Then
+<alignmath>
+\vect{u}\in\spn{\set{\vectorlist{u}{k}}}
+</alignmath>
+if and only if
+<alignmath>
+\vectrep{B}{\vect{u}}\in\spn{\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}}
+</alignmath>
+</p>
 </statement>
 
 <proof>
-<p><implyforward />  Suppose $\vect{u}\in\spn{\set{\vectorlist{u}{k}}}$.  Then there are scalars, $\scalarlist{a}{k}$, such that
+<p><implyforward />  Suppose $\vect{u}\in\spn{\set{\vectorlist{u}{k}}}$.  Then we know there are scalars, $\scalarlist{a}{k}$, such that
 <equation>
 \vect{u}=\lincombo{a}{u}{k}
 </equation>
 </p>
 
-<p>Then,
+<p>Then, by <acroref type="theorem" acro="LTLC" />,
 <alignmath>
 <![CDATA[\vectrep{B}{\vect{u}}&=\vectrep{B}{\lincombo{a}{u}{k}}\\]]>
-<![CDATA[&=a_1\vectrep{B}{\vect{u}_1}+a_2\vectrep{B}{\vect{u}_2}+a_3\vectrep{B}{\vect{u}_3}+\cdots+a_k\vectrep{B}{\vect{u}_k}&&]]>\text{<acroref type="theorem" acro="LTLC" />}
+<![CDATA[&=a_1\vectrep{B}{\vect{u}_1}+a_2\vectrep{B}{\vect{u}_2}+a_3\vectrep{B}{\vect{u}_3}+\cdots+a_k\vectrep{B}{\vect{u}_k}]]>
 </alignmath>
 which says that $\vectrep{B}{\vect{u}}\in\spn{\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}}$.</p>
 
 <![CDATA[\vect{u}&=\lt{I_U}{\vect{u}}&&]]>\text{<acroref type="definition" acro="IDLT" />}\\
 <![CDATA[&=\lt{\left(\compose{\ltinverse{\vectrepname{B}}}{\vectrepname{B}}\right)}{\vect{u}}&&]]>\text{<acroref type="definition" acro="IVLT" />}\\
 <![CDATA[&=\lt{\ltinverse{\vectrepname{B}}}{\lt{\vectrepname{B}}{\vect{u}}}&&]]>\text{<acroref type="definition" acro="LTC" />}\\
-<![CDATA[&=\lt{\ltinverse{\vectrepname{B}}}{b_1\vectrep{B}{\vect{u}_1}+b_2\vectrep{B}{\vect{u}_2}+b_3\vectrep{B}{\vect{u}_3}+\cdots+b_k\vectrep{B}{\vect{u}_k}}\\]]>
+<![CDATA[&=\lt{\ltinverse{\vectrepname{B}}}{b_1\vectrep{B}{\vect{u}_1}+b_2\vectrep{B}{\vect{u}_2}+\cdots+b_k\vectrep{B}{\vect{u}_k}}\\]]>
 <![CDATA[&=b_1\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_1}}+b_2\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_2}}]]>
-+b_3\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_3}}\\
-<![CDATA[&\quad\quad+\cdots]]>
-<![CDATA[+b_k\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_k}}&&]]>\text{<acroref type="theorem" acro="LTLC" />}\\
-<![CDATA[&=b_1\lt{I_U}{\vect{u}_1}+b_2\lt{I_U}{\vect{u}_2}+b_3\lt{I_U}{\vect{u}_3}+\cdots+b_k\lt{I_U}{\vect{u}_k}&&]]>\text{<acroref type="definition" acro="IVLT" />}\\
+<![CDATA[+\cdots+b_k\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_k}}&&]]>\text{<acroref type="theorem" acro="LTLC" />}\\
+<![CDATA[&=b_1\lt{I_U}{\vect{u}_1}+b_2\lt{I_U}{\vect{u}_2}+\cdots+b_k\lt{I_U}{\vect{u}_k}&&]]>\text{<acroref type="definition" acro="IVLT" />}\\
 <![CDATA[&=\lincombo{b}{u}{k}&&]]>\text{<acroref type="definition" acro="IDLT" />}
 </alignmath>
 which says that $\vect{u}\in\spn{\set{\vectorlist{u}{k}}}$.</p>
 </p>
 
 <p>Now, form the subset of $\complex{3}$ that is the result of applying $\vectrepname{B}$ to each element of $D$,
-<equation>
-F=\set{\vectrep{B}{-2-x+3x^2},\,\vectrep{B}{1-2x^2},\,\vectrep{B}{5+4x+x^2}}=
+<alignmath>
+<![CDATA[F&=\set{\vectrep{B}{-2-x+3x^2},\,\vectrep{B}{1-2x^2},\,\vectrep{B}{5+4x+x^2}}\\]]>
+<![CDATA[&=]]>
 \set{
 \colvector{-2\\-1\\3},\,
 \colvector{1\\0\\-2},\,
 \colvector{5\\4\\1}
 }
-</equation>
+</alignmath>
 and ask if $F$ is a linearly independent spanning set for $\complex{3}$.  This is easily seen to be the case by forming a matrix $A$ whose columns are the vectors of $F$, row-reducing $A$ to the identity matrix $I_3$, and then using the nonsingularity of $A$ to assert that $F$ is a basis for $\complex{3}$ (<acroref type="theorem" acro="CNMB" />).  Now, since $F$ is a basis for $\complex{3}$, <acroref type="theorem" acro="CLI" /> and <acroref type="theorem" acro="CSS" /> tell us that $D$ is also a basis for $P_2$.</p>
 
 </example>
 <equation>
 6\colvector{3\\-2\\0\\7\\4\\-3}+2\colvector{-1\\4\\-2\\3\\8\\5}
 </equation>
-which we can compute with the operations of $\complex{6}$ (<acroref type="definition" acro="CVA" />, <acroref type="definition" acro="CVSM" />), to arrive at
+which we can compute with operations in $\complex{6}$ (<acroref type="definition" acro="CVA" />, <acroref type="definition" acro="CVSM" />), to arrive at
 <equation>
 \colvector{16\\-4\\-4\\48\\40\\-8}
 </equation>
 </p>
 
 <p>We are after the result of a computation in $M_{32}$, so we now can apply $\ltinverse{\vectrepname{B}}$ to obtain a $3\times 2$ matrix,
-<equation>
-<![CDATA[16\begin{bmatrix}1&0\\0&0\\0&0\end{bmatrix}+]]>
+<alignmath>
+<![CDATA[&16\begin{bmatrix}1&0\\0&0\\0&0\end{bmatrix}+]]>
 <![CDATA[(-4)\begin{bmatrix}0&0\\1&0\\0&0\end{bmatrix}+]]>
 <![CDATA[(-4)\begin{bmatrix}0&0\\0&0\\1&0\end{bmatrix}+]]>
 <![CDATA[48\begin{bmatrix}0&1\\0&0\\0&0\end{bmatrix}+]]>
 <![CDATA[40\begin{bmatrix}0&0\\0&1\\0&0\end{bmatrix}+]]>
-<![CDATA[(-8)\begin{bmatrix}0&0\\0&0\\0&1\end{bmatrix}]]>
-<![CDATA[=\begin{bmatrix}16&48\\-4&40\\-4&-8\end{bmatrix}]]>
-</equation>
+<![CDATA[(-8)\begin{bmatrix}0&0\\0&0\\0&1\end{bmatrix}]]>\\
+<![CDATA[&\quad\quad=\ \begin{bmatrix}16&48\\-4&40\\-4&-8\end{bmatrix}]]>
+</alignmath>
 which is exactly the matrix we would have computed had we just performed the matrix operations in the first place.  So this was not meant to be an <em>easier</em> way to compute a linear combination of two matrices, just a <em>different</em> way.</p>
 
 </example>