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rbeezer  committed ad83ec5

Line breaks, Chapter VS

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File src/section-S.xml

 \vect{x}+\vect{y}=\colvector{x_1\\x_2\\x_3}+\colvector{y_1\\y_2\\y_3}=
 \colvector{x_1+y_1\\x_2+y_2\\x_3+y_3}
 </equation>
-and we can test this vector for membership in $W$ as follows,
+and we can test this vector for membership in $W$ as follows.  Because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$ and because $\vect{y}\in W$ we know $2y_1-5y_2+7y_3=0$.  Therefore,
 <alignmath>
 2(x_1+y_1)-5(x_2+y_2)+7(x_3+y_3)
 <![CDATA[&=2x_1+2y_1-5x_2-5y_2+7x_3+7y_3\\]]>
 <![CDATA[&=(2x_1-5x_2+7x_3)+(2y_1-5y_2+7y_3)\\]]>
-<![CDATA[&=0 + 0&&\vect{x}\in W,\ \vect{y}\in W\\]]>
+<![CDATA[&=0 + 0\\]]>
 <![CDATA[&=0]]>
 </alignmath>
 and by this computation we see that $\vect{x}+\vect{y}\in W$.  One property down, nine to go.</p>
 
-<p>If $\alpha$ is a scalar and $\vect{x}\in W$, is it always true that $\alpha\vect{x}\in W$?  This is what we need to establish <acroref type="property" acro="SC" />.  Again, the answer is not as obvious as it was when our set of vectors was all of $\complex{3}$.  Let's see.
+<p>If $\alpha$ is a scalar and $\vect{x}\in W$, is it always true that $\alpha\vect{x}\in W$?  This is what we need to establish <acroref type="property" acro="SC" />.  Again, the answer is not as obvious as it was when our set of vectors was all of $\complex{3}$.  Let's see.  First, note that because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$.  Therefore,
 <equation>
 \alpha\vect{x}=\alpha\colvector{x_1\\x_2\\x_3}=\colvector{\alpha x_1\\\alpha x_2\\\alpha x_3}
 </equation>
-and we can test this vector for membership in $W$ with
+and we can test this vector for membership in $W$.  First, note that because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$.  Therefore,
 <alignmath>
 2(\alpha x_1)-5(\alpha x_2)+7(\alpha x_3)
 <![CDATA[&=\alpha(2x_1-5x_2+7x_3)\\]]>
-<![CDATA[&=\alpha 0&&\vect{x}\in W\\]]>
+<![CDATA[&=\alpha 0\\]]>
 <![CDATA[&=0]]>
 </alignmath>
 and we see that indeed $\alpha\vect{x}\in W$.  Always.</p>
 <equation>
 \vect{-x}=\colvector{-x_1\\-x_2\\-x_3}
 </equation>
-and we can test this vector for membership in $W$ with
+and we can test this vector for membership in $W$.  As before, because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$.
 <alignmath>
 2(-x_1)-5(-x_2)+7(-x_3)
 <![CDATA[&=-(2x_1-5x_2+7x_3)\\]]>
-<![CDATA[&=-0&&\vect{x}\in W\\]]>
+<![CDATA[&=-0\\]]>
 <![CDATA[&=0]]>
 </alignmath>
 and we now believe that $\vect{-x}\in W$.</p>
 </statement>
 
 <proof>
-<p>We will examine the three requirements of <acroref type="theorem" acro="TSS" />.  Recall that $\nsp{A}=\setparts{\vect{x}\in\complex{n}}{A\vect{x}=\zerovector}$.</p>
+<p>We will examine the three requirements of <acroref type="theorem" acro="TSS" />.  Recall that <acroref type="definition" acro="NSM" /> can be formulated as $\nsp{A}=\setparts{\vect{x}\in\complex{n}}{A\vect{x}=\zerovector}$.</p>
 
 <p>First, $\zerovector\in\nsp{A}$, which can be inferred as a consequence of <acroref type="theorem" acro="HSC" />.  So $\nsp{A}\neq\emptyset$.</p>
 
 <![CDATA[&=]]>
 (\alpha_1+2\alpha_2)x^4+
 (-4\alpha_1-3\alpha_2)x^3+
-(5\alpha_1-6\alpha_2)x^2+
+(5\alpha_1-6\alpha_2)x^2\\
+<![CDATA[&\quad\quad+]]>
 (-\alpha_1+6\alpha_2)x+
 (-2\alpha_1+4\alpha_2)
 </alignmath>
 </equation>
 in $W$?  To answer this, we want to determine if $\vect{y}$ can be written as a linear combination of the five matrices in $S$.  Can we find scalars, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ so that
 <alignmath>
-\begin{bmatrix}
+<![CDATA[&\begin{bmatrix}]]>
 <![CDATA[9 & 3 \\ 7&3 \\ 10 & -11]]>
-\end{bmatrix}
+\end{bmatrix}\\
 <![CDATA[&=]]>
 \alpha_1
 \begin{bmatrix}
 </equation>
 in $W$?  To answer this, we want to determine if $\vect{x}$ can be written as a linear combination of the five matrices in $S$.  Can we find scalars, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ so that
 <alignmath>
-\begin{bmatrix}
+<![CDATA[&\begin{bmatrix}]]>
 <![CDATA[2 & 1 \\ 3 & 1 \\ 4 & -2]]>
-\end{bmatrix}
+\end{bmatrix}\\
 <![CDATA[&=]]>
 \alpha_1
 \begin{bmatrix}

File src/section-VS.xml

 Feeling better?  Or worse?</p>
 
 <p><acroref type="property" acro="AI" />:  For each vector, $\vect{u}$, we must locate an additive inverse, $\vect{-u}$.  Here it is, $-(x_1,\,x_2)=(-x_1-2,\,-x_2-2)$.  As odd as it may look, I hope you are withholding judgment.  Check:
-<equation>
-\vect{u}+ (\vect{-u})=(x_1,\,x_2)+(-x_1-2,\,-x_2-2)=(x_1+(-x_1-2)+1,\,-x_2+(x_2-2)+1)=(-1,\,-1)=\zerovector
-</equation>
+<alignmath>
+<![CDATA[\vect{u}+ (\vect{-u})&=(x_1,\,x_2)+(-x_1-2,\,-x_2-2)\\]]>
+<![CDATA[&=(x_1+(-x_1-2)+1,\,-x_2+(x_2-2)+1)=(-1,\,-1)=\zerovector]]>
+</alignmath>
 </p>
 
 <p><acroref type="property" acro="SMA" />:
 
 <p><acroref type="property" acro="DVA" />:  If you have hung on so far, here's where it gets even wilder.  In the next two properties we mix and mash the two operations.
 <alignmath>
-<![CDATA[\alpha(\vect{u}+\vect{v})&=\alpha\left((x_1,\,x_2)+(y_1,\,y_2)\right)\\]]>
+<![CDATA[\alpha(\vect{u}&+\vect{v})\\]]>
+<![CDATA[&=\alpha\left((x_1,\,x_2)+(y_1,\,y_2)\right)\\]]>
 <![CDATA[&=\alpha(x_1+y_1+1,\,x_2+y_2+1)\\]]>
 <![CDATA[&=(\alpha(x_1+y_1+1)+\alpha-1,\,\alpha(x_2+y_2+1)+\alpha-1)\\]]>
 <![CDATA[&=(\alpha x_1+\alpha y_1+\alpha+\alpha-1,\,\alpha x_2+\alpha y_2+\alpha+\alpha-1)\\]]>
 
 <p><acroref type="property" acro="DSA" />:
 <alignmath>
-<![CDATA[(\alpha+\beta)\vect{u}&=(\alpha+\beta)(x_1,\,x_2)\\]]>
+<![CDATA[(\alpha&+\beta)\vect{u}\\]]>
+<![CDATA[&=(\alpha+\beta)(x_1,\,x_2)\\]]>
 <![CDATA[&=((\alpha+\beta)x_1+(\alpha+\beta)-1,\,(\alpha+\beta)x_2+(\alpha+\beta)-1)\\]]>
 <![CDATA[&=(\alpha x_1+\beta x_1+\alpha+\beta-1,\,\alpha x_2+\beta x_2+\alpha+\beta-1)\\]]>
 <![CDATA[&=(\alpha x_1+\alpha-1+\beta x_1+\beta-1+1,\,\alpha x_2+\alpha-1+\beta x_2+\beta-1+1)\\]]>