rbeezer avatar rbeezer committed bc60749

Edits to save 6 pages in print

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src/section-EE.xml

 
 <p>This example illustrates the proof of <acroref type="theorem" acro="EMHE" />, so will employ the same notation as the proof <mdash /> look there for full explanations.  It is <em>not</em> meant to be an example of a reasonable computational approach to finding eigenvalues and eigenvectors.  OK, warnings in place, here we go.</p>
 
-<p>Let
-<equation>
-A=\begin{bmatrix}
+<p>Consider the matrix $A$, and choose the vector $\vect{x}$,
+<alignmath>
+A<![CDATA[&=]]>\begin{bmatrix}
 <![CDATA[-7 & -1 & 11 & 0 & -4\\]]>
 <![CDATA[4 & 1 & 0 & 2 & 0\\]]>
 <![CDATA[-10 & -1 & 14 & 0 & -4\\]]>
 <![CDATA[8 & 2 & -15 & -1 & 5\\]]>
 <![CDATA[-10 & -1 & 16 & 0 & -6]]>
 \end{bmatrix}
-</equation>
-and choose
-<equation>
-\vect{x}=\colvector{3\\0\\3\\-5\\4}
-</equation>
+<![CDATA[&]]>
+\vect{x}<![CDATA[&=]]>\colvector{3\\0\\3\\-5\\4}
+</alignmath>
 </p>
 
 <p>It is important to notice that the choice of $\vect{x}$ could be <em>anything</em>, so long as it is <em>not</em> the zero vector.  We have not chosen $\vect{x}$ totally at random, but so as to make our illustration of the theorem as general as possible.  You could replicate this example with your own choice and the computations are guaranteed to be reasonable, provided you have a computational tool that will factor a fifth degree polynomial for you.</p>

src/section-OD.xml

 <definition acro="UTM" index="upper triangular matrix">
 <title>Upper Triangular Matrix</title>
 <p>The $n\times n$ square matrix $A$ is <define>upper triangular</define> if $\matrixentry{A}{ij} =0$ whenever $i>j$.</p>
-
 </definition>
 
 <definition acro="LTM" index="lower triangular matrix">
 <title>Lower Triangular Matrix</title>
 <p>The $n\times n$ square matrix $A$ is <define>lower triangular</define> if $\matrixentry{A}{ij} =0$ whenever
-<![CDATA[$i<j$.]]>
-</p>
-
+<![CDATA[$i<j$.]]></p>
 </definition>
 
 <p>Obviously, properties of a lower triangular matrices will have analogues for upper triangular matrices.  Rather than stating two very similar theorems, we will say that matrices are <q>triangular of the same type</q> as a convenient shorthand to cover both possibilities and then give a proof for just one type.</p>

src/section-PEE.xml

 <![CDATA[0 &  1 &  0 &  0\\]]>
 <![CDATA[0 &  0 &  1 &  0\\]]>
 <![CDATA[0 &  0 &  0 &  1]]>
-\end{bmatrix}\\
-<![CDATA[&=]]>
+\end{bmatrix}
+<![CDATA[=]]>
 \frac{1}{2}
 \begin{bmatrix}
 <![CDATA[1 &  1 &  3 &  3\\]]>

src/section-TSS.xml

 </exercise>
 
 <exercisegroup>
-<p>For Exercises C21<ndash />C28, find the solution set of the given system of linear equations. Identify the values of $n$ and $r$, and compare your answers to the results of the theorems of this section.</p>
+<p>For Exercises C21<ndash />C28, find the solution set of the system of linear equations. Give the values of $n$ and $r$, and interpret your answers in light of the theorems of this section.</p>
 
 <exercise type="C" number="21" rough="3 x 4 system; no solutions">
 <problem contributor="chrisblack"><alignmath>
 <![CDATA[&]]>
 <![CDATA[\matrixentry{B}{3,f_1}&]]>
 </alignmath>
-If you cannot conclude anything about an entry, then say so.  (See <acroref type="exercise" acro="TSS.M46" /> for inspiration.)
+If you cannot conclude anything about an entry, then say so.  (See <acroref type="exercise" acro="TSS.M46" />.)
 </problem>
 </exercise>
 

src/section-VR.xml

 <![CDATA[\begin{bmatrix}0&0\\0&0\\0&1\end{bmatrix}]]>
 }
 </equation>
-and apply $\vectrepname{B}$ to each vector in the linear combination.  This gives us a new computation, now in the vector space $\complex{6}$,
-<equation>
-6\colvector{3\\-2\\0\\7\\4\\-3}+2\colvector{-1\\4\\-2\\3\\8\\5}
-</equation>
-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>
+and apply $\vectrepname{B}$ to each vector in the linear combination.  This gives us a new computation, now in the vector space $\complex{6}$, which we can compute with operations in $\complex{6}$ (<acroref type="definition" acro="CVA" />, <acroref type="definition" acro="CVSM" />),
+<alignmath>
+6\colvector{3\\-2\\0\\7\\4\\-3}+2\colvector{-1\\4\\-2\\3\\8\\5}=\colvector{16\\-4\\-4\\48\\40\\-8}
+</alignmath>
 </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,
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