rbeezer avatar rbeezer committed fab4dfd

Minor edits (David Farmer)
* * *
Mystery

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 ~~~~~~~~~~~~~~~~
 New:  Exercise MM.T12, Theorem HMIP reprised
 Change:  Proof of Theorem OD uses normality of diagonal matrices (Hunter Wills)
+Typo: Miscellaneous edits (David Farmer)
 Typo:  Solution ILT.C29, inner span should be a set (Aysha Orta)
 
 v3.00 2012/12/05

src/archetypes.xml

 <injective>
 <status>No</status>
 <explanation>Since the kernel is nontrivial <acroref type="theorem" acro="KILT" /> tells us that the linear transformation is not injective.  Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 2, just from checking dimensions of the domain and the codomain.  In particular, verify that
-%
 <alignmath>
 <![CDATA[\lt{T}{\colvector{1\\2\\-1\\4\\5}}&=\colvector{38\\24\\-16}&]]>
 <![CDATA[\lt{T}{\colvector{0\\ -3\\ 0\\ 5\\ 6}}&=\colvector{38\\24\\-16}]]>
 </alignmath>
-%
 This demonstration that $T$ is not injective is constructed with the observation that
-%
 <alignmath>
 <![CDATA[\colvector{0\\-3\\0\\5\\6}&=\colvector{1\\2\\-1\\4\\5}+\colvector{-1\\-5\\1\\1\\1}]]>
 \intertext{and}
 <![CDATA[\vect{z}&=\colvector{-1\\-5\\1\\1\\1}\in\krn{T}]]>
 </alignmath>
-%
 so the vector $\vect{z}$ effectively <q>does nothing</q> in the evaluation of $T$.
 </explanation>
 </injective>
 <injective>
 <status>No</status>
 <explanation>Since the kernel is nontrivial <acroref type="theorem" acro="KILT" /> tells us that the linear transformation is not injective.  Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 2, just from checking dimensions of the domain and the codomain.  In particular, verify that
-%
 <alignmath>
 <![CDATA[\lt{T}{\colvector{-3\\1\\-2\\-3\\1}}&=\colvector{6\\19\\6}&]]>
 <![CDATA[\lt{T}{\colvector{-4\\-4\\-2\\-1\\4}}&=\colvector{6\\19\\6}]]>
 </alignmath>
-%
 This demonstration that $T$ is not injective is constructed with the observation that
-%
 <alignmath>
 <![CDATA[\colvector{-4\\-4\\-2\\-1\\4}&=\colvector{-3\\1\\-2\\-3\\1}+\colvector{-1\\-5\\0\\2\\3}]]>
 \intertext{and}
 <![CDATA[\vect{z}&=\colvector{-1\\-5\\0\\2\\3}\in\krn{T}]]>
 </alignmath>
-%
 so the vector $\vect{z}$ effectively <q>does nothing</q> in the evaluation of $T$.
 </explanation>
 </injective>
 <injective>
 <status>No</status>
 <explanation>Since the kernel is nontrivial <acroref type="theorem" acro="KILT" /> tells us that the linear transformation is not injective.  Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 2, just from checking dimensions of the domain and the codomain.  In particular, verify that
-%
 <alignmath>
 <![CDATA[\lt{T}{\colvector{5\\-1\\3}}&=\colvector{-15\\-19\\7\\10\\11}&]]>
 <![CDATA[\lt{T}{\colvector{1\\1\\5}}&=\colvector{-15\\-19\\7\\10\\11}]]>
 </alignmath>
-%
 This demonstration that $T$ is not injective is constructed with the observation that
-%
 <alignmath>
 <![CDATA[\colvector{1\\1\\5}&=\colvector{5\\-1\\3}+\colvector{-4\\2\\2}]]>
 \intertext{and}
 <![CDATA[\vect{z}&=\colvector{-4\\2\\2}\in\krn{T}]]>
 </alignmath>
-%
 so the vector $\vect{z}$ effectively <q>does nothing</q> in the evaluation of $T$.
 </explanation>
 </injective>
 <surjective>
 <status>No</status>
 <explanation>The dimension of the range is 2, and the codomain ($\complex{5}$) has dimension 5.  So the transformation is not onto.  Notice too that since the domain $\complex{3}$ has dimension 3, it is impossible for the range to have a dimension greater than 3, and no matter what the actual definition of the function, it cannot possibly be onto.<br /><br />
-%
 To be more precise, verify that $\colvector{2\\3\\1\\1\\1}\not\in\rng{T}$, by setting the output equal to this vector and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent.  So the preimage, $\preimage{T}{\colvector{2\\3\\1\\1\\1}}$, is empty.  This alone is sufficient to see that the linear transformation is not onto.
 </explanation>
 </surjective>
 <surjective>
 <status>No</status>
 <explanation>The dimension of the range is 3, and the codomain ($\complex{5}$) has dimension 5.  So the transformation is not surjective.  Notice too that since the domain $\complex{3}$ has dimension 3, it is impossible for the range to have a dimension greater than 3, and no matter what the actual definition of the function, it cannot possibly be surjective in this situation.<br /><br />
-%
 To be more precise, verify that $\colvector{2\\1\\-3\\2\\6}\not\in\rng{T}$, by setting the output equal to this vector and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent.  So the preimage, $\preimage{T}{\colvector{2\\1\\-3\\2\\6}}$, is empty.  This alone is sufficient to see that the linear transformation is not onto.
 </explanation>
 </surjective>
 <injective>
 <status>No</status>
 <explanation>Since the kernel is nontrivial <acroref type="theorem" acro="KILT" /> tells us that the linear transformation is not injective.  Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 2, just from checking dimensions of the domain and the codomain.  In particular, verify that
-%
 <alignmath>
 <![CDATA[\lt{T}{\colvector{1\\3\\-1\\2\\4}}&=\colvector{4\\55\\72\\77\\31}&]]>
 <![CDATA[\lt{T}{\colvector{4\\7\\0\\5\\7}}&=\colvector{4\\55\\72\\77\\31}]]>
 </alignmath>
-%
 This demonstration that $T$ is not injective is constructed with the observation that
-%
 <alignmath>
 <![CDATA[\colvector{4\\7\\0\\5\\7}&=\colvector{1\\3\\-1\\2\\4}+\colvector{3\\4\\1\\3\\3}]]>
 \intertext{and}
 <![CDATA[\vect{z}&=\colvector{3\\4\\1\\3\\3}\in\krn{T}]]>
 </alignmath>
-%
 so the vector $\vect{z}$ effectively <q>does nothing</q> in the evaluation of $T$.
 </explanation>
 </injective>
 <surjective>
 <status>No</status>
 <explanation>The dimension of the range is 4, and the codomain ($\complex{5}$) has dimension 5.  So $\rng{T}\neq\complex{5}$ and by <acroref type="theorem" acro="RSLT" /> the transformation is not surjective.<br /><br />
-%
 To be more precise, verify that $\colvector{-1\\2\\3\\-1\\4}\not\in\rng{T}$, by setting the output equal to this vector and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent.  So the preimage, $\preimage{T}{\colvector{-1\\2\\3\\-1\\4}}$, is empty.  This alone is sufficient to see that the linear transformation is not onto.
 </explanation>
 </surjective>
 <injective>
 <status>No</status>
 <explanation>Since the kernel is nontrivial <acroref type="theorem" acro="KILT" /> tells us that the linear transformation is not injective.  Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 1, just from checking dimensions of the domain and the codomain.  In particular, verify that
-%
 <alignmath>
 <![CDATA[\lt{T}{\colvector{2\\1\\3}}&=\begin{bmatrix}1&9\\10&-16\end{bmatrix}]]>
 <![CDATA[&]]>
 <![CDATA[\lt{T}{\colvector{0\\-1\\11}}&=\begin{bmatrix}1&9\\10&-16\end{bmatrix}]]>
 </alignmath>
-%
 This demonstration that $T$ is not injective is constructed with the observation that
-%
 <alignmath>
 <![CDATA[\colvector{0\\-1\\11}&=\colvector{2\\1\\3}+\colvector{-2\\-2\\8}]]>
 \intertext{and}
 <![CDATA[\vect{z}&=\colvector{-2\\-2\\8}\in\krn{T}]]>
 </alignmath>
-%
 so the vector $\vect{z}$ effectively <q>does nothing</q> in the evaluation of $T$.
 </explanation>
 </injective>
 <surjective>
 <status>No</status>
 <explanation>The dimension of the range is 2, and the codomain ($M_{22}$) has dimension 4.  So the transformation is not surjective.  Notice too that since the domain $\complex{3}$ has dimension 3, it is impossible for the range to have a dimension greater than 3, and no matter what the actual definition of the function, it cannot possibly be surjective in this situation.<br /><br />
-%
-<![CDATA[To be more precise, verify that $\begin{bmatrix}2&-1\\1&3\end{bmatrix}\not\in\rng{T}$, by setting the output of $T$ equal to this matrix and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent.  So the preimage, $\preimage{T}{\begin{bmatrix}2&-1\\1&3\end{bmatrix}}$, is empty.   This alone is sufficient to see that the linear transformation is not onto.]]>
+To be more precise, verify that <![CDATA[$\begin{bmatrix}2&-1\\1&3\end{bmatrix}\not\in\rng{T}$]]>, by setting the output of $T$ equal to this matrix and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent.  So the preimage, <![CDATA[$\preimage{T}{\begin{bmatrix}2&-1\\1&3\end{bmatrix}}$]]>, is empty.   This alone is sufficient to see that the linear transformation is not onto.
 </explanation>
 </surjective>
 
 <surjective>
 <status>No</status>
 <explanation>The dimension of the range is 5, and the codomain ($P_5$) has dimension 6.  So the transformation is not surjective.  Notice too that since the domain $P_4$ has dimension 5, it is impossible for the range to have a dimension greater than 5, and no matter what the actual definition of the function, it cannot possibly be surjective in this situation.<br /><br />
-%
 To be more precise, verify that $1+x+x^2+x^3+x^4\not\in\rng{T}$, by setting the output equal to this vector and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent.  So the preimage, $\preimage{T}{1+x+x^2+x^3+x^4}$, is nonempty.  This alone is sufficient to see that the linear transformation is not onto.
 </explanation>
 </surjective>
 <injective>
 <status>No</status>
 <explanation>Since the kernel is nontrivial <acroref type="theorem" acro="KILT" /> tells us that the linear transformation is not injective.  Also, since the rank can not exceed 4, we are guaranteed to have a nullity of at least 2, just from checking dimensions of the domain and the codomain.  In particular, verify that
-%
 <alignmath>
 <![CDATA[\lt{T}{\begin{bmatrix}1&10&-2\\3&-1&1\end{bmatrix}}&=\colvector{-7\\-14\\-1\\-13}]]>
 <![CDATA[&]]>
 <![CDATA[\lt{T}{\begin{bmatrix}5&-3&-1\\5&3&3\end{bmatrix}}&=\colvector{-7\\-14\\-1\\-13}]]>
 </alignmath>
-%
 This demonstration that $T$ is not injective is constructed with the observation that
-%
 <alignmath>
 <![CDATA[\begin{bmatrix}5&-3&-1\\5&3&3\end{bmatrix}]]>
 <![CDATA[&=\begin{bmatrix}1&10&-2\\3&-1&1\end{bmatrix}+\begin{bmatrix}4&-13&1\\2&4&2\end{bmatrix}]]>
 \intertext{and}
 <![CDATA[\vect{z}&=\begin{bmatrix}4&-13&1\\2&4&2\end{bmatrix}\in\krn{T}]]>
 </alignmath>
-%
 so the vector $\vect{z}$ effectively <q>does nothing</q> in the evaluation of $T$.
 </explanation>
 </injective>
 <injective>
 <status>No</status>
 <explanation>Since the kernel is nontrivial <acroref type="theorem" acro="KILT" /> tells us that the linear transformation is not injective.  In particular, verify that
-%
 <alignmath>
 <![CDATA[\lt{T}{\begin{bmatrix}-2&0\\1&-4\end{bmatrix}}&=\begin{bmatrix}115&78\\-38&-35\end{bmatrix}]]>
 <![CDATA[&]]>
 <![CDATA[\lt{T}{\begin{bmatrix}4&3\\-1&3\end{bmatrix}}&=\begin{bmatrix}115&78\\-38&-35\end{bmatrix}]]>
 </alignmath>
-%
 This demonstration that $T$ is not injective is constructed with the observation that
-%
 <alignmath>
 <![CDATA[\begin{bmatrix}4&3\\-1&3\end{bmatrix}]]>
 <![CDATA[&=\begin{bmatrix}-2&0\\1&-4\end{bmatrix}+\begin{bmatrix}6&3\\-2&-1\end{bmatrix}]]>
 \intertext{and}
 <![CDATA[\vect{z}&=\begin{bmatrix}6&3\\-2&-1\end{bmatrix}\in\krn{T}]]>
 </alignmath>
-%
 so the vector $\vect{z}$ effectively <q>does nothing</q> in the evaluation of $T$.
 </explanation>
 </injective>
 <surjective>
 <status>No</status>
 <explanation>The dimension of the range is 3, and the codomain ($M_{22}$) has dimension 5.  So $\rng{T}\neq M_{22}$ and by <acroref type="theorem" acro="RSLT" /> the transformation is not surjective.<br /><br />
-%
 <![CDATA[To be more precise, verify that $\begin{bmatrix}2 & 4\\ 3 & 1\end{bmatrix}\not\in\rng{T}$, by setting the output of $T$ equal to this matrix and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent.  So the preimage, $\preimage{T}{\begin{bmatrix}2 & 4\\ 3 & 1\end{bmatrix}}$, is empty.  This alone is sufficient to see that the linear transformation is not onto.]]>
 </explanation>
 </surjective>
 
 <license>
 <html>
-<!-- FCLA Edits: &mdash; -> <mdash />; &copy; -> <copyright />; <ellipsis /> -> <ellipsis />; -->
+<!-- FCLA Edits: &mdash; -> <mdash />; &copy; -> &#x00A9;; <ellipsis /> -> <ellipsis />; -->
 <![CDATA[<h3 style="text-align: center;">GNU Free Documentation License</h3>
 
 <p style="text-align: center;">Version 1.3, 3 November 2008</p>
 
-<p> Copyright <copyright /> 2000, 2001, 2002, 2007, 2008 Free Software Foundation, Inc.
+<p> Copyright &#x00A9; 2000, 2001, 2002, 2007, 2008 Free Software Foundation, Inc.
      &lt;<a href="http://fsf.org/">http://fsf.org/</a>&gt;
  </p><p>Everyone is permitted to copy and distribute verbatim copies
  of this license document, but changing it is not allowed.</p>

src/section-MM.xml

 </theorem>
 
 <theorem acro="MMA" index="matrix multiplication!associativity">
-<title>Matrix Multiplication is Associative </title>
+<title>Matrix Multiplication is Associative</title>
 <statement>
 <p>Suppose $A$ is an $m\times n$ matrix, $B$ is an $n\times p$ matrix and $D$ is a $p\times s$ matrix.  Then  $A(BD)=(AB)D$.</p>
 

src/section-NM.xml

 <!-- % -->
 <!-- %%%%%%%%%% -->
 <introduction>
-<p>In this section we specialize further and consider matrices with equal numbers of rows and columns, which when considered as coefficient matrices lead to systems with equal numbers of equations and variables.  We will see in the second half of the course (<acroref type="chapter" acro="D" />, <acroref type="chapter" acro="E" /> <acroref type="chapter" acro="LT" />, <acroref type="chapter" acro="R" />) that these matrices are especially important.</p>
+<p>In this section we specialize further and consider matrices with equal numbers of rows and columns, which when considered as coefficient matrices lead to systems with equal numbers of equations and variables.  We will see in the second half of the course (<acroref type="chapter" acro="D" />, <acroref type="chapter" acro="E" />, <acroref type="chapter" acro="LT" />, <acroref type="chapter" acro="R" />) that these matrices are especially important.</p>
 
 </introduction>
 

src/section-RREF.xml

 <![CDATA[-2x_1-4x_2+x_3+11x_4&=-10]]>
 </alignmath>
 </problem>
-<solution contributor="robertbeezer">The augmented matrix row-reduces to\\
-<equation>
+<solution contributor="robertbeezer">The augmented matrix row-reduces to
+<alignmath>
 \begin{bmatrix}
 <![CDATA[\leading{1} &  2 &  0 &  -4 &  2\\]]>
 <![CDATA[0 &  0 &  \leading{1} &  3 &  -6\\]]>
 <![CDATA[0 &  0 &  0 &  0 &  0]]>
 \end{bmatrix}
-</equation>
+</alignmath>
 In the spirit of <acroref type="example" acro="SAA" />, we can express the infinitely many solutions of this system compactly with set notation.  The key is to express certain variables in terms of others.  More specifically, each pivot column number is the index of a variable that can be written in terms of the variables whose indices are non-pivot columns.  Or saying the same thing: for each $i$ in $D$, we can find an expression for $x_i$ in terms of the variables without their index in $D$.  Here $D=\set{1,\,3}$, so
 <alignmath>
 <![CDATA[x_1&=2-2x_2+4x_4\\]]>
 <![CDATA[x_3&=-6\quad\quad-3x_4]]>
 </alignmath>
 As a set, we write the solutions precisely as
-<equation>
+<alignmath>
 \setparts{\colvector{2-2x_2+4x_4\\x_2\\-6-3x_4\\x_4}}{x_2,\,x_4\in\complex{\null}}
-</equation>
+</alignmath>
 </solution>
 </exercise>
 
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