# Commits

committed 78e87c3

LaTeX/XML formatting in Solution LDS.T40 (Vladimir Yelkhimov)

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# File changes.txt

 Typo: Sage RLD, "nest" to "next" (Gavin Tranter)
 Typo: Theorem OSLI, "nozero" to "nonzero"  (Anna Dovzhik)
 Typo: Solution LDS.T55, B should be C at very end (Chris Spalding)
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+Typo: LaTeX/XML formatting in Solution LDS.T40 (Vladimir Yelkhimov)

 v3.10 2013/08/20
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# File src/section-LDS.xml

 </problem>
 <solution contributor="robertbeezer">This is an equality of sets, so <acroref type="definition" acro="SE" /> applies.<br /><br />
 The <q>easy</q> half first.  Show that $X=\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}\subseteq -\spn{\set{\vect{v}_1,\,\vect{v}_2}}=Y$.\\
+\spn{\set{\vect{v}_1,\,\vect{v}_2}}=Y$.<br /><br />  Choose$\vect{x}\in X$. Then $\vect{x}=a_1(\vect{v}_1+\vect{v}_2)+a_2(\vect{v}_1-\vect{v}_2)$for some scalars$a_1$and$a_2. Then,  <alignmath>  <![CDATA[&=(a_1+a_2)\vect{v}_1+(a_1-a_2)\vect{v}_2]]>  </alignmath>  which qualifies\vect{x}$for membership in$Y$, as it is a linear combination of$\vect{v}_1,\,\vect{v}_2$.<br /><br /> -Now show the opposite inclusion,$Y=\spn{\set{\vect{v}_1,\,\vect{v}_2}}\subseteq\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}=X$.\\ +Now show the opposite inclusion,$Y=\spn{\set{\vect{v}_1,\,\vect{v}_2}}\subseteq\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}=X$.<br /><br />  Choose$\vect{y}\in Y$. Then there are scalars$b_1,\,b_2$such that$ \vect{y}=b_1\vect{v}_1+b_2\vect{v}_2 \$.  Rearranging, we obtain,
 <alignmath>
 \vect{y}