1. rbeezer
  2. fcla


rbeezer  committed 3b066ad

Line break, Section LDS

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 <p>First show that $V^\prime\subseteq V$.  Since every vector of $R^\prime$ is in $R$, any vector we can construct in $V^\prime$ as a linear combination of vectors from $R^\prime$ can also be constructed as a vector in $V$ by the same linear combination of the same vectors in $R$.  That was easy, now turn it around.</p>
-<p>Next show that $V\subseteq V^\prime$.  Choose any $\vect{v}$ from $V$.  Then there are scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4$ so that
+<p>Next show that $V\subseteq V^\prime$.  Choose any $\vect{v}$ from $V$.  So there are scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4$ such that