% m211og64.tex  - Review: Dimension of linear sets (new Feb 22)
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\begin{center}\tcr{\rmH Review: Dimensions of Linear Sets} \end{center} %\vspace{0pt}
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\begin{description}  
\item[\tcr{Recall:}] If \mc{V \subset \R^n} is a \tcg{linear space,} then its \tcr{dimension}
is \mc{\dim(V) = |\cal B|,} where \mc{\cal B} is any \tcr{basis} of \mc{V.} 

If  \mc{S \subset \R^n} is a \tcg{linear set,} then \mc{S = \vec a + V,} for some/any \mc{\vec a \in S,} 
and a (unique) subspace \mc{V\subset \R^n.} Its \tcr{dimension} is \mc{\dim(S) := \dim(V).}

\item[\tcr{Computations:}] 1) If \mc{V = \lb\vec v_1,\ldots, \vec v_k\rb \subset \R^n} is a linear space (spanned by \mc{\vec v_1,\ldots, \vec v_k\in\R^n}), put 
$$
\tcb{A \;=\; (\vec v_1|\ldots|\vec v_k),}\quad \mbox{so }\quad \tcb{V \;=\; \mbox{Colsp}(A)}   
$$ 
is the \tcr{column space of} \mc{A.}
Then
$$
 \tcb{\dim(V) = \mbox{rank}(A),} 
$$
and a basis of \mc{V} can be found by using the recipe given in class
(using a \tcr{row echelon form} \mc{R} of \mc{A}).

2) If \mc{S = \mbox{Solsp}(A\vec x = \vec b) \subset \R^n} is the linear set defined by the 
system of equations \mc{A\vec x = \vec b,} then 
$$
\tcb{\dim(S) \;=\; n - \mbox{rank(A)},}
$$  
and a \tcr{vector equation} for \mc{S} can be found by \tcg{row reduction} and \tcg{back-substitution.}

Note that the number \mc{d} of ``free parameters'' \mc{t_1,\ldots, t_d} in the vector equation of \mc{S} 
(obtained by this method) equals the dimension of \mc{S}, i.e., \mc{d = \dim(S).} 
\end{description}
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