% m211oh57.tex - Solution of the Distance Problems
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\begin{center}\tcr{\rmH Solution of the Two Distance Problems} \end{center} \vspace{-12pt}
\huge

\begin{description}  
\item[\tcr{Theorem 7:}] \tcg{(Geometry)}
If \tcb{$\vec{v}_{\scr0} \in V$} 
% = \langle \vec{v}_{\scr1}, \ldots, \vec{v}_{\scr k} \rangle$}
is such that
\beq
\tcb{\vec{v}\cdot(\vec{y} - \vec{v}_{\scr0}) = 0,\quad \mbox{for all }\vec{v} \in V,}
\label{1}
\eeq
(where \tcb{$V = \langle \vec{v}_{\scr1}, \ldots, \vec{v}_{\scr k} \rangle$}), then we have
\beq
\tcb{d(\vec{y}, \vec{v}_{\scr0}) < d(\vec{y}, \vec{v}), \;\; \mbox{for all } \vec{v} \in V, \vec{v} \ne \vec{v}_{\scr0},} \label{2}
\eeq
and hence %In particular, 
\tcb{$P_V(\vec{y}):= \vec{v}_{\scr0}$} is a \tcg{solution} to \tcr{Problem B}.

\item[\tcr{Theorem 8:}]  Let \mc{\vec y\in \R^n} and \tcb{$V = \langle \vec{v}_{\scr1}, \ldots, \vec{v}_{\scr k} \rangle \subset \R^n$} and put \tcb{$A = (\vec{v}_{\scr1} | \ldots | \vec{v}_k)$}.
If  \tcb{$\vec{v}_{\scr1}, \ldots, \vec{v}_{\scr k}$} are \tcr{linearly independent}, 
then \tcb{$A^tA$} is \tcr{invertible}, and the vector
\beq
\tcb{\vec{w}_{\scr0} := (A^tA)^{-1}A^t\vec{y}, \label{4}}
\eeq
has the property that %and hence\vspace{-3pt}
\beq
\label{4a}
\hspace*{20pt}\tcb{||A\vec w_{\scr0} - \vec y|| <  ||A\vec w - \vec y||,}\;\mbox{for all \mc{\vec w\in \R^k} with}
\eeq
\mc{\vec w\ne \vec w_{\scr0}}. Moreover, \mc{\vec w_{\scr0}} is the unique vector of \mc{\R^k}
such that \sref{4a} holds, and hence it is the unique solution of \tcr{Problem A.} Thus 
\mc{A\vec w_{\sc0}} is the unique solution of \tcr{Problem B,} i.e.,     
\beq
\tcb{P_V(\vec{y}) = A\vec{w}_{\scr0} = \underbrace{A(A^tA)^{-1}A^t}_{P_V}\vec{y} \label{5}}\vspace{-3pt}
\eeq
is the \tcg{orthogonal projection} of \tcb{$\vec{y}$} \tcg{onto} \tcb{$V$}.    
\end{description}
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\end{document}                          
