Chapter 6: Q23E (page 331)
Suppose \(A\) is \(m \times n\) with linearly independent columns and \(b\) is in \({\mathbb{R}^m}\). Use the normal equations to produce a formula for \(\hat b\), the projection of b onto \({\rm{Col}}\,A\).
The formula is given by, \(\hat b = A{\left( {{A^T}A} \right)^{ - 1}}{A^T}b\).
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Find the distance between \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right)\) and \({\mathop{\rm z}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\).
In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).
3. \(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{ - {\bf{2}}}\\{ - {\bf{1}}}&{\bf{2}}\\{\bf{0}}&{\bf{3}}\\{\bf{2}}&{\bf{5}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{4}}}\\{\bf{2}}\end{aligned}} \right)\)
In exercises 1-6, determine which sets of vectors are orthogonal.
\(\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right]\), \(\left[ {\begin{align}0\\0\\0\end{align}} \right]\), \(\left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]\)
In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).
4. \(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{\bf{3}}\\{\bf{1}}&{ - {\bf{1}}}\\{\bf{1}}&{\bf{1}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{5}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right)\)
Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).
3.\[y = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{4}}\\{\bf{3}}\end{aligned}} \right]\],\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\)
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