Question 25: Let U be the \(8 \times 4\) matrix in Exercise 36 in Section 6.2. Find the closest point to \({\bf{y}} = \left( {1,1,1,1,1,1,1,1} \right)\) in Col U. Write the keystrokes or commands you use to solve this problem.

Consider \(W\) as a subspace of \({\mathbb{R}^n}\), and assume that y as any vector in \({\mathbb{R}^n}\) and \(\widehat {\bf{y}}\) as the orthogonal projection of y onto \(W\). Then, theclosest point in \(W\) to \({\bf{y}}\) is \(\widehat {\bf{y}}\), such that \(\left\| {{\bf{y}} - \widehat {\bf{y}}} \right\| < \left\| {{\bf{y}} - {\bf{v}}} \right\|\) for every \({\bf{v}}\) in \(W\) distinct from \(\widehat {\bf{y}}\).

Theorem 6states that a \(m \times n\) matrix \(U\) contains orthonormal columnssuch that \({U^T}U = I\).

According to theorem 6, \(U\) contains orthonormal columns because \({U^T}U = {I_4}\).

The orthogonal projection \(\widehat {\bf{y}}\)of \({\bf{y}}\) onto Col U is known as the closest point to y in Col U.

Consider that \(U = \left( {\begin{array}{*{20}{c}}{ - 6}&{ - 3}&6&1\\{ - 1}&2&1&{ - 6}\\3&6&3&{ - 2}\\6&{ - 3}&6&{ - 1}\\2&{ - 1}&2&3\\{ - 3}&6&3&2\\{ - 2}&{ - 1}&2&{ - 3}\\1&2&1&6\end{array}} \right)\) as in Exercise 36.

Use the MATLAB code to compute \(\widehat {\bf{y}} = U{U^T}{\bf{y}}\) by theorem 10 as shown below:

\(\begin{array}{c} > > U = \left( \begin{array}{l} - 6\,\,\, - 3\,\,\,6\,\,\,1;\, - 1\,\,\,2\,\,\,\,1\,\,\, - 6;\,3\,\,\,6\,\,\,3\,\,\, - 2;\,\,6\,\,\, - 3\,\,\,6\,\,\, - 1;\,\\2\,\,\, - 1\,\,\,2\,\,\,\,3;\,\, - 3\,\,\,\,6\,\,\,3\,\,\,2;\,\, - 2\,\,\, - 1\,\,\,2\,\,\, - 3;1\,\,\,2\,\,\,1\,\,\,6\end{array} \right)\\ > > {\bf{y}} = \left( {1\,;\,1;\,\,\,1\,;\,\,1\,;\,\,1;\,\,\,1;\,\,\,1;\,\,\,1} \right)\\ > > \widehat {\bf{y}} = U * U' * {\bf{y}}\end{array}\)

\(\widehat {\bf{y}} = \left( {\begin{array}{*{20}{c}}{1.2}\\{.4}\\{1.2}\\{1.2}\\{.4}\\{1.2}\\{.4}\\{.4}\end{array}} \right)\)

Thus, the closest point is \(\widehat {\bf{y}} = \left( {\begin{array}{*{20}{c}}{1.2}\\{.4}\\{1.2}\\{1.2}\\{.4}\\{1.2}\\{.4}\\{.4}\end{array}} \right)\).