Question

Question 26: Let U be the matrix in Exercise 25. Find the distance from \({\mathop{\rm b}\nolimits} = \left( {1,1,1,1, - 1, - 1, - 1, - 1} \right)\) to Col U.

Short answer

The distance from yto \({\mathop{\rm Col}\nolimits} U\) is \(2.1166\).

Step-by-step solution

The Best Approximation Theorem

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 pointin \(W\) 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 \(\widehat {\bf{y}}\).

Determine the distance from b to Col U

The distance from b to Col U is given by \(\left\| {{\bf{b}} - \widehat {\bf{b}}} \right\|\), with \(\widehat {\bf{b}} = U{U^T}{\bf{b}}\) as:

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 25.

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

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

\(\widehat {\bf{b}} = \left[ {\begin{array}{*{20}{c}}{.2}\\{.92}\\{.44}\\1\\{ - .2}\\{ - .44}\\{.6}\\{ - .92}\end{array}} \right]\)

Use the MATLAB code to compute \({\bf{b}} - \widehat {\bf{b}}\) as shown below:

\(\begin{array}{l} > > {\bf{b}} = \left[ {1\,;\,1;\,\,\,1\,;\,\,1\,;\,\, - 1;\,\,\, - 1;\,\, - 1;\,\, - 1} \right]\\ > > \widehat {\bf{b}} = \left[ {.2;\,\,\,.92;\,\,.44;\,\,\,1;\,\, - .2;\,\, - .44;\,\,.6;\, - .92} \right]\\ > > {\bf{b}} - \widehat {\bf{b}}\end{array}\)

\({\bf{b}} - \widehat {\bf{b}} = \left[ {\begin{array}{*{20}{c}}{.8}\\{.08}\\{.56}\\0\\{ - .8}\\{ - .56}\\{ - 1.6}\\{ - .08}\end{array}} \right]\)

Compute \(\left\| {{\bf{b}} - \widehat {\bf{b}}} \right\|\) as shown below:

\(\begin{array}{c}\left\| {{\bf{b}} - \widehat {\bf{b}}} \right\| = \sqrt {{{\left( {.8} \right)}^2} + {{\left( {.08} \right)}^2} + {{\left( {.56} \right)}^2} + \left( 0 \right) + {{\left( { - .8} \right)}^2} + {{\left( { - .56} \right)}^2} + {{\left( { - 1.6} \right)}^2} + {{\left( { - .08} \right)}^2}} \\ = \frac{{\sqrt {112} }}{5}\end{array}\)

It implies that \(2.1166\) to four decimal places.

Thus, the distance from yto \({\mathop{\rm Col}\nolimits} U\) is \(2.1166\).