Question

Consider the points in Exercise 5 in section 8.1 which of \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), and \({{\bf{p}}_{\bf{3}}}\) are in conv S?

Short answer

Only \({{\bf{p}}_2}\) is in Conv S.

Step-by-step solution

Step 1:Find the projection of \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), and \({{\bf{p}}_{\bf{3}}}\)

As S is an orthogonal set, the projection of \({{\bf{p}}_1}\) is as follows:

\({\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_1} = \frac{{{{\bf{p}}_1} \cdot {{\bf{b}}_1}}}{{{{\bf{b}}_1} \cdot {{\bf{b}}_1}}}{{\bf{b}}_1} + \frac{{{{\bf{p}}_1} \cdot {{\bf{b}}_2}}}{{{{\bf{b}}_2} \cdot {{\bf{b}}_2}}}{{\bf{b}}_2} + \frac{{{{\bf{p}}_1} \cdot {{\bf{b}}_3}}}{{{{\bf{b}}_3} \cdot {{\bf{b}}_3}}}{{\bf{b}}_3}\)

The projection of \({{\bf{p}}_{\bf{2}}}\) is as follows:

\({\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_2} = \frac{{{{\bf{p}}_2} \cdot {{\bf{b}}_1}}}{{{{\bf{b}}_1} \cdot {{\bf{b}}_1}}}{{\bf{b}}_1} + \frac{{{{\bf{p}}_2} \cdot {{\bf{b}}_2}}}{{{{\bf{b}}_2} \cdot {{\bf{b}}_2}}}{{\bf{b}}_2} + \frac{{{{\bf{p}}_2} \cdot {{\bf{b}}_3}}}{{{{\bf{b}}_3} \cdot {{\bf{b}}_3}}}{{\bf{b}}_3}\)

The projection of \({{\bf{p}}_3}\) is as follows:

\({\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_3} = \frac{{{{\bf{p}}_3} \cdot {{\bf{b}}_1}}}{{{{\bf{b}}_1} \cdot {{\bf{b}}_1}}}{{\bf{b}}_1} + \frac{{{{\bf{p}}_3} \cdot {{\bf{b}}_2}}}{{{{\bf{b}}_2} \cdot {{\bf{b}}_2}}}{{\bf{b}}_2} + \frac{{{{\bf{p}}_3} \cdot {{\bf{b}}_3}}}{{{{\bf{b}}_3} \cdot {{\bf{b}}_3}}}{{\bf{b}}_3}\)

Here \({{\bf{b}}_1}\), \({{\bf{b}}_2}\), and \({{\bf{b}}_3}\) are the element of orthogonal set S and\({{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\), \({{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\),and\({{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\).

Find the product in the projection of \({{\bf{p}}_{\bf{1}}}\)

The product for projection\({{\bf{p}}_{\bf{1}}}\) is as follows:

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 19}\\{ - 5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = - 24\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 19}\\{ - 5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 19}\\{ - 5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 90\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

Find the product in the projection of \({{\bf{p}}_{\bf{2}}}\)

The product for projection of \({{\bf{p}}_2}\) is as follows:

\(\begin{array}{c}{{\bf{p}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{ - 1.3}\\{ - 0.5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 1.2\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{ - 1.3}\\{ - 0.5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 2.5\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_2} \cdot {{\bf{b}}_3} = \left[ {\begin{array}{*{20}{c}}{1.5}\\{ - 1.3}\\{ - 0.5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 9.0\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

Find the product in the projection of \({{\bf{p}}_{\bf{3}}}\)

The product for projection of \({{\bf{p}}_3}\) is as follows:

\(\begin{array}{c}{{\bf{p}}_3} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_3} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_3} \cdot {{\bf{b}}_3} = \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

Substitute values in the equation of projections

The projection \({{\bf{p}}_1}\)is as follows:

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_1} = \left( { - \frac{{24}}{6}} \right){{\bf{b}}_1} + \frac{{10}}{5}{{\bf{b}}_2} + \frac{{90}}{{30}}{{\bf{b}}_3}\\ = - 4{{\bf{b}}_1} + 2{{\bf{b}}_2} + 3{{\bf{b}}_3}\end{array}\)

The projection \({{\bf{p}}_2}\) is as follows:

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_2} = \left( {\frac{{1.2}}{6}} \right){{\bf{b}}_1} + \left( {\frac{{2.5}}{5}} \right){{\bf{b}}_2} + \left( {\frac{9}{{30}}} \right){{\bf{b}}_3}\\ = 0.2{{\bf{b}}_1} + 0.5{{\bf{b}}_2} + 0.3{{\bf{b}}_3}\end{array}\)

The projection \({{\bf{p}}_3}\) is as follows:

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_3} = \frac{6}{6}{{\bf{b}}_1} + \frac{5}{5}{{\bf{b}}_2} + \frac{{30}}{{30}}{{\bf{b}}_3}\\ = {{\bf{b}}_1} + {{\bf{b}}_2} + {{\bf{b}}_3}\end{array}\)

Following projections can be made from the projections:

1. For \({{\bf{p}}_1}\), all coefficients are not positive, so \({{\bf{p}}_1} \notin {\rm{conv}}\,S\).
2. For \({{\bf{p}}_2}\), all coefficients are positive but \(\left( {0.2 + 0.5 + 0.3 = 1} \right)\), so \({{\bf{p}}_1} \in {\rm{conv}}\,S\).
3. For \({{\bf{p}}_3}\), all coefficients are positive but \(\left( {1 + 1 + 1 = 3 \ne 1} \right)\), so \({{\bf{p}}_1} \notin {\rm{conv}}\,S\).

Therefore, only \({{\bf{p}}_2}\) is in Conv S.