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

None of the points are 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:

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}3\\8\\4\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 18\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}3\\8\\4\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = - 5\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}3\\8\\4\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

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

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

\(\begin{array}{c}{{\bf{p}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}6\\{ - 3}\\3\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 12\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{1}}} \cdot {{\bf{b}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\1\\1\end{array}} \right]\\ = 6\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}6\\{ - 3}\\3\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 0\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

\(\begin{array}{c}{{\bf{p}}_2} \cdot {{\bf{b}}_3} = \left[ {\begin{array}{*{20}{c}}6\\{ - 3}\\3\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{3}}} \cdot {{\bf{b}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}2\\{ - 5}\\1\end{array}} \right]\\ = 30\end{array}\)

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

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

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

And

\(\begin{array}{c}{{\bf{p}}_3} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}0\\{ - 1}\\{ - 5}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 10\end{array}\), \(\begin{array}{c}{{\bf{b}}_{\bf{2}}} \cdot {{\bf{b}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\ = 5\end{array}\)

And

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

Substitute values in the equation of projections

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

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_1} = \frac{{18}}{6}{{\bf{b}}_1} + \frac{{\left( { - 5} \right)}}{5}{{\bf{b}}_2} + \frac{{\left( { - 30} \right)}}{{30}}{{\bf{b}}_3}\\ = 3{{\bf{b}}_1} - {{\bf{b}}_2} - {{\bf{b}}_3}\end{array}\)

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

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_2} = \left( {\frac{{12}}{6}} \right){{\bf{b}}_1} + \left( {\frac{0}{5}} \right){{\bf{b}}_2} + \left( {\frac{{30}}{{30}}} \right){{\bf{b}}_3}\\ = 2{{\bf{b}}_1} + 0{{\bf{b}}_2} + {{\bf{b}}_3}\end{array}\)

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

\(\begin{array}{c}{\rm{pro}}{{\rm{j}}_w}{{\bf{p}}_3} = \left( { - \frac{6}{6}} \right){{\bf{b}}_1} + \left( {\frac{{10}}{5}} \right){{\bf{b}}_2} + \left( {\frac{0}{{30}}} \right){{\bf{b}}_3}\\ = - {{\bf{b}}_1} + 2{{\bf{b}}_2} + 0{{\bf{b}}_3}\end{array}\)

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 \(\left( {2 + 0 + 1 = 3 \ne 1} \right)\), so \({{\bf{p}}_1} \notin {\rm{conv}}\,S\).
3. For \({{\bf{p}}_3}\), all coefficients are not positive, so \({{\bf{p}}_1} \notin {\rm{conv}}\,S\).

Therefore, none of the points are in conv S.