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

Expert verified

None of the points are in conv S.

Step by step solution

01

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

02

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

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

03

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

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

04

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

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

05

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.

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Most popular questions from this chapter

Question 11: Prove that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\), where P is a \(n \times n\) positive semidefinite matrix with the same rank as A and where Q is an \(n \times n\) orthogonal matrix. (Hint: Use a singular value decomposition, \(A = U\sum {V^T}\), and observe that \(A = \left( {U\sum {U^T}} \right)\left( {U{V^T}} \right)\).) This decomposition is used, for instance, in mechanical engineering to model the deformation of a material. The matrix P describe the stretching or compression of the material (in the directions of the eigenvectors of P), and Q describes the rotation of the material in space.

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

19. \(\left( {\begin{aligned}{{}}3&{ - 2}&4\\{ - 2}&6&2\\4&2&3\end{aligned}} \right)\)

(M) Compute an SVD of each matrix in Exercises 26 and 27. Report the final matrix entries accurate to two decimal places. Use the method of Examples 3 and 4.

27. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{6}}&{ - {\bf{8}}}&{ - {\bf{4}}}&{\bf{5}}&{ - {\bf{4}}}\\{\bf{2}}&{\bf{7}}&{ - {\bf{5}}}&{ - {\bf{6}}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{1}}}&{ - {\bf{8}}}&{\bf{2}}&{\bf{2}}\\{ - {\bf{1}}}&{ - {\bf{2}}}&{\bf{4}}&{\bf{4}}&{ - {\bf{8}}}\end{array}} \right)\)

In Exercises 25 and 26, mark each statement True or False. Justify each answer.

a. An\(n \times n\)matrix that is orthogonally diagonalizable must be symmetric.

b. If\({A^T} = A\)and if vectors\({\rm{u}}\)and\({\rm{v}}\)satisfy\(A{\rm{u}} = {\rm{3u}}\)and\(A{\rm{v}} = {\rm{3v}}\), then\({\rm{u}} \cdot {\rm{v}} = {\rm{0}}\).

c. An\(n \times n\)symmetric matrix has n distinct real eigenvalues.

d. For a nonzero \({\rm{v}}\) in \({\mathbb{R}^n}\) , the matrix \({\rm{v}}{{\rm{v}}^T}\) is called a projection matrix.

Question: 2. Let \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},....,{{\bf{u}}_n}} \right\}\) be an orthonormal basis for \({\mathbb{R}_n}\) , and let \({\lambda _1},....{\lambda _n}\) be any real scalars. Define

\(A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + ..... + {\lambda _n}{\bf{u}}_n^T\)

a. Show that A is symmetric.

b. Show that \({\lambda _1},....{\lambda _n}\) are the eigenvalues of A

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