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Chapter 7: Q-7.3-5E (page 395)

In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).

5. \(Q\left( x \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\).

Short Answer

Expert verified

The required values are:

  1. The maximum value of\(Q\left( {\rm{x}} \right)\)the subject to the constraint\({{\rm{x}}^T}{\rm{x}} = 1\)is\({\lambda _1} = 6\).
  2. A unit vector\({\rm{u}}\)where this maximum is attained is\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{{ - 1} \mathord{\left/
  3. {\vphantom {{ - 1} {\sqrt 2 }}} \right.
  4. \kern-\nulldelimiterspace} {\sqrt 2 }}}\\{{1 \mathord{\left/
  5. {\vphantom {1 {\sqrt 2 }}} \right.
  6. \kern-\nulldelimiterspace} {\sqrt 2 }}}\end{array}} \right]\).
  7. The maximum of\(Q\left( {\rm{x}} \right)\)subject to the constraints\({{\rm{x}}^T}{\rm{x}} = 1\)and \({{\rm{x}}^T}{\rm{u}} = 0\) is\({\lambda _2} = - 4\).

Step by step solution

01

Find the greatest eigenvalue

As per the question, we have:

\(Q\left( {\rm{x}} \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\)

The matrix with its eigenvalues has the form:

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&{ - 5}\\{ - 5}&1\end{array}} \right]\\ \Rightarrow {\lambda _1} = 6\;\;\& \,\,\,{\lambda _2} = - 4\end{array}\)

The maximum value of the given function subjected to constraints\({{\rm{x}}^T}{\rm{x}} = 1\)will be the greatest eigenvalue.

So, we have:

\({\lambda _1} = 6\)

Hence,the maximum value of\(Q\left( {\rm{x}} \right)\)the subject to the constraint\({{\rm{x}}^T}{\rm{x}} = 1\)is\({\lambda _1} = 6\).

02

Find the vector for this greatest eigenvalue

Apply the theorem which states that the value of\({{\rm{x}}^T}A{\rm{x}}\) is maximum when \({\rm{x}}\) is a unit eigenvector \({{\rm{u}}_1}\) corresponding to the greatest eigenvalue \({\lambda _1}\).

Find the eigenvector corresponding to \({\lambda _1} = 6\).

\(\begin{array}{c}A - 6I = \left[ {\begin{array}{*{20}{c}}{1 - 6}&{ - 5}\\{ - 5}&{1 - 6}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\{ - 5}&{ - 5}\end{array}} \right]\end{array}\)

Theorem the null space is \(\left[ {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right]\).

The eigenvector that corresponds to eigenvalue\({\lambda _1} = 6\)is:

\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{{ - 1} \mathord{\left/

{\vphantom {{ - 1} {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\end{array}} \right]\).

Hence,a unit vector\({\rm{u}}\)where this maximum is attained is\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{{ - 1} \mathord{\left/

{\vphantom {{ - 1} {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\end{array}} \right]\).

03

Find the second greatest eigenvalue 

The maximum value of the given function is subjected to constraints\({{\rm{x}}^T}{\rm{x}} = 1\)and\({{\rm{x}}^T}{\rm{u}} = 0\)will be the second-largest eigenvalue. That is:

\({\lambda _2} = - 4\)

Hence,the maximum of\(Q\left( {\rm{x}} \right)\)subject to the constraints\({{\rm{x}}^T}{\rm{x}} = 1\)and \({{\rm{x}}^T}{\rm{u}} = 0\) is\({\lambda _2} = - 4\).

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Most popular questions from this chapter

Question: 13. Exercises 12鈥�14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Suppose the equation\(A{\rm{x}} = {\rm{b}}\)is consistent, and let\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\). By Exercise 23 in Section 6.3, there is exactly one vector\({\rm{p}}\)in Row\(A\)such that\(A{\rm{p}} = {\rm{b}}\). The following steps prove that\({{\rm{x}}^ + } = {\rm{p}}\)and\({{\rm{x}}^ + }\)is the minimum length solution of\(A{\rm{x}} = {\rm{b}}\).

  1. Show that \({{\rm{x}}^ + }\) is in Row \(A\). (Hint: Write \({\rm{b}}\) as \(A{\rm{x}}\) for some \({\rm{x}}\), and use Exercise 12.)
  2. Show that\({{\rm{x}}^ + }\)is a solution of\(A{\rm{x}} = {\rm{b}}\).
  3. Show that if \({\rm{u}}\) is any solution of \(A{\rm{x}} = {\rm{b}}\), then \(\left\| {{{\rm{x}}^ + }} \right\| \le \left\| {\rm{u}} \right\|\), with equality only if \({\rm{u}} = {{\rm{x}}^ + }\).

Question: [M] The covariance matrix below was obtained from a Landsat image of the Columbia River in Washington, using data from three spectral bands. Let \({x_1},{x_2},{x_3}\) denote the spectral components of each pixel in the image. Find a new variable of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2} + {c_3}{x_3}\) that has maximum possible variance, subject to the constraint that \(c_1^2 + c_2^2 + c_3^2 = 1\). What percentage of the total variance in the data is explained by \({y_1}\)?

\[S = \left[ {\begin{array}{*{20}{c}}{29.64}&{18.38}&{5.00}\\{18.38}&{20.82}&{14.06}\\{5.00}&{14.06}&{29.21}\end{array}} \right]\]

(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.

26. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{18}}}&{{\bf{13}}}&{ - {\bf{4}}}&{\bf{4}}\\{\bf{2}}&{{\bf{19}}}&{ - {\bf{4}}}&{{\bf{12}}}\\{ - {\bf{14}}}&{{\bf{11}}}&{ - {\bf{12}}}&{\bf{8}}\\{ - {\bf{2}}}&{{\bf{21}}}&{\bf{4}}&{\bf{8}}\end{array}} \right)\)

Classify the quadratic forms in Exercises 9鈥�18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

14. \({\bf{3}}x_{\bf{1}}^{\bf{2}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}}\)

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.
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