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Chapter 7: Q-7.3-8E (page 395)

Let \(Q\left( x \right) = 7x_1^2 + x_2^2 + 7x_3^2 - 8x_1^{}x_2^{} - 4x_1^{}x_3^{} - 8x_2^{}x_3^{}\). Find a unit vector \(x{\rm{ in }}{\mathbb{R}^3}\) at which \(Q\left( x \right)\) is maximized, subject to \({x^T}x = 1\).

[Hint: The eigenvalues of the matrix of the quadratic form \(Q\) are \(9,{\rm{ and }} - 3\).]

Short Answer

Expert verified

A unit vector \({\rm{x}}\) in \({\mathbb{R}^3}\) at which \(Q\left( {\rm{x}} \right)\) is maximized, subject to \({{\rm{x}}^T}{\rm{x}} = 1\) is \({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right]\).

Step by step solution

01

Find the eigenvalues 

As per the question, we have:

\(Q\left( {\rm{x}} \right) = 7x_1^2 + x_2^2 + 7x_3^2 - 8x_1^{}x_2^{} - 4x_1^{}x_3^{} - 8x_2^{}x_3^{}\)

And the eigenvalues as:

\(\begin{array}{l}{\lambda _1} = 9\\{\lambda _2} = - 3\end{array}\)

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

So, we have:

\({\lambda _1} = 9\)

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 \({\lambda _1}\).

The eigenvector corresponding to \({\lambda _1} = 9\)are a linear combination of\(\left[ {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right]\)and \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right]\)

The unit eigenvector that corresponds to the eigenvalue\({\lambda _1} = 9\)is:

\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right]\).

Hence, aunit vector \({\rm{x}}\) in \({\mathbb{R}^3}\) at which \(Q\left( {\rm{x}} \right)\) is maximized, subject to \({{\rm{x}}^T}{\rm{x}} = 1\) is \({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right]\).

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Most popular questions from this chapter

In Exercises 17鈥�24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) ,

\(\)

where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) 鈥渄iagonal鈥� matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix \(P\) and a diagonal matrix \(D\). To save you time, the eigenvalues in Exercises 17鈥�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.

21. \(\left( {\begin{aligned}{{}}4&3&1&1\\3&4&1&1\\1&1&4&3\\1&1&3&4\end{aligned}} \right)\)

Question: 14. 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 \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Given any \({\rm{b}}\) in \({\mathbb{R}^m}\), adapt Exercise 13 to show that \({A^ + }{\rm{b}}\) is the least-squares solution of minimum length. [Hint: Consider the equation \(A{\rm{x}} = {\rm{b}}\), where \(\mathop {\rm{b}}\limits^\^ \) is the orthogonal projection of \({\rm{b}}\) onto Col \(A\).

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

Classify the quadratic forms in Exercises 9-18. 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.

10. \({\bf{2}}x_{\bf{1}}^{\bf{2}} + {\bf{6}}{x_{\bf{1}}}{x_{\bf{2}}} - {\bf{6}}x_{\bf{2}}^{\bf{2}}\)
See all solutions

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