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Chapter 7: Q-7.3-10E (page 395)

Find the maximum value of \(Q\left( {\rm{x}} \right) = - 3x_1^2 + 5x_2^2 - 2x_1^{}x_2^{}\), subject to the constraint \(x_1^2 + x_2^2 = 1\). (Do not go on to find a vector where the maximum is attended.)

Short Answer

Expert verified

The required maximum value is at: \({\lambda _1} = 1 + \sqrt {17} \).

Step by step solution

01

Find the eigenvalues

As per the question, we have:

\(Q\left( {\rm{x}} \right) = - 3x_1^2 + 5x_2^2 - 2x_1^{}x_2^{}\)

This implies that the corresponding matrix is\(A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 1}\\{ - 1}&5\end{array}} \right]\).

Find the eigenvalues by solving \(\det \left( {A - \lambda I} \right) = 0\).

\(\begin{array}{c}\det \left( {\left[ {\begin{array}{*{20}{c}}{ - 3 - \lambda }&{ - 1}\\{ - 1}&{5 - \lambda }\end{array}} \right]} \right) = 0\\\left( { - 3 - \lambda } \right)\left( {5 - \lambda } \right) - \left( { - 1} \right)\left( { - 1} \right) = 0\\ - 15 + 3\lambda - 5\lambda + {\lambda ^2} - 1 = 0\\{\lambda ^2} - 2\lambda - 16 = 0\\\lambda = \frac{{2 \pm \sqrt {68} }}{2}\\ = 1 \pm \sqrt {17} \end{array}\)

Thus, the eigenvalues are \({\lambda _1} = 1 + \sqrt {17} \) and \({\lambda _2} = 1 - \sqrt {17} \).

02

Find the maximum value

The maximum value of the given function subjected to constraints\(x_1^2 + x_2^2 = 1\)will be the greatest eigenvalue.

So, we have:

\({\lambda _1} = 1 + \sqrt {17} \)

The required maximum value is at:\({\lambda _1} = 1 + \sqrt {17} \).

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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 \({\rm{u}} = {{\rm{x}}^ + }\).

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.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

Question: If A is \(m \times n\), then the matrix \(G = {A^T}A\) is called the Gram matrix of A. In this case, the entries of G are the inner products of the columns of A. (See Exercises 9 and 10).

9. Show that the Gram matrix of any matrix A is positive semidefinite, with the same rank as A. (See the Exercises in Section 6.5.)

Question: Mark Each statement True or False. Justify each answer. In each part, A represents an \(n \times n\) matrix.

  1. If A is orthogonally diagonizable, then A is symmetric.
  2. If A is an orthogonal matrix, then A is symmetric.
  3. If A is an orthogonal matrix, then \(\left\| {A{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).
  4. The principal axes of a quadratic from \({{\bf{x}}^T}A{\bf{x}}\) can be the columns of any matrix P that diagonalizes A.
  5. If P is an \(n \times n\) matrix with orthogonal columns, then \({P^T} = {P^{ - {\bf{1}}}}\).
  6. If every coefficient in a quadratic form is positive, then the quadratic form is positive definite.
  7. If \({{\bf{x}}^T}A{\bf{x}} > {\bf{0}}\) for some x, then the quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is positive definite.
  8. By a suitable change of variable, any quadratic form can be changed into one with no cross-product term.
  9. The largest value of a quadratic form \({{\bf{x}}^T}A{\bf{x}}\), for \(\left\| {\bf{x}} \right\| = {\bf{1}}\) is the largest entery on the diagonal A.
  10. The maximum value of a positive definite quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is the greatest eigenvalue of A.
  11. A positive definite quadratic form can be changed into a negative definite form by a suitable change of variable \({\bf{x}} = P{\bf{u}}\), for some orthogonal matrix P.
  12. An indefinite quadratic form is one whose eigenvalues are not definite.
  13. If P is an \(n \times n\) orthogonal matrix, then the change of variable \({\bf{x}} = P{\bf{u}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form whose matrix is \({P^{ - {\bf{1}}}}AP\).
  14. If U is \(m \times n\) with orthogonal columns, then \(U{U^T}{\bf{x}}\) is the orthogonal projection of x onto ColU.
  15. If B is \(m \times n\) and x is a unit vector in \({\mathbb{R}^n}\), then \(\left\| {B{\bf{x}}} \right\| \le {\sigma _{\bf{1}}}\), where \({\sigma _{\bf{1}}}\) is the first singular value of B.
  16. A singular value decomposition of an \(m \times n\) matrix B can be written as \(B = P\Sigma Q\), where P is an \(m \times n\) orthogonal matrix and \(\Sigma \) is an \(m \times n\) diagonal matrix.
  17. If A is \(n \times n\), then A and \({A^T}A\) have the same singular values.

Question 8: Use Exercise 7 to show that if A is positive definite, then A has a LU factorization, \(A = LU\), where U has positive pivots on its diagonal. (The converse is true, too).
See all solutions

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