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Chapter 7: Q19E (page 395)

What is the largest possible value of the quadratic form \({\bf{5}}x_{\bf{1}}^{\bf{2}} + {\bf{8}}x_{\bf{2}}^{\bf{2}}\) if \({\bf{x}} = \left( {{x_{\bf{1}}},{x_{\bf{2}}}} \right)\) and \({{\bf{x}}^T}{\bf{x = 1}}\), that is, if \(x_{\bf{1}}^{\bf{2}} + x_{\bf{2}}^{\bf{2}} = {\bf{1}}\)? (Try some examples of \({\bf{x}}\).

Short Answer

Expert verified

The largest possible value of the given quadratic form is \(8\).

Step by step solution

01

Find the matrix form of the given quadratic form

Consider the quadratic form \(5x_1^2 + 8x_2^2\),

\(\begin{aligned}{}5x_1^2 + 8x_2^2 &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}5&0\\0&8\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= {{\bf{x}}^T}A{\bf{x}}\end{aligned}\)

02

Step 2: Find the eigen vector

As from the diagonal matrix, the eigenvalues are \(5\) and \(8\). Therefore, eigenvectors are shown below:

\(\lambda = 8:\left( {\begin{aligned}{{}}0\\{ \pm 1}\end{aligned}} \right)\)and \(\lambda = 5:\left( {\begin{aligned}{{}}{ \pm 1}\\0\end{aligned}} \right)\)

Since the vectors are in normalized form thus, the largest possible value of the given quadratic form is \(8\).

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

In Exercises 17–24, \(A\) 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 \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

22. Show that if \(A\) is an \(n \times n\) positive definite matrix, then an orthogonal diagonalization \(A = PD{P^T}\) is a singular value decomposition of \(A\).

Let \(A = PD{P^{ - {\bf{1}}}}\), where P is orthogonal and D is diagonal, and let \(\lambda \) be an eigenvalue of A of multiplicity k. Then \(\lambda \) appears k times on the diagonal of D.Explain why the dimension of the eigenspace for \(\lambda \) is k.

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.

22. \(\left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\)

Question: 14. Exercises 12–14 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 \(A\).
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