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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

Suppose\(A = PR{P^{ - {\bf{1}}}}\), where P is orthogonal and R is upper triangular. Show that if A is symmetric, then R is symmetric and hence is actually a diagonal matrix.

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.

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a. \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\)

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