Find the maximum value of \(Q\left( {\rm{x}} \right) = 7x_1^2 + 3x_2^2 - 2x_1^{}x_2^{}\), 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: \({\lambda _1} = 5 + \sqrt 5 \).

Step by step solution

01

Find the eigenvalues

As per the question, we have:

\(Q\left( {\rm{x}} \right) = 7x_1^2 + 3x_2^2 - 2x_1^{}x_2^{}\)

This implies that corresponding matrix is\(A = \left[ {\begin{array}{*{20}{c}}7&{ - 1}\\{ - 1}&3\end{array}} \right]\).

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

\(\begin{array}{c}\det \left( {\left[ {\begin{array}{*{20}{c}}{7 - \lambda }&{ - 1}\\{ - 1}&{3 - \lambda }\end{array}} \right]} \right) = 0\\\left( {7 - \lambda } \right)\left( {3 - \lambda } \right) - \left( { - 1} \right)\left( { - 1} \right) = 0\\21 - 7\lambda - 3\lambda + {\lambda ^2} - 1 = 0\\{\lambda ^2} - 10\lambda + 20 = 0\\\lambda = \frac{{10 \pm \sqrt {100 - 80} }}{2}\\ = 5 \pm \sqrt 5 \end{array}\)

Thus, the eigenvalues are \({\lambda _1} = 5 + \sqrt 5 \) and \({\lambda _2} = 5 - \sqrt 5 \).

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:

\({\lambda _1} = 5 + \sqrt 5 \)

Hence, the required maximum value is at: \({\lambda _1} = 5 + \sqrt 5 \).

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Most popular questions from this chapter

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\).

a. \(3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}\)

b. \(6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\)

Question: 11. Given multivariate data \({X_1},................,{X_N}\) (in \({\mathbb{R}^p}\)) in mean deviation form, let \(P\) be a \(p \times p\) matrix, and define \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).

  1. Show that \({Y_1},................,{Y_N}\) are in mean-deviation form. (Hint: Let \(w\) be the vector in \({\mathbb{R}^N}\) with a 1 in each entry. Then \(\left( {{X_1},................,{X_N}} \right)w = 0\) (the zero vector in \({\mathbb{R}^p}\)).)
  2. Show that if the covariance matrix of \({X_1},................,{X_N}\) is \(S\), then the covariance matrix of \({Y_1},................,{Y_N}\) is \({P^T}SP\).

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.

20. Show that if\(A\)is an orthogonal\(m \times m\)matrix, then \(PA\) has the same singular values as \(A\).

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.

9. \({\bf{4}}x_{\bf{1}}^{\bf{2}} - {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{4}}x_{\bf{2}}^{\bf{2}}\)

Suppose A is invertible and orthogonally diagonalizable. Explain why \({A^{ - {\bf{1}}}}\) is also orthogonally diagonalizable.

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