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

[Hint: The eigenvalues of the matrix of the quadratic form \(Q\) are \(2, - 1,{\rm{ and }} - 4\).]

Short Answer

Expert verified

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 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{array}} \right]\).

Step by step solution

01

Determine the eigenvalues

As per the question, we have:

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

And the eigenvalues as:

\(\begin{array}{l}{\lambda _1} = 2\\{\lambda _2} = - 1\\{\lambda _3} = - 4\end{array}\)

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

So, we have:

\({\lambda _1} = 2\)

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 \({\lambda _1}\).

Find the eigenvector corresponding to \({\lambda _1} = 2\).

\({\rm{v}} = \left[ {\begin{array}{*{20}{c}}{1/2}\\1\\1\end{array}} \right]\)

The unit eigenvector that corresponds to the eigenvalue\({\lambda _1} = 2\)is:

\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\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 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{array}} \right]\).

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Most popular questions from this chapter

A certain experiment produce the data \(\left( {1,7.9} \right),\left( {2,5.4} \right)\) and \(\left( {3, - .9} \right)\). Describe the model that produces a least-squares fit of these points by a function of the form

\(y = A\cos x + B\sin x\)

In Exercises 9-12, find a unit vector in the direction of the given vector.

10. \(\left( {\begin{aligned}{*{20}{c}}{ - 6}\\4\\{ - 3}\end{aligned}} \right)\)

Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)—the sum of the squares of the “regression term.” Denote this number by .

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)—the sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)—the “total” sum of the squares of the \(y\)-values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of -values.

19. Justify the equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\). (Hint: Use a theorem, and explain why the hypotheses of the theorem are satisfied.) This equation is extremely important in statistics, both in regression theory and in the analysis of variance.

Find a \(QR\) factorization of the matrix in Exercise 11.

A simple curve that often makes a good model for the variable costs of a company, a function of the sales level \(x\), has the form \(y = {\beta _1}x + {\beta _2}{x^2} + {\beta _3}{x^3}\). There is no constant term because fixed costs are not included.

a. Give the design matrix and the parameter vector for the linear model that leads to a least-squares fit of the equation above, with data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\).

b. Find the least-squares curve of the form above to fit the data \(\left( {4,1.58} \right),\left( {6,2.08} \right),\left( {8,2.5} \right),\left( {10,2.8} \right),\left( {12,3.1} \right),\left( {14,3.4} \right),\left( {16,3.8} \right)\) and \(\left( {18,4.32} \right)\), with values in thousands. If possible, produce a graph that shows the data points and the graph of the cubic approximation.

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