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Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and

a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

Short Answer

Expert verified

a. For \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\)

b.For \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 185\)

c.For \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 16\)

Step by step solution

01

Find \({x^T}Ax\)

(a)

The given matrix is\(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\), and the given vector is \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\).

Find \({{\bf{x}}^T}A{\bf{x}}\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= {\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{5{x_1} + \frac{1}{3}{x_2}}\\{\frac{1}{3}{x_1} + {x_2}}\end{aligned}} \right)\\ &= {x_1}\left( {5{x_1} + \frac{1}{3}{x_2}} \right) + {x_2}\left( {\frac{1}{3}{x_1} + {x_2}} \right)\\ &= 5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2{\rm{ }}\left( 1 \right)\end{aligned}\)

Hence, the expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\).

02

Find \({x^T}Ax\) for \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

(b)

As the expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) when \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\).

So, for \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\), simplify \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) by substituting \({x_1} = 6,{x_2} = 1\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 5{\left( 6 \right)^2} + \left( {\frac{2}{3}} \right)\left( 6 \right)\left( 1 \right) + {\left( 1 \right)^2}\\ &= 185\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is 185.

03

Find \({x^T}Ax\) for  \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

(c)

For \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\), simplify \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) by substituting \({x_1} = 1,{x_2} = 3\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 5{\left( 1 \right)^2} + \left( {\frac{2}{3}} \right)\left( 1 \right)\left( 3 \right) + {\left( 3 \right)^2}\\ &= 16\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is 16.

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Most popular questions from this chapter

(M) Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

39. \(\left( {\begin{aligned}{{}}{.{\bf{31}}}&{.{\bf{58}}}&{.{\bf{08}}}&{.{\bf{44}}}\\{.{\bf{58}}}&{ - .{\bf{56}}}&{.{\bf{44}}}&{ - .{\bf{58}}}\\{.{\bf{08}}}&{.{\bf{44}}}&{.{\bf{19}}}&{ - .{\bf{08}}}\\{ - .{\bf{44}}}&{ - .{\bf{58}}}&{ - .{\bf{08}}}&{.{\bf{31}}}\end{aligned}} \right)\)

Determine which of the matrices in Exercises 1โ€“6 are symmetric.

3. \(\left( {\begin{aligned}{{}}2&{\,\,3}\\{\bf{2}}&4\end{aligned}} \right)\)

In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).

4. \(Q\left( x \right) = 3x_1^2 + 3x_2^2 + 5x_3^2 + 6x_1^{}x_2^{} + 2x_1^{}x_3^{} + 2x_2^{}x_3^{}\).

Question:Repeat Exercise 7 for the data in Exercise 2.

Question: 2. Let \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},....,{{\bf{u}}_n}} \right\}\) be an orthonormal basis for \({\mathbb{R}_n}\) , and let \({\lambda _1},....{\lambda _n}\) be any real scalars. Define

\(A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + ..... + {\lambda _n}{\bf{u}}_n^T\)

a. Show that A is symmetric.

b. Show that \({\lambda _1},....{\lambda _n}\) are the eigenvalues of A

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