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Chapter 7: Q12E (page 395)

Classify the quadratic forms in Exercises 9鈥�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.

12.\({\bf{ - }}x_{\bf{1}}^{\bf{2}}{\bf{ - 2}}{x_{\bf{1}}}{x_{\bf{2}}} - x_{\bf{2}}^{\bf{2}}\)

Short Answer

Expert verified

The new quadratic form is \(Q\left( {\rm{y}} \right) = - 2y_1^2\).The matrix \(P\)is \(P = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\).

Step by step solution

01

Step 1: Find the coefficient matrix of the quadratic form

Consider \( - x_1^2 - 2{x_1}{x_2} - x_2^2\).

As the quadratic form is:

\({{\rm{x}}^T}A{\rm{x}} = \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 1}&{ - 1}\\{ - 1}&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

Therefore, thecoefficient matrix of the quadratic form is\(A = \left( {\begin{aligned}{{}}{ - 1}&{ - 1}\\{ - 1}&{ - 1}\end{aligned}} \right)\).

Now Find the characteristic polynomial.

\(\begin{aligned}{}\left| {\begin{aligned}{{}}{ - 1 - \lambda }&{ - 1}\\{ - 1}&{ - 1 - \lambda }\end{aligned}} \right| & = 0\\\left( { - 1 - \lambda } \right)\left( { - 1 - \lambda } \right) - 1 &= 0\\\left( { - 1 - \lambda } \right)\left( { - 1 - \lambda } \right) &= 1\\\lambda &= - 2,0\end{aligned}\)

02

Find the basis for eigen value \({\bf{ - 2}}\)

\(\begin{aligned}{}\left( {A - \lambda I} \right){{\rm{u}}_1} &= 0\\\left( {\begin{aligned}{{}}1&{ - 1}\\{ - 1}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) &= 0\\{x_1} - {x_2} &= 0\\ - {x_1} + {x_2} &= 0\end{aligned}\)

Thus, the eigenvector is \({{\rm{u}}_1} = \left( {\begin{aligned}{{}}1\\1\end{aligned}} \right)\).

03

Find the basis for eigen value \({\bf{0}}\)

\(\begin{aligned}{}\left( {A - \lambda I} \right){{\rm{u}}_2} &= 0\\\left( {\begin{aligned}{{}}{ - 1}&{ - 1}\\{ - 1}&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) &= 0\\ - {x_1} - {x_2} &= 0\\ - {x_1} - {x_2} &= 0\end{aligned}\)

Thus, the eigenvector is \({{\rm{u}}_2} = \left( {\begin{aligned}{{}}{ - 1}\\1\end{aligned}} \right)\).

04

Find the orthonormal basis for eigen value \({\bf{ - 2}}\)

\(\begin{aligned}{}{P_1} &= \frac{1}{{\left\| {{{\rm{u}}_1}} \right\|}}{{\rm{u}}_1}\\ &= \frac{1}{{\sqrt {1 + 1} }}\left( {\begin{aligned}{{}}1\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\end{aligned}\)

05

Find the orthonormal basis for eigen value \({\bf{3}}\)

\(\begin{aligned}{}{P_1} &= \frac{1}{{\left\| {{{\rm{u}}_2}} \right\|}}{{\rm{u}}_2}\\ &= \frac{1}{{\sqrt {1 + 1} }}\left( {\begin{aligned}{{}}{ - 1}\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 1 }}}\\{\sqrt 2 }\end{aligned}} \right)\end{aligned}\)

Therefore, the matrix\(P\)and\(D\)are shown below:

\(\begin{aligned}{}P &= \left( {\begin{aligned}{{}}{{P_1}}&{{P_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\end{aligned}\)

\(D = \left( {\begin{aligned}{{}}{ - 2}&0\\0&0\end{aligned}} \right)\)

Now write the new quadratic form by using the transformation\({\rm{x}} = P{\rm{y}}\).

\(\begin{aligned}{}Q\left( {\rm{x}} \right) &= {{\rm{x}}^T}A{\rm{x}}\\ &= {\left( {P{\rm{y}}} \right)^T}A\left( {P{\rm{y}}} \right)\\ &= {{\rm{y}}^T}{P^T}AP{\rm{y}}\\ &= {{\rm{y}}^T}D{\rm{y}}\\ &= Q\left( {\rm{y}} \right)\end{aligned}\)

Therefore,

\(\begin{aligned}{}{{\rm{y}}^T}D{\rm{y}} &= \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 2}&0\\0&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{ - 2{y_1}}&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ &= - 2y_1^2\end{aligned}\)

Thus, the new quadratic form is \(Q\left( {\rm{y}} \right) = - 2y_1^2\).

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Most popular questions from this chapter

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.

37. \(\left( {\begin{aligned}{{}}{\bf{6}}&{\bf{2}}&{\bf{9}}&{ - {\bf{6}}}\\{\bf{2}}&{\bf{6}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{9}}&{ - {\bf{6}}}&{\bf{6}}&{\bf{2}}\\{\bf{6}}&{\bf{9}}&{\bf{2}}&{\bf{6}}\end{aligned}} \right)\)

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.

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

Determine which of the matrices in Exercises 7鈥�12 are orthogonal. If orthogonal, find the inverse.

12. \(P = \left( {\begin{aligned}{{}}{.5}&{.5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{.5}\\{.5}&{ - .5}&{ - .5}&{.5}\\{.5}&{ - .5}&{.5}&{ - .5}\end{aligned}} \right)\)

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix \(P\) and a diagonal matrix \(D\). To save you time, the eigenvalues in Exercises 17鈥�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.

16. \(\left( {\begin{aligned}{{}}{\,6}&{ - 2}\\{ - 2}&{\,\,\,9}\end{aligned}} \right)\)

In Exercises 1 and 2,find the change of variable \({\rm{x}} = P{\rm{y}}\) that transforms the quadratic form \({{\rm{x}}^T}A{\rm{x}}\) into \({{\rm{y}}^T}D{\rm{y}}\) as shown.

2. \(3x_1^2 + 3x_2^2 + 5x_3^2 + 6x_1^{}x_2^{} + 2x_1^{}x_3^{} + 2x_2^{}x_3^{} = 7y_1^2 + 4y_2^2\).
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