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Chapter 7: Q17E (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.

17. \({\bf{11}}x_{\bf{1}}^{\bf{2}}{\bf{ + 11}}x_{\bf{2}}^{\bf{2}}{\bf{ + 11}}x_{\bf{3}}^{\bf{2}}{\bf{ + 11}}x_{\bf{4}}^{\bf{2}}{\bf{ + 16}}{x_{\bf{1}}}{x_{\bf{2}}}{\bf{ - 12}}{x_{\bf{1}}}{x_{\bf{4}}}{\bf{ + 12}}{x_{\bf{2}}}{x_{\bf{3}}}{\bf{ + 16}}{x_{\bf{3}}}{x_{\bf{4}}}\)

Short Answer

Expert verified

The new quadratic form is \(Q\left( {\rm{y}} \right) = 21y_1^2 + 21y_2^2 + y_3^2 + y_4^2\).

Step by step solution

01

Step 1: Find the coefficient matrix of the quadratic form

Consider \(11x_1^2 + 11x_2^2 + 11x_3^2 + 11x_4^2 + 16{x_1}{x_2} - 12{x_1}{x_4} + 12{x_2}{x_3} + 16{x_3}{x_4}\).

As the quadratic form is:

\({{\rm{x}}^T}A{\rm{x}} = \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{11}&8&0&{ - 6}\\8&{11}&6&0\\0&6&{11}&8\\{ - 6}&0&8&{11}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{aligned}} \right)\)

Therefore, the coefficient matrix of the quadratic form is \(A = \left( {\begin{aligned}{{}}{11}&8&0&{ - 6}\\8&{11}&6&0\\0&6&{11}&8\\{ - 6}&0&8&{11}\end{aligned}} \right)\).

02

Find the eigen values

Using the MATLAB command\({\rm{eigs}}\left( A \right)\), the eigenvalues are as:

\(\begin{aligned}{}{\lambda _1} = 21,\\{\lambda _2} = 21,\\{\lambda _3} = 1,\\{\lambda _4} = 1\end{aligned}\)

The eigenvectors are:

\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}4\\5\\3\\0\end{aligned}} \right)\), \({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{ - 5}\\{ - 4}\\0\\3\end{aligned}} \right)\), \({{\rm{v}}_3} = \left( {\begin{aligned}{{}}4\\{ - 5}\\3\\0\end{aligned}} \right)\), \({{\rm{v}}_4} = \left( {\begin{aligned}{{}}5\\{ - 4}\\0\\3\end{aligned}} \right)\)

03

Find the matrix \(P\) and \(D\)

Write the normalization of the vectors.

\({{\bf{u}}_1} = \left( {\begin{aligned}{{}}{\frac{{2\sqrt 2 }}{5}}\\{\frac{{\sqrt 2 }}{2}}\\{\frac{{3\sqrt 2 }}{{10}}}\\0\end{aligned}} \right)\),\({{\bf{u}}_2} = \left( {\begin{aligned}{{}}{ - \frac{{3\sqrt 2 }}{{10}}}\\0\\{\frac{{2\sqrt 2 }}{5}}\\{\frac{{\sqrt 2 }}{2}}\end{aligned}} \right)\),\({{\bf{u}}_3} = \left( {\begin{aligned}{{}}{\frac{{2\sqrt 2 }}{5}}\\{\frac{{ - \sqrt 2 }}{2}}\\{\frac{{3\sqrt 2 }}{{10}}}\\0\end{aligned}} \right)\),\({{\bf{u}}_4} = \left( {\begin{aligned}{{}}{\frac{{3\sqrt 2 }}{{10}}}\\0\\{ - \frac{{2\sqrt 2 }}{5}}\\{\frac{{\sqrt 2 }}{2}}\end{aligned}} \right)\)

Then the matrix\(P\)and\(D\)is shown below:

\(P = \left( {\begin{aligned}{{}}{\frac{{2\sqrt 2 }}{5}}&{ - \frac{{3\sqrt 2 }}{{10}}}&{\frac{{2\sqrt 2 }}{5}}&{\frac{{3\sqrt 2 }}{{10}}}\\{\frac{{\sqrt 2 }}{2}}&0&{ - \frac{{\sqrt 2 }}{2}}&0\\{\frac{{3\sqrt 2 }}{{10}}}&{\frac{{2\sqrt 2 }}{5}}&{\frac{{3\sqrt 2 }}{{10}}}&{ - \frac{{2\sqrt 2 }}{5}}\\0&{\frac{{\sqrt 2 }}{2}}&0&{\frac{{\sqrt 2 }}{2}}\end{aligned}} \right)\)and\(D = \left( {\begin{aligned}{{}}{21}&0&0&0\\0&{21}&0&0\\0&0&1&0\\0&0&0&1\end{aligned}} \right)\)

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

\(\begin{aligned}{}Q\left( {\rm{y}} \right) &= {{\rm{y}}^T}D{\rm{y}}\\ & = \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}&{{y_3}}&{{y_4}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{21}&0&0&0\\0&{21}&0&0\\0&0&1&0\\0&0&0&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\\{{y_3}}\\{{y_4}}\end{aligned}} \right)\\ & = 21y_1^2 + 21y_2^2 + y_3^2 + y_4^2\end{aligned}\)

Thus, the new quadratic form is \(Q\left( {\rm{y}} \right) = 21y_1^2 + 21y_2^2 + y_3^2 + y_4^2\).

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Most popular questions from this chapter

Question: Let \({x_1}\,,{x_2}\) denote the variables for the two-dimensional data in Exercise 1. Find a new variable \({y_1}\) of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2}\), with\(c_1^2 + c_2^2 = 1\), such that \({y_1}\) has maximum possible variance over the given data. How much of the variance in the data is explained by \({y_1}\)?

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.

40. \(\left( {\begin{aligned}{{}}{\bf{8}}&{\bf{2}}&{\bf{2}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{2}}&{\bf{8}}&{\bf{2}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{2}}&{\bf{2}}&{\bf{8}}&{ - {\bf{6}}}&{\bf{9}}\\{ - {\bf{6}}}&{ - {\bf{6}}}&{ - {\bf{6}}}&{{\bf{24}}}&{\bf{9}}\\{\bf{9}}&{\bf{9}}&{\bf{9}}&{\bf{9}}&{ - {\bf{21}}}\end{aligned}} \right)\)

(M) Compute an SVD of each matrix in Exercises 26 and 27. Report the final matrix entries accurate to two decimal places. Use the method of Examples 3 and 4.

27. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{6}}&{ - {\bf{8}}}&{ - {\bf{4}}}&{\bf{5}}&{ - {\bf{4}}}\\{\bf{2}}&{\bf{7}}&{ - {\bf{5}}}&{ - {\bf{6}}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{1}}}&{ - {\bf{8}}}&{\bf{2}}&{\bf{2}}\\{ - {\bf{1}}}&{ - {\bf{2}}}&{\bf{4}}&{\bf{4}}&{ - {\bf{8}}}\end{array}} \right)\)

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.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)

Question: In Exercises 15 and 16, construct the pseudo-inverse of \(A\). Begin by using a matrix program to produce the SVD of \(A\), or, if that is not available, begin with an orthogonal diagonalization of \({A^T}A\). Use the pseudo-inverse to solve \(A{\rm{x}} = {\rm{b}}\), for \({\rm{b}} = \left( {6, - 1, - 4,6} \right)\) and let \(\mathop {\rm{x}}\limits^\^ \)be the solution. Make a calculation to verify that \(\mathop {\rm{x}}\limits^\^ \) is in Row \(A\). Find a nonzero vector \({\rm{u}}\) in Nul\(A\), and verify that \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\| < \left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\), which must be true by Exercise 13(c).

15. \(A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 3}&{ - 6}&6&{\,\,1}\\{ - 1}&{ - 1}&{ - 1}&1&{ - 2}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\end{array}} \right]\)
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