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Q8E

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Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Let A be the matrix of the quadratic form

\({\bf{9}}x_{\bf{1}}^{\bf{2}} + {\bf{7}}x_{\bf{2}}^{\bf{2}} + {\bf{11}}x_{\bf{3}}^{\bf{2}} - {\bf{8}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{8}}{x_{\bf{1}}}{x_{\bf{3}}}\)

It can be shown that the eigenvalues of A are 3,9, and 15. Find an orthogonal matrix P such that the change of variable \({\bf{x}} = P{\bf{y}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form which no cross-product term. Give P and the new quadratic form.

The matrix Pis \(P = \left( {\begin{aligned}{{}}{ - \frac{2}{3}}&{ - \frac{1}{3}}&{\frac{2}{3}}\\{ - \frac{2}{3}}&{\frac{2}{3}}&{ - \frac{1}{3}}\\{\frac{1}{3}}&{\frac{2}{3}}&{\frac{2}{3}}\end{aligned}} \right)\).

The new quadratic form is \(3y_1^2 + 9y_2^2 + 15y_3^2\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the eigenvalues of the coefficient matrix of the quadratic equation

The coefficient matrix for the equation \(9x_1^2 + + 7x_2^2 + 11x_3^2 - 8{x_1}{x_2} + 8{x_1}{x_3}\) is shown as:

\(A = \left( {\begin{aligned}{{}}9&{ - 4}&4\\{ - 4}&7&0\\4&0&{11}\end{aligned}} \right)\)

The characteristic equation of A can be written as:

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) &= 0\\\left| {\begin{aligned}{{}}{9 - \lambda }&{ - 4}&4\\{ - 4}&{7 - \lambda }&0\\4&0&{11 - \lambda }\end{aligned}} \right| &= 0\\\left( {\lambda - 3} \right)\left( {\lambda - 9} \right)\left( {\lambda - 15} \right) &= 0\\\lambda &= 3,9,15\end{aligned}\)

Step 2: Find the eigen vector of matrix A

Find the eigenvector for \(\lambda = 3\):

\(\begin{aligned}{}\left( {A - 3I} \right)X & = 0\\\left( {\begin{aligned}{{}}6&{ - 4}&4\\{ - 4}&4&0\\4&0&8\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\{x_1} + 2{x_3} = 0\\{x_2} + 2{x_3} & = 0\end{aligned}\)

Thus, the general solution of the equation is:

\(\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}{ - 2}\\{ - 2}\\1\end{aligned}} \right)\)

Find the eigenvector for \(\lambda = 9\):

\(\begin{aligned}{}\left( {A - 9I} \right)X & = 0\\\left( {\begin{aligned}{{}}1&0&{\frac{1}{2}}\\0&1&{ - 1}\\0&0&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}}0\\0\\0\end{aligned}} \right)\\{x_1} + \frac{1}{2}{x_3} & = 0\\{x_2} - {x_3} & = 0\end{aligned}\)

Thus, the general solution of the equation is\(\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}{ - \frac{1}{2}}\\1\\1\end{aligned}} \right)\).

Find the eigenvector for \(\lambda = 15\):

\(\begin{aligned}{}\left( {A - 15I} \right)X & = 0\\\left( {\begin{aligned}{{}}{ - 6}&{ - 4}&4\\{ - 4}&{ - 8}&0\\4&0&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}}0\\0\\0\end{aligned}} \right)\\ - 6{x_1} - 4{x_2} + 4{x_3} & = 0\\ - 4{x_1} - 8{x_2} & = 0\\4{x_1} - 3{x_2} & = 0\end{aligned}\)

Thus, the general solution of the equation is\(\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}2\\{ - 1}\\2\end{aligned}} \right)\).

Step 3: Find normalized eigen vectors of A

The normalized eigenvectors are shown below:

\(\begin{aligned}{}{{\bf{u}}_1} & = \frac{1}{{\sqrt {\left( { - 2} \right) + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2}} }}\left( {\begin{aligned}{{}}{ - 2}\\{ - 2}\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{ - \frac{2}{3}}\\{ - \frac{2}{3}}\\{\frac{1}{3}}\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{u}}_2} &= \frac{1}{{\sqrt {{{\left( { - \frac{1}{2}} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }}\left( {\begin{aligned}{{}}{ - \frac{1}{2}}\\1\\1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{ - \frac{1}{3}}\\{\frac{2}{3}}\\{\frac{2}{3}}\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{u}}_3} &= \frac{1}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} }}\left( {\begin{aligned}{{}}2\\{ - 1}\\2\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{2}{3}}\\{ - \frac{1}{3}}\\{\frac{2}{3}}\end{aligned}} \right)\end{aligned}\)

Step 4: Write the matrix P and D

Write matrix P using the normalized eigenvectors:

\(\begin{aligned}{}P &= \left( {\begin{aligned}{{}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}&{{{\bf{u}}_3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{ - \frac{2}{3}}&{ - \frac{1}{3}}&{\frac{2}{3}}\\{ - \frac{2}{3}}&{\frac{2}{3}}&{ - \frac{1}{3}}\\{\frac{1}{3}}&{\frac{2}{3}}&{\frac{2}{3}}\end{aligned}} \right)\end{aligned}\)

Write matrix D using the eigenvalues of A.

\(D = \left( {\begin{aligned}{{}}3&0&0\\0&9&0\\0&0&{15}\end{aligned}} \right)\)

Step 5: Find the new quadratic form

Consider the expression \({{\bf{x}}^T}A{\bf{x}}\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= {\left( {P{\bf{y}}} \right)^T}A\left( {P{\bf{y}}} \right)\\ &= {{\bf{y}}^T}{P^T}AP{\bf{y}}\\ &= {{\bf{y}}^r}D{\bf{y}}\\ &= \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}&{{y_3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}3&0&0\\0&9&0\\0&0&{15}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\\{{y_3}}\end{aligned}} \right)\\ &= 3y_1^2 + 9y_2^2 + 15y_3^2\end{aligned}\)

Thus, the new quadratic form is \(3y_1^2 + 9y_2^2 + 15y_3^2\).

Most popular questions for Math Textbooks

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix \(P\) and a diagonal matrix \(D\). To save you time, the eigenvalues in Exercises 17–22 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. \(\left( {\begin{aligned}{{}}{\,6}&{ - 2}\\{ - 2}&{\,\,\,9}\end{aligned}} \right)\)

25. Let \({\bf{T:}}{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{m}}}\) be a linear transformation. Describe how to find a basis \(B\) for \({\mathbb{R}^n}\) and a basis \(C\) for \({\mathbb{R}^m}\) such that the matrix for \(T\) relative to \(B\) and \(C\) is an \(m \times n\) “diagonal” matrix.

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.

24. Using the notation of Exercise 23, show that \({A^T}{u_j} = {\sigma _j}{v_j}\) for \({\bf{1}} \le {\bf{j}} \le {\bf{r}} = {\bf{rank}}\;{\bf{A}}\)

Question: 4. Let A be an \(n \times n\) symmetric matrix.

a. Show that \({({\rm{Col}}A)^ \bot } = {\rm{Nul}}A\). (Hint: See Section 6.1.)

b. Show that each y in \({\mathbb{R}^n}\) can be written in the form \(y = \hat y + z\), with \(\hat y\) in \({\rm{Col}}A\) and z in \({\rm{Nul}}A\).

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.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).
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