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Q7E

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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

Make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form \(x_{\bf{1}}^{\bf{2}} + {\bf{10}}{x_{\bf{1}}}{x_{\bf{2}}} + x_{\bf{2}}^{\bf{2}}\) into a quadratic form with no cross-product term. Give P and the new quadratic form.

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

The new quadratic form is \(6y_1^2 - 4y_2^2\).

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 \(x_1^2 + 10{x_1}{x_2} + x_2^2\) is shown below:

\(A = \left( {\begin{aligned}{{}{}}1&5\\5&1\end{aligned}} \right)\)

The characteristic equation of A can be written as:

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) &= 0\\\left| {\begin{aligned}{{}{}}{1 - \lambda }&5\\5&{1 - \lambda }\end{aligned}} \right| &= 0\\{\left( {1 - \lambda } \right)^2} - 25 &= 0\\\lambda &= 6, - 4\end{aligned}\)

Step 2: Find the eigen vector of matrix A

Find the eigenvector for \(\lambda = 6\):

\(\begin{aligned}{}\left( {A - 6I} \right)X &= 0\\\left( {\begin{aligned}{{}{}}{ - 5}&5\\5&{ - 5}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}{}}0\\0\end{aligned}} \right)\\ - 5{x_1} + 5{x_2} &= 0\\{x_1} - {x_2} &= 0\end{aligned}\)

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

Find the eigenvector for \(\lambda = - 4\):

\(\begin{aligned}{}\left( {A + 4I} \right)X &= 0\\\left( {\begin{aligned}{{}{}}5&5\\5&5\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}{}}0\\0\end{aligned}} \right)\\5{x_1} + 5{x_2} &= 0\\{x_1} + {x_2} &= 0\end{aligned}\)

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

Step 3: Find normalized eigenvectors of A

The normalized eigenvectors are:

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

And,

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

Step 4: Write the matrix P and D

Write matrix P using the normalized eigenvectors:

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

Write matrix D using the eigenvalues of A as:

\(D = \left( {\begin{aligned}{{}{}}6&0\\0&{ - 4}\end{aligned}} \right)\)

Step 5: Find the new quadratic form

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

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= {\left( {P{\bf{y}}} \right)^T}A\left( {P{\bf{y}}} \right)\\ &= {{\bf{y}}^T}{P^T}AP{\bf{y}}\\ &= {{\bf{y}}^r}D{\bf{y}}\\ &= \left( {\begin{aligned}{{}{}}{{{\bf{y}}_1}}&{{{\bf{y}}_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}6&0\\0&{ - 4}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{{\bf{y}}_1}}\\{{{\bf{y}}_2}}\end{aligned}} \right)\\ &= 6y_1^2 - 4y_2^2\end{aligned}\)

Thus, the new quadratic form is \(6y_1^2 - 4y_2^2\).

Most popular questions for Math Textbooks

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.

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

Question: 5. Show that if v is an eigenvector of an \(n \times n\) matrix A and v corresponds to a nonzero eigenvalue of A, then v is in Col A. (Hint: Use the definition of an eigenvector.)

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.

22. \(\left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\)

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

Question: Let \({\bf{X}}\) denote a vector that varies over the columns of a \(p \times N\) matrix of observations, and let \(P\) be a \(p \times p\) orthogonal matrix. Show that the change of variable \({\bf{X}} = P{\bf{Y}}\) does not change the total variance of the data. (Hint: By Exercise 11, it suffices to show that \(tr\left( {{P^T}SP} \right) = tr\left( S \right)\). Use a property of the trace mentioned in Exercise 25 in Section 5.4.)
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