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Found in: Page 395

### 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

# 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 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 $$3y_1^2 + 9y_2^2 + 15y_3^2$$.