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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

# Classify the quadratic forms in Exercises 9-18. 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 7.1.9. $${\bf{4}}x_{\bf{1}}^{\bf{2}} - {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{4}}x_{\bf{2}}^{\bf{2}}$$

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).

The new quadratic form is $$2y_1^2 + 6y_2^2$$.

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 $$4x_1^2 - 4{x_1}{x_2} + 4x_2^2$$.

A = \left( {\begin{aligned}{{}}4&{ - 2}\\{ - 2}&4\end{aligned}} \right)

The characteristic equation of A can be written as:

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

## Step 2: Find the eigen vector of matrix A

Find the eigenvector for $$\lambda = 2$$:

\begin{aligned}{}\left( {A - 2I} \right)X &= 0\\\left( {\begin{aligned}{{}}2&{ - 2}\\{ - 2}&2\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\2{x_1} - 2{x_2} & = 0\\ - 2{x_1} + 2{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 $$\lambda = 6$$:

\begin{aligned}{}\left( {A - 6I} \right)X & = 0\\\left( {\begin{aligned}{{}}{ - 2}&{ - 2}\\{ - 2}&{ - 2}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\ - 2{x_1} - 2{x_2} = 0\\ - 2{x_1} - 2{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 eigen vectors of A

The normalized eigenvectors are:

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

D = \left( {\begin{aligned}{{}}2&0\\0&6\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}}\end{aligned}} \right)\left( {\begin{aligned}{{}}2&0\\0&6\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ & = 2y_1^2 + 6y_2^2\end{aligned}

Thus, the new quadratic form is $$2y_1^2 + 6y_2^2$$.