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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

# 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 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 $$6y_1^2 - 4y_2^2$$.