Question

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.

10. \({\bf{2}}x_{\bf{1}}^{\bf{2}} + {\bf{6}}{x_{\bf{1}}}{x_{\bf{2}}} - {\bf{6}}x_{\bf{2}}^{\bf{2}}\)

Short answer

The matrix P is \(P = \left( {\begin{aligned}{{}}{ - \frac{1}{{\sqrt {10} }}}&{\frac{3}{{\sqrt {10} }}}\\{\frac{3}{{\sqrt {10} }}}&{\frac{1}{{\sqrt {10} }}}\end{aligned}} \right)\).

The new quadratic form is \( - 7y_1^2 + 3y_2^2\).

Step-by-step solution

Find the eigenvalues of the coefficient matrix of the quadratic equation

The coefficient matrix for the equation \(2x_1^2 + 6{x_1}{x_2} - 6x_2^2\) is shown below:

\(A = \left( {\begin{aligned}{{}}2&3\\3&{ - 6}\end{aligned}} \right)\)

The characteristic equation of A can be written as:

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) &= 0\\\left| {\begin{aligned}{{}}{2 - \lambda }&3\\3&{ - 6 - \lambda }\end{aligned}} \right| &= 0\\\left( {2 - \lambda } \right)\left( { - 6 - \lambda } \right) - 9 &= 0\\\lambda &= - 7,\,3\end{aligned}\)

Find the eigen vector of matrix A

Find the eigenvector for \(\lambda = - 7\):

\(\begin{aligned}{}\left( {A - 2I} \right)X &= 0\\\left( {\begin{aligned}{{}}0&3\\1&{ - 6}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\3{x_2} & = 0\\{x_1} - 6{x_2} & = 0\end{aligned}\)

Thus, the general solution of the equation is:

\(\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}{ - 1}\\3\end{aligned}} \right)\)

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

\(\begin{aligned}{}\left( {A - 3I} \right)X & = 0\\\left( {\begin{aligned}{{}}{ - 1}&3\\3&{ - 9}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\ - {x_1} + 3{x_2} & = 0\\3{x_1} - 9{x_2} & = 0\end{aligned}\)

Thus, the general solution of the equation is\(\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}3\\1\end{aligned}} \right)\).

Find normalized eigen vectors of A

The normalized eigen vectors are:

\(\begin{aligned}{}{{\bf{u}}_1} &= \frac{1}{{\sqrt {{1^2} + {{\left( { - 3} \right)}^2}} }}\left( {\begin{aligned}{{}}{ - 1}\\3\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{ - \frac{1}{{\sqrt {10} }}}\\{\frac{3}{{\sqrt {10} }}}\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{u}}_2} & = \frac{1}{{\sqrt {{3^2} + {1^2}} }}\left( {\begin{aligned}{{}}3\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{\frac{3}{{\sqrt {10} }}}\\{\frac{1}{{\sqrt {10} }}}\end{aligned}} \right)\end{aligned}\)

Write the matrix P and D

Write matrix Pusing the normalized eigenvectors:

\(\begin{aligned}{}P &= \left( {\begin{aligned}{{}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{ - \frac{1}{{\sqrt {10} }}}&{\frac{3}{{\sqrt {10} }}}\\{\frac{3}{{\sqrt {10} }}}&{\frac{1}{{\sqrt {10} }}}\end{aligned}} \right)\end{aligned}\)

Write matrix D using the eigenvalues of A.

\(D = \left( {\begin{aligned}{{}}{ - 7}&0\\0&3\end{aligned}} \right)\)

Find the new quadratic form

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

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= {\left( {P{\bf{y}}} \right)^T}A\left( {P{\bf{y}}} \right)\\ & = {{\bf{y}}^T}{P^T}AP{\bf{y}}\\ & = {{\bf{y}}^r}D{\bf{y}}\\ & = \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 7}&0\\0&3\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ & = - 7y_1^2 + 3y_2^2\end{aligned}\)

Thus, the new quadratic form is \( - 7y_1^2 + 3y_2^2\).