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.

13. \({\bf{ - }}x_{\bf{1}}^{\bf{2}}{\bf{ - 6}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{9}}x_{\bf{2}}^{\bf{2}}\)

Short answer

The new quadratic form is \({{\rm{x}}^T}A{\rm{x}} = 10y_1^2\).

Step-by-step solution

Step 1: Find the coefficient matrix of the quadratic form

Consider \(x_1^2 - 6{x_1}{x_2} + 9x_2^2\).

As the quadratic form is:

\({{\rm{x}}^T}A{\rm{x}} = \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}1&{ - 3}\\{ - 3}&9\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

Therefore, thecoefficient matrix of the quadratic form is\(A = \left( {\begin{aligned}{{}}1&{ - 3}\\{ - 3}&9\end{aligned}} \right)\).

Now Find the characteristic polynomial.

\(\begin{aligned}{}\left| {\begin{aligned}{{}}{1 - \lambda }&{ - 3}\\{ - 3}&{9 - \lambda }\end{aligned}} \right| &= 0\\\left( {1 - \lambda } \right)\left( {9 - \lambda } \right) - 9 &= 0\\{\lambda ^2} - 10\lambda &= 0\end{aligned}\)

Find the basis for eigen value \({\bf{10}}\)

\(\begin{aligned}{}\left( {A - 10I} \right){\bf{x}} &= 0\\\left( {\begin{aligned}{{}}{ - 9}&{ - 3}\\{ - 3}&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}}x\\y\end{aligned}} \right) &= \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\ - 3x &= y\\\frac{x}{1} &= \frac{y}{{ - 3}}\end{aligned}\)

Thus, the eigenvector is \({{\rm{v}}_1} = \left( {\begin{aligned}{{}}1\\{ - 3}\end{aligned}} \right)\).

Find the basis for eigenvalue \({\bf{0}}\)

\(\begin{aligned}{}\left( {A - 0 \cdot I} \right){\bf{x}} &= 0\\\left( {\begin{aligned}{{}}1&{ - 3}\\{ - 3}&9\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\x &= 3y\\\frac{x}{3} &= \frac{y}{1}\end{aligned}\)

Thus, the eigenvector is \({{\rm{v}}_2} = \left( {\begin{aligned}{{}}3\\1\end{aligned}} \right)\).

Find the matrix \(P\) and \(D\)

Write the normalize of the vectors.

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

Then the matrix\(P\)and\(D\)is shown below:

\(P = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt {10} }}}&{\frac{3}{{\sqrt {10} }}}\\{ - \frac{3}{{\sqrt {10} }}}&{\frac{1}{{\sqrt {10} }}}\end{aligned}} \right)\)and\(D = \left( {\begin{aligned}{{}}{10}&0\\0&0\end{aligned}} \right)\)

Now write the new quadratic form by using the transformation\({\rm{x}} = P{\rm{y}}\).

\(\begin{aligned}{}Q\left( {\rm{x}} \right) &= {{\rm{x}}^T}A{\rm{x}}\\ &= {\left( {P{\rm{y}}} \right)^T}A\left( {P{\rm{y}}} \right)\\ &= {{\rm{y}}^T}{P^T}AP{\rm{y}}\\ &= {{\rm{y}}^T}D{\rm{y}}\\ &= 10y_1^2\end{aligned}\)

Therefore,

\(\begin{aligned}{}{{\rm{y}}^T}D{\rm{y}} &= \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 2}&0\\0&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{ - 2{y_1}}&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ &= - 2y_1^2\end{aligned}\)

Thus, the new quadratic form is \({{\rm{x}}^T}A{\rm{x}} = 10y_1^2\).