91Ó°ÊÓ

Consider \(3x_1^2 + 4{x_1}{x_2}\).

As the quadratic form is:

\({{\rm{x}}^T}A{\rm{x}} = \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}3&2\\2&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

Therefore, thecoefficient matrix of the quadratic form is\(A = \left( {\begin{aligned}{{}}3&2\\2&0\end{aligned}} \right)\).

Now find the characteristic polynomial.

\(\begin{aligned}{}\left| {\begin{aligned}{{}}{3 - \lambda }&2\\2&{ - \lambda }\end{aligned}} \right| &= 0\\\left( {3 - \lambda } \right)\left( { - \lambda } \right) - 4 &= 0\\{\lambda ^2} - 3\lambda - 4 &= 0\\\lambda & = - 1,4\end{aligned}\)

\(\begin{aligned}{}\left( {A - I} \right){\bf{x}} &= 0\\\left( {\begin{aligned}{{}}4&2\\2&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\4{x_1} + 2{x_2} &= 0\\2{x_1} + {x_2} &= 0\end{aligned}\)

Thus, the eigenvector is \({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\2\end{aligned}} \right)\).

\(\begin{aligned}{}\left( {A - 0 \cdot I} \right){\bf{x}} &= 0\\\left( {\begin{aligned}{{}}{ - 1}&2\\2&{ - 4}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\ - {x_1} + 2{x_2} &= 0\\2{x_1} - 4{x_2} &= 0\end{aligned}\)

Thus, the eigenvector is \({{\rm{v}}_2} = \left( {\begin{aligned}{{}}2\\1\end{aligned}} \right)\).

Write the normalization of the vectors.

\(\begin{aligned}{}{{\bf{u}}_1} &= \frac{1}{{\sqrt {1 + 4} }}\left( {\begin{aligned}{{}}{ - 1}\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{ - \frac{1}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}\end{aligned}} \right)\end{aligned}\)

and

\(\begin{aligned}{}{{\bf{u}}_2} & = \frac{1}{{\sqrt {1 + 4} }}\left( {\begin{aligned}{{}}2\\1\end{aligned}} \right)\\ & = \left( \begin{aligned}{l}\frac{2}{{\sqrt 5 }}\\\frac{1}{{\sqrt 5 }}\end{aligned} \right)\end{aligned}\)

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

\(P = \left( {\begin{aligned}{{}}{ - \frac{1}{{\sqrt 5 }}}&{\frac{2}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}}\end{aligned}} \right)\)and\(D = \left( {\begin{aligned}{{}}{ - 1}&0\\0&4\end{aligned}} \right)\)

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,

\(\begin{aligned}{}{{\rm{y}}^T}D{\rm{y}} & = \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 1}&0\\0&4\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{ - {y_1}}&{4{y_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ & = - y_1^2 + 4y_2^2\end{aligned}\)

Thus, the new quadratic form is \(Q\left( {\rm{y}} \right) = - y_1^2 + 4y_2^2\).