Question

In Exercises 1 and 2,find the change of variable \({\rm{x}} = P{\rm{y}}\) that transforms the quadratic form \({{\rm{x}}^T}A{\rm{x}}\) into \({{\rm{y}}^T}D{\rm{y}}\) as shown.

1. \(5x_1^2 + 6x_2^2 + 7x_3^2 + 4x_1^{}x_2^{} - 4x_2^{}x_3^{} = 9y_1^2 + 6y_2^2 + 3y_3^2\)

Short answer

The requiredchange of variable is:

\(P = \left( {\begin{aligned}{{}}{\frac{1}{3}}&{\frac{2}{3}}&{ - \frac{2}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}&{\frac{2}{3}}\\{ - \frac{2}{3}}&{\frac{2}{3}}&{\frac{1}{3}}\end{aligned}} \right)\).

Step-by-step solution

Symmetric Matrices and Quadratic Forms

When any Symmetric Matrix \(A\) is diagonalized orthogonally as \(PD{P^{ - 1}}\) we have:

\(\begin{aligned}{}{x^T}Ax = {y^T}Dy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as }}x = Py} \right\}\\{\rm{and}}\\\left\| x \right\| = \left\| {Py} \right\| = \left\| y \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\forall y \in \mathbb{R}} \right\}\end{aligned}\)

Find the Change of Variables

As per the question, we have:

\(5x_1^2 + 6x_2^2 + 7x_3^2 + 4x_1^{}x_2^{} - 4x_2^{}x_3^{} = 9y_1^2 + 6y_2^2 + 3y_3^2\)

The matrix of thequadratic formwill be:

\(A = \left( {\begin{aligned}{{}}5&2&0\\2&6&{ - 2}\\0&{ - 2}&7\end{aligned}} \right)\)

And for each eigenvalue of\(A\)defined in the right-hand side of the equation will be expressed as:

\(\begin{aligned}{}{\lambda _1} = 9 \Rightarrow {P_1} = \left( {\begin{aligned}{{}}{\frac{1}{3}}\\{\frac{2}{3}}\\{ - \frac{2}{3}}\end{aligned}} \right)\\{\lambda _2} = 6 \Rightarrow {P_2} = \left( {\begin{aligned}{{}}{\frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{aligned}} \right)\\{\lambda _3} = 3 \Rightarrow {P_3} = \left( {\begin{aligned}{{}}{ - \frac{2}{3}}\\{\frac{2}{3}}\\{\frac{1}{3}}\end{aligned}} \right)\end{aligned}\)

Hence,therequired change of variable is:

\(P = \left( {\begin{aligned}{{}}{\frac{1}{3}}&{\frac{2}{3}}&{ - \frac{2}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}&{\frac{2}{3}}\\{ - \frac{2}{3}}&{\frac{2}{3}}&{\frac{1}{3}}\end{aligned}} \right)\).