Here's an example of finding a suitable coordinate system for the equation

For this equation, M = \pmatrix{3&-5\\-5&3} , d = {14 \choose -2}.

The first step is to find the eigenvalues and eigenvetors of matrix M which means solving the following equation

Evaluating the determinant gives (3 - \lambda)^2 - (-5)^2 = \lambda^2 - 6\lambda - 16 = 0. This is a quadratic equation with two real roots \lambda_1 = 8, \lambda_2 = -2. Since substituting either into the matrix M - \lambda I renders its determinant zero, the rows of the matrix are then linearly dependent giving us a freedom of choosing an eigenvector. Thus when solving (M - \lambda I){{x}\choose{y}}=0 we need only consider one row. For example, for \lambda = 8, the first row becomes -5x - 5y = 0 giving a solution {1}\choose{-1} which we normalize to {{{1}/\sqrt{2}}\choose{{-1}/\sqrt{2}}}. For the second eigenvalue \lambda = -2, the first row is 5x - 5y = 0 resulting in a solution {1 \choose 1}. This is normalized to {1/ \sqrt 2 \choose 1/ \sqrt 2}. Use the two eigenvector to compose the transform matrix

As the consequence of the normalization, |P| = 1 meaning that the matrix is orthogonal and the induced transformtion is a rotation around the origin. Naturally,

as expected. Thus the substitution {x \choose y} = P{x' \choose y'} leads to the equation

Completing the equare (or rather *squares*, in this case) we obtain

which, more conventionally, is written as

This is an equation of a hyperbola. In the new coordinate system, its center is at {x \choose y} = {-1/\sqrt{2} \choose 3/\sqrt{2}}. In the original coordinate system, the hyperbola is centered at

**References**

- D. A. Brannan
*et al*,*Geometry*Cambridge University Press, 2002