Quadratics in the plane  This application uses the notions of  eigenvalues, eigenvectors, and the process of matrix diagonalization.  We refer the reader to the application to Fibonacci numbers where these notions were reviewed.

Definition In the plane (x,y), a quadratic form in the variables x and y is a function of the form where A, B and C are real numbers. The above form can also be written using matrix multiplication: or where .

Note that A is a symmetric matrix.

Example Consider the quadratic form which can be written as We would like to give a geometric description of the plane curve given by the above equation. To this end, we “diagonalize” the matrix The characteristic polynomial of A is The eigenvalues of A are then 1/2 and 3/2, and one can show that the vector is a basis for the eigenspace of A corresponding to the eigenvalue 3/2 and that is a basis for the eigenspace of A corresponding to the eigenvalue 1/2. Let (the constant in front of the matrix is there to make its columns of length 1) then where is the “diagonal form” of A. Note that so equation (*) above becomes But and So, .

Now, let us consider the following change of variables: so the above equation can be written as: In the new system , the equation of the curve is: This is clearly the equation of an ellipse with axes √(2/3) and √2.

We conclude that is the equation of a rotated ellipse with foci (1, 1), (-1, 1) and axes √(2/3), √2.

Graphically, the following diagram represents the curve: Another question that is important to answer in this example is what is the rotation angle that will get rid of the product xy in the curve equation ?

To answer this question, we look again at the “orthogonal” matrix (orthogonal in the sense that P-1 =PT). Note that ,

so P  can be written as with θ=π/4. Such a matrix is called a rotation matrix and is always orthogonal. So, if one makes a rotation of 45ο of the original axes, one obtains an ellipse in  “standard” form in the new system.

Let us look at another example.

Example Consider The quadratic form then its corresponding matrix is The eigenvalues of A are .

One can easily verify that the vector is a basis for the eigenspace of A corresponding to the eigenvalue √5/2 and that is a basis for the eigenspace of A corresponding to the eigenvalue -√5/2. Let be the rotation matrix in this case, then with respect to a new system (x1, y1) obtained from (x, y) by a rotation of an angle the quadratic equation is ,

which is one of a hyperbola. Graphically, the following diagram represents the curve: The above two examples are particular cases of the following theorem that gives a general analysis of the quadratic forms in the plane.

Theorem Consider a quadratic form in the plane.

1.      There is a counterclockwise rotation of the coordinate system about the origin such that, in the new coordinate system, f(x,y) has no cross term xy.

2.      The graph of f(x,y)=D is an ellipse if C2-4AB<0 and a hyperbola if C2-4AB>0.

In the first example, C2-4AB=1-4(1)(1)=-3<0. This is why the curve was an ellipse, and in the second C2-4AB=1-4(-1)(1)=5 yielding a hyperbola.