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:
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
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.