Note that if (the identity matrix) then the sesquilinear form reduces to the standard complex inner product. Also, it can be easily seen that this form is `linear' in the first component and `conjugate linear' in the second component. Also, if we want then the matrix need to be an Hermitian matrix. Note that if and , then the sesquilinear form reduces to a bilinear form.
The expression is called the quadratic form and the Hermitian form. We generally write and in place of and , respectively. It can be easily shown that for any choice of the Hermitian form is a real number.
Therefore, in matrix notation, for a Hermitian matrix , the Hermitian form can be rewritten as
The main idea is to express as sum of squares and hence determine the possible values that it can take. Note that if we replace by where is any complex number, then simply gets multiplied by and hence one needs to study only those for which i.e., is a normalised vector.
From Exercise 126.96.36.199 one knows that if ( is Hermitian) then there exists a unitary matrix such that ( with 's the eigenvalues of the matrix which we know are real). So, taking (i.e., choosing 's as linear combination of 's with coefficients coming from the entries of the matrix ), one gets
Equation (6.4.1) gives one method of writing as a sum of absolute squares of linearly independent linear forms. One can easily show that there are more than one way of writing as sum of squares. The question arises, ``what can we say about the coefficients when has been written as sum of absolute squares".
This question is answered by `Sylvester's law of inertia' which we state as the next lemma.
where are linearly independent linear forms in and the integers and depend only on
Hence, let us assume on the contrary that there exist positive integers
Now, this can hold only if which gives a contradiction. Hence
Similarly, the case can be resolved. height6pt width 6pt depth 0pt
Note: The integer is the rank of the matrix and the number is sometimes called the inertial degree of
We complete this chapter by understanding the graph of
for We first look at the following example.
The eigenpairs for are Thus,
Therefore, the given graph represents an ellipse with the principal axes and That is, the principal axes are
The eccentricity of the ellipse is the foci are at the points and and the equations of the directrices are
is called the quadratic form associated with the given conic.
We now consider the general conic. We obtain conditions on the eigenvalues of the associated quadratic form to characterise the different conic sections in (endowed with the standard inner product).
Prove that this conic represents
As is a symmetric matrix, by Corollary 6.3.7, the eigenvalues of are both real, the corresponding eigenvectors are orthonormal and is unitarily diagonalisable with
and the equation of the conic section in the -plane, reduces to
Now, depending on the eigenvalues we consider different cases:
Also, observe that implies that the That is,
In this case, we have the following:
The terms on the left can be written as product of two factors as Thus, in this case, the given equation represents a pair of intersecting straight lines in the -plane.
This equation represents a hyperbola in the -plane, with principal axes
we now consider the following cases:
This equation represents an ellipse in the -plane, with principal axes
implies that the principal axes of the conic are functions of the eigenvectors and
As a last application, we consider the following problem that helps us in understanding the quadrics. Let
So, the equation of the quadric in standard form is
where the point is the centre. The calculation of the planes of symmetry is left as an exercise to the reader.
A K Lal 2007-09-12