in reply to Help with Matrix math!
If you set aside for a moment the constraint on the vector x, then the problem of maximizing
x_{i}D_{ij}x_{j}
(a sum over repeated indices is implied) translates, upon taking the derivative with respect to x_{i},
into the equation
D_{ij}x_{j} = 0.
This is a special form of the
eigenvalue equation for the matrix D, with the
eigenvalue being 0. In order for nontrivial solutions, the matrix D cannot have an inverse, so it's
determinant must vanish. Finding the eigenvectors can
be done with PDL; see
Mastering Algorithms with Perl for a discussion. The
Math::Cephes::Matrix module can also do this for real symmetric matrices. Generally, the components of the
eigenvectors found in this way are not completely determined; by convention, the normalization x_{i}x_{i} = 1 is imposed.
Update: The preceding needs to be augmented by a
test on the
Hessian of the matrix to determine what type of
critical point is present.
Re^2: Help with Matrix math! by pc88mxer (Vicar) on Dec 17, 2007 at 05:04 UTC 
In this case you have to take into account the constraint on the vector x. The quadratic form x_{i}D_{ij}x_{j} is unbounded on R^{n}, so a derivative test (even with an Hessian test) is only going to find local extreme points, and there is no guarantee that those points will lie on the surface of interest (i.e. x'x = 1.) The difference is that for a point to be a local maximum when x is constrained, the derivative only needs to be zero when evaluated in the tangent space of the constraining surface at that point, not zero in every direction.
The standard way to include the surface constraint is to use Lagrangian multipliers.
 [reply] 

I was thinking of the problem differently 
find the extremal points of
x^{†}Dx, which leads to the eigenvalue
equation Dx = 0. This equation doesn't determine
x^{†}x completely, so one is free to impose, for example, x^{†}x = 1 as a
normalization condition. But you're right that if
x^{†}x = 1 is intended as a true constraint, then a method like Lagrange multipliers should be used.
 [reply] 
