Hadamard-Matrices

[Paul C. Leopardi]

Hi all,
At the risk of making a fool of myself, I'll try answering this, below.
Best regards

On Mon, 28 Mar 2005 09:44 pm, Annette.Warlich at t-online.de wrote:
> By matching a method from another area of math I came across
> the problem of Hadamard-matrices, in the sense of the maximum
> determinant problem.
> In factor analysis I am experimenting with some rotation criteria -
> say with a start of a matrix A to find some orthogonal rotation-
> matrix, which maximizes (or minimizes) some criteria for that matrix.
>
> Since Hadamard-Matrices H are orthogonal they can be rotated to a
> n * I -matrix by
>
>     H * T = n*I
>
> so must a n*I-matrix be rotatable to H by
>
>     H = n*I * T'

I don't know what definition of Hadamard matrix you are using, but using
http://mathworld.wolfram.com/HadamardMatrix.html
we have the definition that a Hadamard matrix is a (-1,1) matrix with 
H * H' = n * I.
Therefore Hadamard matrices are NOT orthogonal, they are sqrt(n) times an 
orthogonal matrix.

The Mathworld web page says that you can obtain a 2^n * 2^n Hadamard matrix by 
using a complete set of Walsh functions, and you can use the Kronecker 
product to get large Hadamard matrices from small ones.

> From  factor-analysis the "quartimax"- and "varimax" rotation are known,
> which maximize the variance of the squares of the entries of the columns.
>
> One form some other descriptions of Hadamard-matrix besides the
> maximum-determinant criterion is, that the variance of the squares
> is zero.

Presumably this is because each entry is +/- 1, so each square is always 1.

> Changing the Quartimax-criterion to a minimum criterion should
> produce a Hadamard-matrix just by some simple canned rotation-
> procedure of a factor-analysis-program.
>
> something like
>    n = 4
>    A = sqrt(n) * Unit(n)
>    H4 = rotate(A,"-quartimax")   // the *minimizing" criterion of quartimax

Here you are using sqrt(n) * I, where above you had n*I. In fact sqrt(n)*I is 
right, in that the Hadamard matrix is sqrt(n) times an orthogonal matrix.

> (With n=16 a certain precondition helped improving convergence.)
>
> In fact, I just find the n=4,N=8,n=12 and n=16 - solutions by
> using that criterion.
> I was not able to detect such of higher order - already with the
> n=16 case the convergence indicated the possibility of ending
> in local minima.

I am not familiar with this, and so cannot answer, except to ask, how do you 
know that the problem does not have local minima? Is there a theorem which 
says that any local minimum must be a global minimum? Or a counterexample?

> I would like to understand this problem better - mainly: how can
> the iterative process terminate without reaching its minimum.
> I just tried it with higher numeric precision (up to 512 bits), but
> it doesnt seem, that this is the problem.

What is the iterative process? Can you spell it out? Is it a nonlinear 
optimization algorithm? Can you supply it with a Jacobian?