Sentencia en contra de exintegrantes del Bloque Libertadores

Let A= (aij)_1≤i,j≤n be a real n× n matrix. The following question seems quite natural in a course in advanced vector calculus or real analysis. It is not solved in general and is called the Hadamard maximum determinant problem.²

1The books by Assmus and Key [AK] and Conway and Sloane [CS1] do into more detail in these matters and are highly recommended.

2Hadamard determined the maximum value of |det(A)|, where the entries of A range over all complex numbers|aij| ≤ 1, to be n^n/2and that this maximum was attained by the Vandermonde matrices of the nth roots of unity.

D. Joyner, J.-L. Kim, Selected Unsolved Problems in Coding Theory, Applied and Numerical Harmonic Analysis,

DOI10.1007/978-0-8176-8256-9_3, © Springer Science+Business Media, LLC 2011

47

Open Problem 11 What is the maximum value of|det(A)|, where the entries of A range over all real numbers|aij| ≤ 1?

From vector calculus we know that the absolute value of the determinant of a real square matrix equals the volume of the parallelepiped spanned by the row (or column) vectors of the matrix. The volume of a parallelepiped with sides of a fixed length depends on the angles the row vectors make with each other. This volume is maximized when the row vectors are mutually orthogonal, i.e., when the paral-lelepiped is a cube inRⁿ. Suppose now that the row vectors of A are all orthogonal.

The row vectors of A,|aij| ≤ 1, are longest when each aij= ±1, which implies that the length of each row vector is√

n. Suppose, in addition, that the row vectors of A are all of length√

n. Such a matrix is called a Hadamard matrix of order n. Then

|det(A)| = (√

n)ⁿ= n^n/2, since the cube has n sides of length√

n. Now, if A is any matrix as in the above question, then we must have|det(A)| ≤ n^n/2. This inequality is called Hadamard’s determinant inequality.

Jacques Hadamard (1865–1963) was a prolific mathematician who worked in many areas, but he is most famous for giving one of the first proofs of the prime number theorem (in 1896). (The prime number theorem, “known” to Gauss though not proven, roughly states that the number of prime numbers less than N is about N/log(N ) as N grows to infinity.)

The above question is unsolved for arbitrary n in the sense that it is not yet (as of this writing) known for which n Hadamard matrices exist. Moreover, if n≡ 3 (mod 4) (for such n, it is known that Hadamard matrices cannot exist), then the best possible upper bound for|det(A)|, as A ranges over all n × n (−1, 1)-matrices, is not known at the time of this writing.

Open Problem 12 (Hadamard conjecture) For each n which is a multiple of 4, a Hadamard matrix exists.

(This conjecture might, in fact, be due to Raymond Paley, a brilliant mathemati-cian who contributed to several areas of mathematics, including two constructions of Hadamard matrices using the theory of finite fields. Sadly, he died in 1933 at the age of 26 in a skiing accident.) At the time of this writing, the smallest order for which no Hadamard matrix is presently known is 668.

What might be surprising at first sight is that there does not always exist a Hadamard matrix—for some n, they exist, and for other n, they do not. For ex-ample, there is a 2× 2 Hadamard matrix but not a 3 × 3 one. What is perhaps even more surprising is that, in spite of the fact that the above question arose from an analytic perspective, Hadamard matrices are related more to coding theory, number theory, and combinatorics [vLW], [Ho]! In fact, a linear code constructed from a Hadamard matrix was used in the 1971 Mariner 9 mission to Mars.

Example 59 Let

This is a Hadamard matrix of order 12. A different Hadamard matrix of order 12 can be computed usingSAGE:

SAGE

HereSAGEis calling a native Python program, which implements some methods for constructing Hadamard matrices. Another way to useSAGEto obtain a Hadamard matrix is to look up the file name on Sloane’s database http://www.research.

att.com/~njas/hadamard/, for example,had.16.2.txt, and use theSAGE com-mand

SAGE

sage: hadamard_matrix_www(’had.16.2.txt’)

[ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]

[ 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1]

[ 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1]

[ 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1]

[ 1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1]

[ 1 -1 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1]

[ 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 1 1]

[ 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1]

[ 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1]

[ 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1]

[ 1 1 -1 -1 1 -1 1 -1 -1 -1 1 1 -1 1 -1 1]

[ 1 1 -1 -1 -1 1 -1 1 -1 -1 1 1 1 -1 1 -1]

[ 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 1 1 -1]

[ 1 -1 1 -1 -1 1 1 -1 -1 1 -1 1 1 -1 -1 1]

[ 1 -1 -1 1 1 1 -1 -1 -1 1 1 -1 -1 -1 1 1]

[ 1 -1 -1 1 -1 -1 1 1 -1 1 1 -1 1 1 -1 -1]

This command assumes that you are running SAGE from a computer with an internet connection since it actually fetches the file had.16.2.txt from http://www research.att.com/~njas/hadamard/and parses that file to return the ma-trix given above.

Some easy to prove facts:

(a) if you swap two rows or columns of a Hadamard matrix, you will get another Hadamard matrix;

(b) if you multiply any row or column of a Hadamard matrix by−1, you will get another Hadamard matrix;

(c) if you multiply any Hadamard matrix on the left by a signed permutation matrix (that is, a matrix with exactly one±1 per row and column), you will get another Hadamard matrix;

(d) if you multiply any Hadamard matrix on the left by a signed permutation matrix (that is, a matrix with exactly one±1 per row and column), you will get another Hadamard matrix.

Definition 60 Let A, B be two Hadamard matrices of order n. Call A and B left equivalent if there is an n× n signed permutation matrix P such that A = P B. Let Abe a Hadamard matrix of order n. Let Aut(A) denote the group of all n× n signed permutation matrices Q such that A is left equivalent to AQ. This object Aut(A) is called the automorphism group of A.

Mathieu groups, discovered in the 1800’s, are now know to arise natuarally in many fields of mathematics (see Conway-Sloane [CS1] for an excellent treatment).

The following result is just one indication of the unique role of Mathieu groups in mathematics.

Theorem 61 It is known that any two 12× 12 Hadamard matrices are equivalent, i.e., that there is only one Hadamard matrix of order 12, up to equivalence. Let A be a 12× 12 Hadamard matrix. Then Aut(A) ∼= M12.

The proof can be found in Assmus and Mattson [AM1] or in Sect. 7.4 in Assmus and Key [AK] (see in particular their Theorem 7.4.3, which also discusses more general codes associated to Hadamard matrices). Kantor [Kan] is another excellent paper on this topic.