Hadamard's inequality

In mathematics, Hadamard's inequality, first published by Jacques Hadamard in 1893,^[1] is a bound on the determinant of a matrix whose entries are complex numbers in terms of the lengths of its column vectors. In geometrical terms, when restricted to real numbers, it bounds the volume in Euclidean space of n dimensions marked out by n vectors v_i for 1 ≤ i ≤ n in terms of the lengths of these vectors ||v_i||.

Specifically, Hadamard's inequality states that if N is the matrix having columns^[2] v_i, then

$|\det(N)|\le \prod_{i=1}^n \|v_i\|.$

If the n vectors are linearly independent, equality in Hadamard's inequality is achieved if and only if the vectors are orthogonal.

Alternate forms and corollaries

A corollary is that if the entries of an n by n matrix N are bounded by B, so |N_ij|≤B for all i and j, then

$|\det(N)|\le B^nn^{n/2}.$

In particular, if the entries of N are +1 and −1 only then^[3]

$|\det(N)|\le n^{n/2}.$

In combinatorics, matrices N for which equality holds, i.e. those with orthogonal columns, are called Hadamard matrices.

A positive-semidefinite matrix P can be written as N^*N, where N^* denotes the conjugate transpose of N (see Cholesky decomposition). Then

$\det(P)=\det(N)^2 \le \prod_{i=1}^n \|v_i\|^2 = \prod_{i=1}^n p_{ii}.$

So, the determinant of a positive definite matrix is less than or equal to the product of its diagonal entries. Sometimes this is also known as Hadamard's inequality.^[1]

Proof

The result is trivial if the matrix N is singular, so assume the columns of N are linearly independent. By dividing each column by its length, it can be seen that the result is equivalent to the special case where each column has length 1, in other words if e_i are unit vectors and M is the matrix having the e_i as columns then

$|\det M|\le 1,$

and equality is achieved if and only if the vectors are an orthogonal set, that is when the matrix is unitary. The general result now follows:

$|\det N| = \bigg (\prod_{i=1}^n \|v_i\| \bigg) |\det M| \leq \prod_{i=1}^n \|v_i\|.$

For the positive definite case, let P =M^*M and let the eigenvalues of P be λ₁, λ₂, … λ_n. By assumption, each entry in the diagonal of P is 1, so the trace of P is n. Applying the inequality of arithmetic and geometric means,

$\det P=\prod_{i=1}^n \lambda_i \le \bigg({1 \over n}\sum_{i=1}^n \lambda_i\bigg)^n = \left({1 \over n} \mathrm{tr} P\right)^n = 1^n = 1,$

so

$|\det M| = \sqrt{\det P} \le 1.$

If there is equality then each of the λ_i's must all be equal and their sum is n, so they must all be 1. The matrix P is Hermitian, therefore diagonalizable, so it is the identity matrix—in other words the columns of M are an orthonormal set and the columns of N are an orthogonal set.^[4]

Many other proofs can be found in the literature.

References

1. Maz'ya & Shaposhnikova
2. The result is sometimes stated in terms of row vectors. That this is equivalent is seen by applying the transpose.
3. Garling
4. Proof follows, with minor modifications, the second proof given in Maz'ya & Shaposhnikova. See also Proof of Hadamard's inequality at PlanetMath.org..

• Maz'ya, Vladimir; Shaposhnikova, T. O. (1999). Jacques Hadamard: A Universal Mathematician. AMS. pp. 383ff. ISBN 0-8218-1923-2.
• Garling, D. J. H. (2007). Inequalities: A Journey into Linear Analysis. Cambridge. p. 233. ISBN 0-521-69973-8.
• Riesz, Frigyes; Szőkefalvi-Nagy, Béla (1990). Functional Analysis. Dover. p. 176. ISBN 0-486-66289-6.
• Weisstein, Eric W., "Hadamard's Inequality", MathWorld.

Further reading

• Beckenbach, Edwin F; Bellman, Richard Ernest (1965). Inequalities. Springer. p. 64.