Hadamard's inequality

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

${\displaystyle \left|\det(N)\right|\leq B^{n}n^{n/2}.}$

In particular, if the entries of N are +1 and −1 only then[4]

${\displaystyle \left|\det(N)\right|\leq n^{n/2}.}$

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 Decomposition of a semidefinite matrix). Then

${\displaystyle \det(P)=\det(N)^{2}\leq \prod _{i=1}^{n}\|v_{i}\|^{2}=\prod _{i=1}^{n}p_{ii}.}$

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.[2][5]

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

${\displaystyle \left|\det M\right|\leq 1,}$

(1)

and equality is achieved if and only if the vectors are an orthogonal set. The general result now follows:

${\displaystyle \left|\det N\right|={\bigg (}\prod _{i=1}^{n}\|v_{i}\|{\bigg )}\left|\det M\right|\leq \prod _{i=1}^{n}\|v_{i}\|.}$

To prove (1), consider P =M^*M and let the eigenvalues of P be λ₁, λ₂, … λ_n. Since the length of each column of M is 1, each entry in the diagonal of P is 1, so the trace of P is n. Applying the inequality of arithmetic and geometric means,

${\displaystyle \det P=\prod _{i=1}^{n}\lambda _{i}\leq {\bigg (}{1 \over n}\sum _{i=1}^{n}\lambda _{i}{\bigg )}^{n}=\left({1 \over n}\operatorname {tr} P\right)^{n}=1^{n}=1,}$

so

${\displaystyle \left|\det M\right|={\sqrt {\det P}}\leq 1.}$

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.[6] Many other proofs can be found in the literature.

• Fischer's inequality

• 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 978-0-521-69973-0.
• Riesz, Frigyes; Szőkefalvi-Nagy, Béla (1990). Functional Analysis. Dover. p. 176. ISBN 0-486-66289-6.
• Weisstein, Eric W. "Hadamard's Inequality". MathWorld.

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