Inmathematics,Hadamard's inequality (also known asHadamard's theorem on determinants^[1]) is a result first published byJacques Hadamard in 1893.^[2] It is abound on thedeterminant of amatrix whose entries arecomplex numbers in terms of the lengths of its column vectors. Ingeometrical terms, when restricted toreal numbers, it bounds thevolume inEuclidean space ofn dimensions marked out byn vectorsv_i for 1 ≤i ≤n in terms of the lengths of these vectors ||v_i||.

Specifically, Hadamard's inequality states that ifN is the matrix having columns^[3]v_i, then

${\displaystyle \left|det(N)\right|\le \prod _{i=1}^n\Vert v_i\Vert .}$

If then vectors are non-zero, equality in Hadamard's inequality is achievedif and only if the vectors areorthogonal.

Acorollary is that if the entries of ann byn matrixN are bounded byB, so |N_ij| ≤B for alli andj, then

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

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

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

Incombinatorics, matricesN for which equality holds, i.e. those with orthogonal columns, are calledHadamard matrices.

More generally, suppose thatN is acomplex matrix of ordern, whose entries are bounded by |N_ij| ≤ 1, for eachi,j between 1 andn. Then Hadamard's inequality states that

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

Equality in this bound is attained for a real matrixNif and only ifN is a Hadamard matrix.

Apositive-semidefinite matrixP can be written asN^*N, whereN^* denotes theconjugate transpose ofN (seeDecomposition of a semidefinite matrix). Then

${\displaystyle det(P)=det(N{)}^2\le \prod _{i=1}^n\Vert v_i{\Vert }^2=\prod _{i=1}^n{p}_{ii}.}$

So, the determinant of apositive 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 matrixN issingular, so assume the columns ofN arelinearly independent. By dividing each column by its Euclidean norm, it can be seen that the result is equivalent to the special case where each column has norm 1, in other words ife_i areunit vectors andM is the matrix having thee_i as columns then

${\displaystyle \left|detM\right|\le 1,}$

1

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

${\displaystyle \left|detN\right|=(\prod _{i=1}^n\Vert v_i\Vert )\left|detM\right|\le \prod _{i=1}^n\Vert v_i\Vert .}$

Toprove(1), considerP =M^*M whereM^* is the conjugate transpose ofM, and let theeigenvalues ofP be λ₁, λ₂, … λ_n. Since the length of each column ofM is 1, each entry in the diagonal ofP is 1, so thetrace ofP isn. Applying theinequality of arithmetic and geometric means,

${\displaystyle detP=\prod _{i=1}^n{\lambda }_i\le {\left(\frac{1}{n}\sum _{i=1}^n{\lambda }_i\right)}^n={\left(\frac{1}{n}\mathrm{tr}P\right)}^n=1^n=1,}$

so

${\displaystyle \left|detM\right|=\sqrt{detP}\le 1.}$

If there is equality then each of theλ_i's must all be equal and their sum isn, so they must all be 1. The matrixP isHermitian, thereforediagonalizable, so it is theidentity matrix—in other words the columns ofM are an orthonormal set and the columns ofN 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.

• Inequalities (mathematics)
• Determinants

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