Abstract. The aim of this paper is to prove a generalized version of Hadamard’s inequality in inner product spaces with assumptions signifi-cantly weaker than the ones found in existing variants. 1. Introduction. Several different results are known in the literature as Hadamard’s inequal-ity. The most basic version for real square matrices A = [aij]i,j=1,...,n takes the following form

Kulczycki, Marcin

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UNIVERSITATIS IAGELLONICAE ACTA MATHEMATICA, FASCICULUS XL2002HADAMARD’S INEQUALITY IN INNER PRODUCT SPACESby Marcin KulczyckiAbstract. The aim of this paper is to prove a generalized version ofHadamard’s inequality in inner product spaces with assumptions signifi-cantly weaker than the ones found in existing variants.1. Introduction.Several different results are known in the literature as Hadamard’s inequal-ity. The most basic version for real square matrices A = [aij ]i,j=1,...,n takes thefollowing form:(detA)2 ≤n∏j=1(n∑i=1a2ij)It is, though, but a simple collorary of a more advanced theorem.From now on, given a unitary space, we shall consider norms and scalarproducts on the exterior products of this vector space to be given by a natu-ral extrapolation of a canonical construction presented in [1]. This standardmethod is presented for real spaces only. It is possible, however, to modifyit to fit the case of complex spaces. The following result is often cited asHadamard’s inequality.Theorem 1.1. For each finite-dimensional real unitary space V and foreach pair of vectors ξ ∈ ∧p V, η ∈ ∧q V , if at least one of the vectors ξ or η issimple, then|ξ ∧ η| ≤ |ξ||η|The simplicity of ξ∧η is also a sufficient condition for the above inequalityto hold, as was shown by Anna Wach in [3].Author wishes to express his gratitude to prof. dr hab. Piotr Tworzewski forsuggesting the problem and many stimulatng conversations and to the referee for pointingout an incompatibility with a quoted reference.114Is is also known that a somewhat weaker inequality is true for any pair ofvectors ξ ∈ ∧p V, η ∈ ∧q V|ξ ∧ η| ≤ (p+qp ) 12 |ξ||η|.See [1] for details.Our objective is to prove the ineqality in a much more general setting. Theabove results will turn out to be particular cases of the main result.2. Definitions and notation. Define(ab)= 0 for b > a. For 1 ≤ p ≤ nletΛ(n, p) = {λ : {1, . . . , p} → {1, . . . , n} strictly increasing }Let V be a finite-dimensional inner product space. If e1, . . . , en is a givenorthonormal basis of V , then for every λ ∈ Λ(n, p) define eλ ∈∧p V byeλ = eλ(1) ∧ . . . ∧ eλ(p)For p ≥ 0, we say that η ∈ ∧p V is of rank r iff there exist v1, . . . , vp−r ∈ V ,v˜ ∈ ∧r V such that η = v1 ∧ . . . ∧ vp−r ∧ v˜ and r is the least number havingthis property.Remark 2.1. It is a simple modification of a known result that {eλ}λ∈Λ(n,p)is a basis of∧p V .Definition 2.2. We introduce an inner product on∧p V by defining theaforementioned basis to be an orthonormal basis for∧p V .Remark 2.3. The above definition only depends on the scalar product onV rather than the original choice of the basis e1, . . . , en for V . In fact, anequivalent construction can be done without need for a particular basis of Vto be fixed first, thus guaranteeing the uniqueness of a product (see [1] forsuch construction for R case). In the proof of theorem 3.1, however, it is moreconvenient to have the scalar product on∧p V introduced in the above way.3. Hadamard’s inequality. The following theorem is the main result ofthis paper.Theorem 3.1. For every pair of vectors ξ ∈ ∧p V, η ∈ ∧q V such that ξ isof rank not greater than r, η is of rank not greater than s, and ξ ∧ η is of ranknot greater than t we have|ξ ∧ η| ≤ min{(r+sr ) 12 , (n+t−p−qt ) 12 } · |ξ||η|Remark 3.2. Since a vector is simple iff it is of rank 0, all previouslymentioned versions of Hadamard’s inequality follow from the aboveresult.115Proof of theorem 3.1. The cases p = 0, q = 0 or p+ q > n are elemen-tary. Let p, q > 0, p+ q ≤ n. It is enough to prove(A) (ξ ∧ η) • (ξ ∧ η) ≤ (r+sr )(ξ • ξ)(η • η)(B) (ξ ∧ η) • (ξ ∧ η) ≤ (n+t−p−qt )(ξ • ξ)(η • η),where • is the scalar product on the exterior products.Ad A. It is a known result that by the ordinary orthonormalization we canget two orthonormal bases of V : e1, . . . , en and e˜1, . . . , e˜n such thatξ = e1 ∧ . . . ∧ ep−r ∧ ξ˜, ξ˜ ∈∧r V,η = e˜1 ∧ . . . ∧ e˜q−s ∧ η˜, η˜ ∈∧s V,ξ˜ ∧ η =∑λ∈Λ(n,r+q) aλeλ, aλ ∈ C,ξ˜ ∧ η˜ =∑λ∈Λ(n,r+s) bλe˜λ, bλ ∈ C,ξ˜ =∑λ∈Λ(n,r) cλeλ, cλ ∈ C,η˜ =∑λ∈Λ(n,s) dλeλ, dλ ∈ C.Then(ξ ∧ η) • (ξ ∧ η) = (e1 ∧ . . . ∧ ep−r ∧ ξ˜ ∧ η) • (e1 ∧ . . . ∧ ep−r ∧ ξ˜ ∧ η)= (e1 ∧ . . . ∧ ep−r ∧ (∑λ∈Λ(n,r+q)aλeλ)) • (e1 ∧ . . . ∧ ep−r ∧ (∑λ∈Λ(n,r+q)aλeλ))≤∑λ∈Λ(n,r+q)aλaλ = (ξ˜ ∧ η) • (ξ˜ ∧ η)= ((−1)r(q−s)e˜1 ∧ . . . ∧ e˜q−s ∧ ξ˜ ∧ η˜) • ((−1)r(q−s)e˜1 ∧ . . . ∧ e˜q−s ∧ ξ˜ ∧ η˜)= (e˜1 ∧ . . . ∧ e˜q−s ∧ (∑λ∈Λ(n,r+s)bλe˜λ)) • (e˜1 ∧ . . . ∧ e˜q−s ∧ (∑λ∈Λ(n,r+s)bλe˜λ))≤∑λ∈Λ(n,r+s)bλbλ = (ξ˜ ∧ η˜) • (ξ˜ ∧ η˜) =∑µ∈Λ(n,r+s)(∑eλ∧eκ=±eµ±cλdκ)(∑eλ∧eκ=±eµ±cλdκ)≤∑µ∈Λ(n,r+s)(∑eλ∧eκ=±eµ|cλ||dκ|)2 ≤∑µ∈Λ(n,r+s)(r+sr)(∑eλ∧eκ=±eµ|cλ|2|dκ|2)≤ (r+sr )( ∑λ∈Λ(n,r)|cλ|2)(∑κ∈Λ(n,s)|dκ|2) =(r+sr)(ξ˜ • ξ˜)(η˜ • η˜) = (r+sr )(ξ • ξ)(η • η).Ad B. It is again possible to obtain, by orthonormalization, an orthonormal116basis e1, . . . , en of V such thatξ ∧ η = e1 ∧ . . . ∧ ep+q−t ∧ e˜, e˜ ∈∧t V,ξ =∑λ∈Λ(n,p) xλeλ, xλ ∈ C,η =∑λ∈Λ(n,q) yλeλ, yλ ∈ C.Define w = e1 ∧ . . . ∧ ep+q−t. Then(ξ ∧ η) • (ξ ∧ η) =∑µ∈Λ(n+t−p−q,t)(∑eλ∧eκ=±w∧eµ+p+q−t±xλyκ)(∑eλ∧eκ=±w∧eµ+p+q−t±xλyκ)≤∑µ∈Λ(n+t−p−q,t)(∑eλ∧eκ=±w∧eµ+p+q−t|xλ||yκ|)2 ≤ by the Schwartz inequality≤∑µ∈Λ(n+t−p−q,t)(∑eλ∧eκ=±w∧eµ+p+q−txλxλ)(∑eλ∧eκ=±w∧eµ+p+q−tyκyκ)≤∑µ∈Λ(n+t−p−q,t)(∑λ∈Λ(n,p)xλxλ)(∑κ∈Λ(n,q)yκyκ) =(n+t−p−qt)(ξ • ξ)(η • η).References1. Federer H., Geometric Measure Theory, Springer, Berlin, 1996.2. Voyevodin V.V., Linear Algebra, Mir Publishers, Moscow, 1983.3. Wach A., A note on Hadamard’s inequality, Univ. Iagel. Acta Math., XXXI (1994).Received March 15, 1999Jagiellonian UniversityInstitute of MathematicsReymonta 4Krako´w 30-059, Polande-mail : kulczyck@im.uj.edu.pl

Hadamard's inequality in inner product spaces

Marcin Kulczycki

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HADAMARD’S INEQUALITY IN INNER PRODUCT SPACES

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