We propose a generalization of the parallelogram identity in any dimension N  65 2, establishing the ratio of the quadratic mean of the diagonals to the quadratic mean of the faces of a parallelotope. The proof makes use of simple properties of the exterior product of vectors

Fonda, Alessandro

English

We propose a generalization of the parallelogram identity in any dimension N ≥ 2, establishing the ratio of the quadratic mean of the diagonals to the quadratic mean of the faces of a parallelotope. The proof makes use of simple properties of the exterior product of vectors

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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)ARS MATHEMATICA CONTEMPORANEA 16 (2019) 411–417https://doi.org/10.26493/1855-3974.1704.34c(Also available at http://amc-journal.eu)A generalization of the parallelogram law tohigher dimensionsAlessandro Fonda ∗Dipartimento di Matematica e Geoscienze, Universita` di Trieste,P.le Europa 1, I-34127 Trieste, ItalyReceived 24 May 2018, accepted 9 July 2018, published online 8 February 2019AbstractWe propose a generalization of the parallelogram identity in any dimension N ≥ 2,establishing the ratio of the quadratic mean of the diagonals to the quadratic mean of thefaces of a parallelotope. The proof makes use of simple properties of the exterior productof vectors.Keywords: Parallelogram law, parallelotope.Math. Subj. Class.: 51M041 Introduction and statement of the resultThe well known parallelogram law states:For any parallelogram, the sum of the squares of the lengths of its two diago-nals is equal to the sum of the squares of the lengths of its four sides.Figure 1: The two diagonals of a parallelogram.∗I would probably never have written this paper without the support of my son Marcello. I warmly thank him,in particular, for helping me finding the formula for the three-dimensional case.E-mail address: a.fonda@units.it (Alessandro Fonda)cb This work is licensed under https://creativecommons.org/licenses/by/4.0/412 Ars Math. Contemp. 16 (2019) 411–417Equivalently: given two vectors a and b, one has‖a+ b‖2 + ‖a− b‖2 = 2(‖a‖2 + ‖b‖2) .This identity holds in any inner product space, but, since the two vectors belong to thesame plane, we can see it as being of a two-dimensional nature. The aim of this paper is toprovide a generalization to higher dimensions.The parallelogram law has a natural geometric interpretation, involving the areas of thesquares constructed on the sides and on the diagonals of the parallelogram. In particular,when ‖a + b‖ = ‖a − b‖, it reduces to the Pythagorean theorem. In this paper, however,we will look at the parallelogram law from a rather unusual point of view: writing it as‖a+ b‖2 + ‖a− b‖22= 2‖a‖2 + ‖b‖2 + ‖a‖2 + ‖b‖24,and taking the square roots, we can state it in the following equivalent form.For any parallelogram, the ratio of the quadratic mean of the lengths of itsdiagonals to the quadratic mean of the lengths of its sides is equal to√2 .Now, instead of a parallelogram, we will consider an N -dimensional parallelotope,and our goal will be to prove that the same type of proposition holds in this general case.Indeed, our result can be stated as follows.Theorem 1.1. For any N -dimensional parallelotope, the ratio of the quadratic mean ofthe (N − 1)-dimensional measures of its diagonals to the quadratic mean of the (N − 1)-dimensional measures of its faces is equal to√2.For N = 2, the 1-dimensional measure is the length, and we recover the parallelogramlaw. In the general case, we first need to specify what a diagonal should be, and indeedthis will be clarified in the following sections. For example, if N = 3, the diagonals of aparallelepiped are precisely the parallelograms obtained joining the opposite edges of theparallelepiped (see Figure 2 below), so that the 2-dimensional measures of the diagonalsare the areas of these parallelograms.Notice that our definition of a diagonal is not the same as the one given in [1, 2],where a different generalization of the parallelogram law has been proposed; in the three-dimensional case, e.g., their diagonals are triangles. We believe that our definition is some-what more natural, since here the diagonals share the same geometrical shape of the faces.We provide the proof of our main theorem in Section 3. However, for the reader’sconvenience, we thought it useful to first explain its proof in detail in the more familiarthree-dimensional case. This is what we are going to do next.2 The three-dimensional caseTo start with, let us consider a three-dimensional parallelepiped P , and see how to extendthe parallelogram law to this case. Instead of the lengths of the four sides of the parallel-ogram, we would like to take the areas of the six faces of the parallelepiped. On the otherhand, the lengths of the two diagonals of the parallelogram should naturally be replacedby the areas of the six diagonals of the parallelepiped, i.e., the six parallelograms obtainedjoining the opposite edges of the parallelepiped. In this case, Theorem 1.1 can be rephrasedas follows.A. Fonda: A generalization of the parallelogram law to higher dimensions 413For any three-dimensional parallelepiped, the sum of the squares of the areasof its six diagonals is equal to twice the sum of the squares of the areas of itssix faces.Figure 2: One of the six diagonals of a parallelepiped.In order to prove this statement, assume the parallelepiped to be generated by the fol-lowing three vectors:a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3).By this we mean that P is the set of points obtained as linear combinations of these threevectors, with coefficients in the interval [0, 1]:P = {αa+ βb+ γc : α, β, γ ∈ [0, 1]}.The six faces of P are defined asF−1 = {αa+ βb+ γc ∈ P : α = 0},F+1 = {αa+ βb+ γc ∈ P : α = 1},F−2 = {αa+ βb+ γc ∈ P : β = 0},F+2 = {αa+ βb+ γc ∈ P : β = 1},F−3 = {αa+ βb+ γc ∈ P : γ = 0},F+3 = {αa+ βb+ γc ∈ P : γ = 1}.So,F−1 is generated by b and c,F−2 is generated by a and c,F−3 is generated by a and b,while F+k is congruent to F−k , for each k = 1, 2, 3.414 Ars Math. Contemp. 16 (2019) 411–417The six diagonals of P are defined asD11,2 = {αa+ βb+ γc ∈ P : α = β},D21,2 = {αa+ βb+ γc ∈ P : α+ β = 1},D11,3 = {αa+ βb+ γc ∈ P : α = γ},D21,3 = {αa+ βb+ γc ∈ P : α+ γ = 1},D12,3 = {αa+ βb+ γc ∈ P : β = γ},D22,3 = {αa+ βb+ γc ∈ P : β + γ = 1}.So,D11,2 is generated by a+ b and c,D11,3 is generated by a+ c and b,D12,3 is generated by b+ c and a,whileD21,2 is congruent to the set generated by a− b and c,D21,3 is congruent to the set generated by a− c and b,D22,3 is congruent to the set generated by b− c and a.Our proposition is thus translated into the following identity:‖(a+ b)× c‖2 + ‖(a− b)× c‖2+ ‖(a+ c)× b‖2 + ‖(a− c)× b‖2+ ‖(b+ c)× a‖2 + ‖(b− c)× a‖2 = 4(‖b× c‖2 + ‖a× c‖2 + ‖a× b‖2).Here, we have used the vector product, so that, e.g.,‖a× b‖2 =∣∣∣∣a2 a3b2 b3∣∣∣∣2 + ∣∣∣∣a3 a1b3 b1∣∣∣∣2 + ∣∣∣∣a1 a2b1 b2∣∣∣∣2= (a2b3 − b2a3)2 + (a3b1 − b3a1)2 + (a1b2 − b1a2)2.In order to prove the above identity, we just notice that, by the parallelogram law,‖(a+ b)× c‖2 + ‖(a− b)× c‖2 == ‖(a× c) + (b× c)‖2 + ‖(a× c)− (b× c)‖2= 2(‖a× c‖2 + ‖b× c‖2),and similarly‖(a+ c)× b‖2 + ‖(a− c)× b‖2 = 2(‖a× b‖2 + ‖c× b‖2),‖(b+ c)× a‖2 + ‖(b− c)× a‖2 = 2(‖b× a‖2 + ‖c× a‖2).Summing up the three formulas, our identity is proved.A. Fonda: A generalization of the parallelogram law to higher dimensions 415Remark 2.1. There surely are several ways to extend the parallelogram law to higherdimensions. Just to mention one of these, in the three-dimensional case we have‖a+ b+ c‖2 + ‖a+ b− c‖2+ ‖a− b+ c‖2 + ‖a− b− c‖2 = 4(‖a‖2 + ‖b‖2 + ‖c‖2).We acknowledge the referee for pointing out this identity. It is proved directly (by the useof the classical parallelogram law) and can be easily extended to any dimension.3 Proof of the main theoremWe now provide a proof for the general N -dimensional case. Let P be the parallelotopegenerated by the vectors a1, . . . ,aN , i.e.,P ={N∑k=1ckak : ck ∈ [0, 1], for k = 1, . . . , N}.Its 2N faces are defined byF−n ={N∑k=1ckak ∈ P : cn = 0}, F+n ={N∑k=1ckak ∈ P : cn = 1},with n = 1, . . . , N . Each F−n is generated by the vectors a1, . . . , ân, . . . ,aN , where,as usual, ân means that an is missing, while F+n is a translation of F−n , for every n =1, . . . , N .Concerning the diagonals, they are defined asD1i,j ={N∑k=1ckak ∈ P : ci = cj}, D2i,j ={N∑k=1ckak ∈ P : ci + cj = 1},with indices i < j varying from 1 to N . There are N(N −1) of them. Hence, we have thatD1i,j is generated by ai + aj and a1, . . . , âi, . . . , âj , . . . ,aN ,whileD2i,j is a translation of the set generated by ai − aj and a1, . . . , âi, . . . , âj , . . . ,aN .In order to compute the (N−1)-dimensional measures of the faces and the diagonals ofour parallelotope, we make use of the following proposition involving the exterior productof vectors in RN . (See, e.g., [3] for the definition and the main properties of the exteriorproduct.)Proposition 3.1. The M -dimensional measure of a parallelotope generated by M vectorsv1, . . . ,vM in RN , with 1 ≤M ≤ N , is given by ‖v1 ∧ · · · ∧ vM‖.Proof. If v1, . . . ,vM are linearly dependent, the M -dimensional measure of the parallelo-tope generated by v1, . . . ,vM is equal to zero, hence coincides with ‖v1 ∧ · · · ∧ vM‖.416 Ars Math. Contemp. 16 (2019) 411–417Assume now that the vectors v1, . . . ,vM are linearly independent, and let V be the sub-space generated by them. Choose an orthonormal basis e1, . . . , eM of V , and writev1 = v11e1 + · · ·+ v1MeM ,...vM = vM1e1 + · · ·+ vMMeM .Then,v1 ∧ · · · ∧ vM = det v11 · · · v1M... ...vM1 · · · vMM e1 ∧ · · · ∧ eM ,so that‖v1 ∧ · · · ∧ vM‖ =∣∣∣∣∣∣∣det v11 · · · v1M... ...vM1 · · · vMM∣∣∣∣∣∣∣ ,which is indeed the M -dimensional measure of the parallelotope generated by the vectorsv1, . . . ,vM .Hence, the (N − 1)-dimensional measures of the faces F±n are given by‖a1 ∧ · · · ∧ ân ∧ · · · ∧ aN‖,while the (N − 1)-dimensional measures of the diagonals D1i,j are equal to‖(ai + aj) ∧∧k 6=i,j ak‖,and those of the diagonals D2i,j are equal to‖(ai − aj) ∧∧k 6=i,j ak‖.Choosing any couple i < j, by the parallelogram law we have that‖(ai + aj) ∧∧k 6=i,j ak‖2 + ‖(ai − aj) ∧∧k 6=i,j ak‖2 == 2(‖a1 ∧ · · · ∧ âj ∧ · · · ∧ aN‖2 + ‖a1 ∧ · · · ∧ âi ∧ · · · ∧ aN‖2).We now want to take the sum of all these equalities, with i < j varying form 1 to N . Weclaim that, for any n = 1, . . . , N , when performing such a sum, in the right hand side,the term 2‖a1 ∧ · · · ∧ ân ∧ · · · ∧ aN‖2 will appear N − 1 times.Indeed, this term may appear with j = n, while i varies from 1 to n − 1, or with i = n,while j varies from n + 1 to N , and there are exactly N − 1 of such possibilities. Hence,summing all the equalities, we have that∑i<j(‖(ai + aj) ∧∧k 6=i,j ak‖2 + ‖(ai − aj) ∧∧k 6=i,j ak‖2)== (N − 1)N∑n=12‖a1 ∧ · · · ∧ ân ∧ · · · ∧ aN‖2.So, we have proved the following.A. Fonda: A generalization of the parallelogram law to higher dimensions 417For any N -dimensional parallelotope, the sum of the squares of the (N − 1)-dimensional measures of its N(N − 1) diagonals is equal to N − 1 times thesum of the squares of the (N − 1)-dimensional measures of its 2N faces.The proof of the theorem is now easily completed, dividing each of the two sums bythe number of their addends and taking the square roots.Remark 3.2. Since our result is valid in any dimension N , it would be interesting toinvestigate whether it could be extended also to some infinite-dimensional vector spaces.This seems to be a remarkable problem which could lead to further insight on the nature ofthese identities.References[1] M. Khosravi and M. D. Taylor, The wedge product and analytic geometry, Amer. Math. Monthly115 (2008), 623–644, doi:10.1080/00029890.2008.11920573.[2] A. Nash, A generalized parallelogram law, Amer. Math. Monthly 110 (2003), 52–57, doi:10.2307/3072345.[3] I. R. Shafarevich and A. O. Remizov, Linear Algebra and Geometry, Springer, Heidelberg, 2013,doi:10.1007/978-3-642-30994-6, translated from the 2009 Russian original by D. Kramer andL. Nekludova.

A generalization of the parallelogram law to higher dimensions

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A generalization of the parallelogram law to higher dimensions

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