In this paper is defined ann-inner product of typeha1

http://jipam.vu.edu.au/

Volume 7, Issue 2, Article 53, 2006

ON A GENERALIZED n−INNER PRODUCT AND THE CORRESPONDING CAUCHY-SCHWARZ INEQUALITY

KOSTADIN TREN ˇCEVSKI AND RISTO MAL ˇCESKI INSTITUTE OFMATHEMATICS

STS. CYRIL ANDMETHODIUSUNIVERSITY

P.O. BOX162, 1000 SKOPJE

MACEDONIA

kostatre@iunona.pmf.ukim.edu.mk

FACULTY OFSOCIALSCIENCES

ANTONPOPOV,B.B., 1000 SKOPJE

MACEDONIA

rmalcheski@yahoo.com

Received 22 November, 2004; accepted 27 February, 2006 Communicated by C.P. Niculescu

ABSTRACT. In this paper is defined ann-inner product of typeha₁, . . . ,a_n|b₁· · ·b_niwhere a1, . . . ,an,b1, . . . ,bnare vectors from a vector spaceV. This definition generalizes the defini- tion of Misiak ofn-inner product [5], such that in special case if we consider only such pairs of sets{a1, . . . ,a1}and{b1· · ·bn}which differ for at most one vector, we obtain the definition of Misiak. The Cauchy-Schwarz inequality for this general type ofn-inner product is proved and some applications are given.

Key words and phrases: Cauchy-Schwarz inequality,n-inner product,n-norm.

2000 Mathematics Subject Classification. 46C05, 26D20.

1. INTRODUCTION

A. Misiak [5] has introduced ann-inner product by the following definition.

Definition 1.1. Assume that n is a positive integer and V is a real vector space such that dimV ≥nand(•,•| •, . . . ,•

| {z }

n−1

)is a real function defined onV ×V × · · · ×V

| {z }

n+1

such that:

i) (x₁,x₁|x₂, . . . ,x_n)≥0, for anyx₁,x₂, . . . ,x_n ∈V and(x₁,x₁|x₂, . . . ,x_n) = 0if and only ifx₁,x₂, . . . ,x_nare linearly dependent vectors;

ii) (a,b|x₁, . . . ,xn−1) = (ϕ(a), ϕ(b)|π(x₁), . . . , π(xn−1)), for any a,b,x₁, . . . ,xn−1 ∈ V and for any bijections

π:{x₁, . . . ,x_n−1} → {x₁, . . . ,x_n−1} and ϕ:{a,b} → {a,b};

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

219-04

iii) If n > 1, then(x₁,x₁|x₂, . . . ,x_n) = (x₂,x₂|x₁,x₃, . . . ,x_n), for anyx₁,x₂, . . . ,x_n ∈ V;

iv) (αa,b|x₁, . . . ,xn−1) = α(a,b|x₁, . . . ,xn−1), for anya,b,x₁, . . . ,xn−1 ∈ V and any scalarα∈R;

v) (a+a₁,b|x₁, . . . ,xn−1) = (a,b|x₁, . . . ,xn−1) + (a₁,b|x₁, . . . ,xn−1), for anya,b,a₁, x₁, . . . ,xn−1 ∈V.

Then(•,•| •, . . . ,•

| {z }

n−1

)is called then-inner product and(V,(•,•| •, . . . ,•

| {z }

n−1

))is called then-prehilbert space.

Ifn = 1, then Definition 1.1 reduces to the ordinary inner product.

Thisn-inner product induces ann-norm ([5]) by kx1, . . . ,xnk=p

(x1,x1|x2, . . . ,xn).

In the next section we introduce a more general and more convenient definition ofn-inner prod- uct and prove the corresponding Cauchy-Schwarz inequality. In the last section some related results are given.

Although in this paper we only consider real vector spaces, the results of this paper can easily be generalized for the complex vector spaces.

2. n-INNER PRODUCT AND THECAUCHY-SCHWARZ INEQUALITY

First we give the following definition ofn-inner products.

Definition 2.1. Assume thatnis a positive integer,V is a real vector space such thatdimV ≥n andh•, . . . ,•|•, . . . ,•iis a real function onV²ⁿsuch that

i)

(2.1) ha₁, . . . ,a_n|a₁, . . . ,a_ni>0 ifa₁, . . . ,a_nare linearly independent vectors,

ii)

(2.2) ha₁, . . . ,a_n|b₁, . . . ,b_ni=hb₁, . . . ,b_n|a₁, . . . ,a_ni for any a₁, . . . ,a_n,b₁, . . . ,b_n ∈V,

iii)

(2.3) hλa₁, . . . ,a_n|b₁, . . . ,b_ni=λha₁, . . . ,a_n|b₁, . . . ,b_ni for any scalarλ∈Rand any a₁, . . . ,a_n,b₁, . . . ,b_n∈V,

iv)

(2.4) ha₁, . . . ,a_n|b₁, . . . ,b_ni=−ha_σ(1), . . . ,a_σ(n)|b₁, . . . ,b_ni

for any odd permutationσin the set{1, . . . , n}and anya₁, . . . ,a_n,b₁, . . . ,b_n ∈V, v)

(2.5) ha₁+c,a₂, . . . ,a_n|b₁, . . . ,b_ni

=ha1,a2, . . . ,an|b1, . . . ,bni+hc,a2, . . . ,an|b1, . . . ,bni for anya₁, . . . ,a_n,b₁, . . . ,b_n,c∈V,

vi) if

(2.6) ha₁,b₁, . . . ,bi−1,b_i+1, . . . ,b_n|b₁, . . . ,b_ni= 0 for eachi∈ {1,2, . . . , n}, then

(2.7) ha1, . . . ,an|b1, . . . ,bni= 0 for arbitrary vectorsa₂, . . . ,a_n.

Then the function h•, . . . ,•|•, . . . ,•i is called an n-inner product and the pair (V,h•, . . . ,•|

•, . . . ,•i)is called ann-prehilbert space.

We give some consequences from the conditions i) – vi) of Definition 2.1.

From (2.4) it follows that if two of the vectors a₁, . . . ,a_n are equal, then ha₁, . . . ,a_n| b₁, . . . ,b_ni= 0.

From (2.3) it follows that

ha₁, . . . ,a_n|b₁, . . . ,b_ni= 0 if there existsisuch thata_i = 0.

From (2.4) and (2.2) it follows more generally that iv’)

ha₁, . . . ,a_n|b₁, . . . ,b_ni= (−1)sgn(π)+sgn(τ)ha_π(1), . . . ,a_π(n)|b_τ(1), . . . ,b_τ(n)i for any permutationsπandτ on{1, . . . , n}and a₁, . . . ,a_n,b₁, . . . ,b_n ∈V. From (2.3), (2.4) and (2.5) it follows that

ha₁, . . . ,a_n|b₁, . . . ,b_ni= 0

ifa₁, . . . ,a_nare linearly dependent vectors. Thus i) can be replaced by

i’) ha₁, . . . ,a_n|a₁, . . . ,a_ni ≥ 0for anya₁, . . . ,a_n ∈V andha₁, . . . ,a_n|a₁, . . . ,a_ni= 0if and only ifa₁, . . . ,a_nare linearly dependent vectors.

Note that then-inner product onV induces ann-normed space by kx₁, . . . ,x_nk=p

hx₁, . . . ,x_n|x₁, . . . ,x_ni, and it is the same norm induced by Definition 1.1.

In the special case if we consider only such pairs of sets a1, . . . ,a1 and b1, . . . ,bn which differ for at most one vector, for examplea₁ =a, b₁ =banda₂ =b₂ = x₁, . . . ,a_n =b_n = x_n−1, then by putting

(a,b|x₁, . . . ,xn−1) = ha,x₁, . . . ,xn−1|b,x₁, . . . ,xn−1i

we obtain ann-inner product according to Definition 1.1 of Misiak. Indeed, the conditions i), iv) and v) are triavially satisfied. The condition ii) is satisfied for an arbitrary permutationπ, because according to iv⁰)

ha1, . . . ,an|b1, . . . ,bni=ha1,a_π(2), . . . ,a_π(n)|b1,b_π(2), . . . ,b_π(n)i

for any permutationπ :{2,3, . . . , n} → {2,3, . . . , n}. Similarly the condition iii) is satisfied.

Moreover, in this special case of Definition 2.1 we do not have any restriction of Definition 1.1.

For example, then the condition vi) does not say anything. Namely, ifa₁ ∈ {b/ ₁, . . . ,b_n}, then the vectors a₂, . . . ,a_n must be from the set {b₁, . . . ,b_n}, and (2.6⁰⁰) is satisfied because the assumption(2.6⁰)is satisfied. Ifa₁ ∈ {b₁, . . . ,b_n}, for examplea₁ =b_j, then(2.6⁰)implies thathb_j,b₁, . . . ,bj−1,b_j+1, . . . ,b_n|b₁, . . . ,b_ni = 0, and it is possible only if b₁, . . . ,b_n are linearly dependent vectors. However, then(2.6⁰⁰)is satisfied. Thus Definition 2.1 generalizes Definition 1.1.

Now we give the following example of then-inner product.

Example 2.1. We refer to the classical known example, as an n-inner product according to Definition 2.1. LetV be a space with inner producth·|·i. Then

ha₁, . . . ,a_n|b₁, . . . ,b_ni=

ha₁|b₁i ha₁|b₂i · · · ha₁|b_ni ha₂|b₁i ha₂|b₂i · · · ha₂|b_ni

·

·

·

ha_n|b₁i ha_n|b₂i · · · ha_n|b_ni

satisfies the conditions i) - vi) and hence it defines ann-inner product onV. The conditions i) - v) are trivial, and we will prove vi). Ifb₁, . . . ,b_nare linearly independent vectors and

ha1,b1, . . . ,bi−1,bi+1, . . . ,bn|b1, . . . ,bni

≡(−1)ⁱ⁻¹hb1, . . . ,bi−1,a1,bi+1, . . . ,bn|b1, . . . ,bni

≡(−1)ⁱ⁻¹

hb₁|b₁i hb₁|b₂i · · · hb₁|b_ni hb₂|b₁i hb₂|b₂i · · · hb₂|b_ni

· · · · ha₁|b₁i ha₁|b₂i · · · ha₁|b_ni

· · · · hbn|b1i hbn|b2i · · · hbn|bni

= 0,

then the vector

(ha₁|b₁i,ha₁|b₂i, . . . ,ha₁|b_ni)∈Rⁿ is a linear combination of

(hb₁|b₁i, . . . ,hb₁|b_ni), . . . ,(hbi−1|b₁i, . . . ,hbi−1|b_ni), (hbi+1|b1i, . . . ,hbi+1|bni), . . . ,(hbn|b1i, . . . ,hbn|bni).

Since this is true for each i ∈ {1,2, . . . , n}, it must be that ha₁|b₁i = · · · = ha₁|b_ni = 0.

Hence

ha₁, . . . ,a_n|b₁, . . . ,b_ni= 0 for arbitrarya₂, . . . ,a_n.

Note that the inner product defined by

ha₁∧ · · · ∧an|b₁∧ · · · ∧bni=

ha₁|b₁i ha₁|b₂i · · · ha₁|b_ni ha₂|b₁i ha₂|b₂i · · · ha₂|b_ni

·

·

·

han|b1i han|b2i · · · han|bni

can uniquely be extended to ordinary inner products over the spaceΛ_n(V)ofn-forms overV [4]. Indeed, if{e_i}i∈I,I an index set, is an orthonormal basis of(V,h∗|∗i), then

he_i1 ∧ · · · ∧e_in|e_j1 ∧ · · · ∧e_jni=δⁱ_j^1···in

1···j_n

where the expression δ_jⁱ¹₁^···i_···jⁿ_n is equal to 1 or -1 if {i₁, . . . , i_n} = {j₁, . . . , j_n} with different i₁, . . . , i_nand additionally the permutation _j^i1i2···in

1j2···jn

is even or odd respectively, and where the above expression is 0 otherwise. It implies an inner product overΛn(V).

Before we prove the next theorem, we give the following remarks assuming thatdimV > n.

Letb₁, . . . ,b_nbe linearly independent vectors. If a vectorais such that ha,b₁, . . . ,bi−1,b_i+1, . . . ,b_n|b₁, . . . ,b_ni= 0, (1≤i≤n)

then we say that the vectorais orthogonal to the subspace generated byb₁, . . . ,b_n. Note that the set of orthogonal vectors to thisn-dimensional subspace is a vector subspace ofV, and the orthogonality ofato the considered vector subspace is invariant of the base vectorsb₁, . . . ,b_n. Ifxis an arbitrary vector, then there exist uniqueλ₁, . . . , λ_n ∈ Rsuch thatx−λ₁b₁ − · · · − λ_nb_nis orthogonal to the vector subspace generated byb₁, . . . ,b_n. Namely, the orthogonality conditions

hb₁, . . . ,bi−1,x−λ₁b₁− · · · −λ_nb_n,b_i+1, . . . ,b_n|b₁, . . . ,b_ni= 0, (1≤i≤n) have unique solutions

λ_i = hb₁, . . . ,bi−1,x,b_i+1, . . . ,b_n|b₁, . . . ,b_ni

hb₁, . . . ,b_n|b₁, . . . ,b_ni , (1≤i≤n).

Hence each vector x can uniquely be decomposed as x = λ₁b₁ + · · ·+λ_nb_n + c, where the vector c is orthogonal to the vector subspace generated by b₁, . . . ,b_n. According to this definition, the condition vi) of Definition 2.1 says that if the vectora₁is orthogonal to the vector subspace generated byb₁, . . . ,b_n, then(2.6⁰⁰)holds for arbitrary vectorsa₂, . . . ,a_n.

Now we prove the Cauchy-Schwarz inequality as a consequence of Definition 2.1.

Theorem 2.1. Ifh•, . . . ,•|•, . . . ,•i is ann-inner product onV, then the following inequality (2.8) ha1, . . . ,an|b1, . . . ,bni² ≤ ha1, . . . ,an|a1, . . . ,anihb1, . . . ,bn|b1, . . . ,bni,

is true for any vectorsa₁, . . . ,a_n,b₁, . . . ,b_n ∈ V. Moreover, equality holds if and only if at least one of the following conditions is satisfied

i) the vectorsa1,a2, . . . ,anare linearly dependent, ii) the vectorsb₁,b₂, . . . ,b_nare linearly dependent,

iii) the vectorsa₁,a₂, . . . ,a_nandb₁,b₂, . . . ,b_n generate the same vector subspace of di- mensionn.

Proof. Ifa₁, . . . ,a_nare linearly dependent vectors orb₁, . . . ,b_nare linearly dependent vectors, then both sides of (2.8) are zero and hence equality holds. Thus, suppose thata1, . . . ,an and alsob1, . . . ,bnare linearly independent vectors. Note that the inequality (2.8) does not depend on the choice of the basisa₁, . . . ,a_nof the subspace generated by thesenvectors. Indeed, each vector row operation preserves the inequality (2.8), because both sides are invariant or both sides are multiplied by a positive real scalar after any elementary vector row operation. We assume thatdimV > n, because ifdimV =n, then the theorem is obviously satisfied.

LetΣbe a space generated by the vectorsa₁, . . . ,a_n andΣ^∗ be the orthogonal subspace to Σ. Let us decompose the vectorsb_i asb_i =c_i+d_i wherec_i ∈Σandd_i ∈Σ^∗. Thus

b_i =

n

X

j=1

P_ija_j +d_i, (1≤i≤n)

ha₁, . . . ,a_n|b₁, . . . ,b_ni=

*

a₁, . . . ,a_n

n

X

j1=1

P_1j1a_j1 +d₁, . . . ,

n

X

jn=1

P_njna_jn+d_n +

=

n

X

j1=1

· · ·

n

X

jn=1

P_1j1P_2j2· · ·P_njnha₁, . . . ,a_n|a_j1, . . . ,a_jni

=

n

X

j1=1

· · ·

n

X

jn=1

P_1j1P_2j2· · ·P_njn(−1)^sgnσha₁, . . . ,a_n|a₁, . . . ,a_ni

=detP · ha₁, . . . ,a_n|a₁, . . . ,a_ni

where we used the conditions ii) - vi) from Definition 2.1 and we denoted byP the matrix with entriesP_ij, andσ= _j^{1 2···n}

1j2···jn

.

IfdetP = 0, then the left side of (2.8) is 0, the right side is positive and hence the inequality (2.8) is true. So, let us suppose that P is a non-singular matrix and Q = P⁻¹. Now the inequality (2.8) is equivalent to

(detP)²ha₁, . . . ,a_n|a₁, . . . ,a_ni² ≤ ha₁, . . . ,a_n|a₁, . . . ,a_nihb₁, . . . ,b_n|b₁, . . . ,b_ni,

(2.9) ha1, . . . ,an|a1, . . . ,ani ≤ hb⁰₁, . . . ,b⁰_n|b⁰₁, . . . ,b⁰_ni, whereb⁰_i =Pn

j=1Q_ijb_j, (1≤i≤n). Note thatb⁰_idecomposes as b⁰_i =

n

X

j=1

Q_ij

n

X

l=1

P_jla_l+d_j

!

=a_i+d⁰_i whered⁰_i =Pn

j=1Q_ijd_j ∈Σ^∗. Now we will prove (2.9), i.e.

(2.10) ha₁, . . . ,a_n|a₁, . . . ,a_ni ≤ ha₁+d⁰₁, . . . ,a_n+d⁰_n|a₁+d⁰₁, . . . ,a_n+d⁰_ni

and equality holds if and only if b⁰₁ = a₁, . . ., b⁰_n = a_n, i.e., d⁰₁ = · · · = d⁰_n = 0. More precisely, we will prove that (2.10) is true for at least one basisa₁, . . . ,a_nofΣ.

Using (2.5) and (2.2) we obtain

ha₁ +d⁰₁, . . . ,a_n+d⁰_n|a₁+d⁰₁, . . . ,a_n+d⁰_ni

=ha₁,a₂+d⁰₂, . . . ,a_n+d⁰_n|a₁,a₂+d⁰₂, . . . ,a_n+d⁰_ni

+hd⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_n|d⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_ni + 2ha₁,a₂+d⁰₂, . . . ,a_n+d⁰_n|d⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_ni

=ha1,a2,a3+d⁰₃, . . . ,an+d⁰_n|a1,a2,a3+d⁰₃, . . . ,an+d⁰_ni

+ha₁,d⁰₂,a₃+d⁰₃, . . . ,a_n+d⁰_n|a₁,d⁰₂,a₃+d⁰₃, . . . ,a_n+d⁰_ni +hd⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_n|d⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_ni + 2ha₁,a₂+d⁰₂, . . . ,a_n+d⁰_n|d⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_ni + 2ha₁,a₂,a₃+d⁰₃, . . . ,a_n+d⁰_n|a₁,d⁰₂,a₃+d⁰₃, . . . ,a_n+d⁰_ni

=· · ·

=ha₁, . . . ,a_n|a₁, . . . ,a_ni+ha₁, . . . ,an−1,d⁰_n|a₁, . . . ,an−1,d⁰_ni +· · ·+ha1,d⁰₂, . . . ,an+d⁰_n|a1,d⁰₂, . . . ,an+d⁰_ni

+hd⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_n|d⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_ni+S, where

S = 2ha₁,a₂+d⁰₂, . . . ,a_n+d⁰_n|d⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_ni

+ 2ha₁,a₂,a₃+d⁰₃, . . . ,a_n+d⁰_n|a₁,d⁰₂,a₃+d⁰₃, . . . ,a_n+d⁰_ni

+· · ·+ 2ha₁,a₂, . . . ,an−1,a_n|a₁,a₂, . . . ,an−1,d⁰_ni. We can change the basisa₁, . . . ,a_n ofΣsuch that the sum S vanishes. Indeed, if we replace a₁ byλa₁ we can choose almost always a scalar λ such that S = 0. The other cases can be considered by another analogous linear transformations. Thus without loss of generality we can putS = 0.

According to i⁰) the inequality (2.10) is true and equality holds if and only if the following sets of vectors{a₁, . . . ,an−1,d⁰_n},. . .,{a₁,d⁰₂,a₃+d⁰₃, . . . ,a_n+d⁰_n},{d⁰₁,a₂+d⁰₂, . . . ,a_n+d⁰_n}

are linearly dependent. This is satisfied if and only ifd⁰₁ =d⁰₂ =· · · =d⁰_n= 0, i.e. if and only

ifb⁰₁ =a₁,. . .,b⁰_n=a_n.

3. SOME APPLICATIONS

LetΣ₁andΣ₂ be two subspaces ofV of dimensionn. We define the angleϕbetweenΣ₁and Σ₂by

(3.1) cosϕ = ha₁, . . . ,a_n|b₁, . . . ,b_ni ka₁, . . . ,a_nk · kb₁, . . . ,b_nk,

where a₁, . . . ,a_n are linearly independent vectors of Σ₁, b₁, . . . ,b_n are linearly independent vectors ofΣ₂ and

ka₁, . . . ,a_nk=p

ha₁, . . . ,a_n|a₁, . . . ,a_ni, kb₁, . . . ,b_nk=p

hb₁, . . . ,b_n|b₁, . . . ,b_ni.

The angleϕdoes not depend on the choice of the basesa₁, . . . ,a_nandb₁, . . . ,b_n.

Note that anyn-inner product induces an ordinary inner product over the vector spaceΛ_n(V) ofn-forms onV as follows. Let{e_α}, be a basis ofV. Then we define

* X

i1,...,in

a_i1···ine_i1 ∧ · · · ∧e_in

X

j1,...,jn

b_j1···jne_j1 ∧ · · · ∧e_jn +

= X

i1,...,in,j1,...,jn

a_i1···inb_j1···jnhe_i1, . . . ,e_in|e_j1, . . . ,e_jni.

The first requirement for the inner product is a consequence of Theorem 2.1. For example, if w=pei1 ∧ · · · ∧ein−qej1 ∧ · · · ∧ejn,

then

hw|wi=p²he_i1, . . . ,e_in|e_i1, . . . ,e_ini+q²he_j1, . . . ,e_jn|e_j1, . . . ,e_jni

−2pqhe_i1, . . . ,e_in|e_j1, . . . ,e_jni ≥0 and moreover, the last expression is 0 if and only if

he_i1, . . . ,e_in|e_j1, . . . ,e_jni=p

he_i1, . . . ,e_in|e_i1, . . . ,e_ini q

he_j1, . . . ,e_jn|e_j1, . . . ,e_jni which means thate_i1, . . . ,e_inande_j1, . . . ,e_jngenerate the same subspace, andpe_i1∧· · ·∧e_in = qe_j1∧· · ·∧e_jn, i.e. if and only ifw= 0. The other requirements for inner products are obviously satisfied. Hence we obtain an induced ordinary inner product on the vector spaceΛ_n(V)ofn- forms onV.

Remark 3.1. Note that the inner product onΛn(V)introduced in Example 2.1 is only a special case of an inner product onΛ_n(V)and alson-inner product. It is induced via the existence of an ordinary inner product onV.

The angle between subspaces defined by (3.1) coincides with the angle between twon-forms in the vector spaceΛ_n(V). Since the angle between two "lines" in any vector space with ordi- nary inner product can be considered as a distance, we obtain that

(3.2) ϕ= arccos ha₁, . . . ,a_n|b₁, . . . ,b_ni ka₁, . . . ,a_nk · kb₁, . . . ,b_nk

determines a metric among the n-dimensional subspaces of V. Indeed, it induces a metric on the Grassmann manifold G_n(V), which is compatible with the ordinary topology of the

Grassman manifoldG_n(V). This metric over Grassmann manifolds appears natural and appears convenient also for the infinite dimensional vector spacesV.

Further, we shall consider a special case of ann-inner product for which there exists a ba- sis {e_α} of V such that the vectore_i is orthogonal to the subspace generated by the vectors e_i1, . . . ,e_in for different values ofi, i₁, . . . , i_n. For such ann-inner product we have

(3.3) he_i1, . . . ,e_in|e_j1, . . . ,e_jni=C_i1···inδ_j^i1···in

1···j_n

where δ_j^i1···in

1···j_n is equal to 1 or -1 if {i₁, . . . , i_n} = {j₁, . . . , j_n} with different i₁, . . . , i_n, the permutation _j^i1i2···in

1j2···jn

is even or odd respectively, the expression is 0 otherwise, and where C_i1···in > 0. Moreover, one can verify that the previous formula induces an n-inner product, i.e. the six conditions i) - vi) are satisfied if and only if all the coefficients C_i1···in are equal to a positive constant C > 0. Moreover, we can assume that C = 1, because otherwise we can consider the basis{eα/C^1/2n} instead of the basis{eα} ofV. Hence this special case of n-inner product reduces to then-inner product given by the Example 2.1. Indeed, the ordinary inner product is uniquely defined such that{e_α}has an orthonormal system of vectors.

If the dimension of V is finite, for example dimV = m > n, then the previous n-inner product induces a dual(m−n)-inner product onV which is induced by

(3.4) he_i1, . . . ,e_im−n|e_j1, . . . ,e_jm−ni^∗ =δ_j^{i1···im−n}

1···jm−n.

The dual(m−n)-inner product is defined using the "orthonormal basis"{e_α}ofV. If we have chosen another "orthonormal basis", the result will be the same. Further we prove the following theorem.

Theorem 3.2. LetV be a finite dimensional vector space and let then-inner product onV be defined as in Example 2.1. Then

ϕ(Σ₁,Σ₂) = ϕ(Σ^∗₁,Σ^∗₂),

whereΣ₁ andΣ₂ are arbitraryn-dimensional subspaces ofV andΣ^∗₁ andΣ^∗₂are their orthog- onal subspaces inV.

Proof. LetΣ₁ = hω₁i, Σ₂ = hω₂i, Σ^∗₁ = hω₁^∗i, Σ^∗₂ = hω^∗₂i, where kω₁k = kω₂k = kω^∗₁k = kω₂^∗k= 1. We will prove that

ω1·ω2 =±ω^∗₁ ·ω₂^∗.

Indeed,ω₁·ω₂ =ω₁^∗·ω₂^∗ ifω₁∧ω₂ andω^∗₁∧ω₂^∗ have the same orientation inV andω₁·ω₂ =

−ω₁^∗·ω^∗₂ ifω1∧ω2 andω₁^∗∧ω^∗₂ have the opposite orientations inV.

Assume that the dimension ofV ism. Without loss of generality we can assume that ω₁ =e₁∧e₂∧ · · · ∧e_n and ω₁^∗ =e_n+1∧e_n+2∧ · · · ∧e_m.

Without loss of generality we can assume that

ω₂ =a₁∧a₂∧ · · · ∧a_n and ω₂^∗ =a_n+1∧a_n+2∧ · · · ∧a_m,

wherea₁, . . . ,a_m is an orthonormal system. Suppose thata_i = (a_i1, . . . , a_im) (1 ≤ i ≤ m), and let us introduce an orthogonalm×mmatrix

A=































a₁₁ · · · a_1n a_1,n+1 · · · a_1m

·

·

·

a_n1 · · · a_nn a_n,n+1 · · · a_nm a_n+1,1 · · · a_n+1,n a_n+1,n+1 · · · a_n+1,m

·

·

·

a_m1 · · · a_mn a_m,n+1 · · · a_mm





























 .

We denote byA_i1···in (1 ≤ i₁ < i₂ < · · · < i_n ≤ m), then×n submatrix ofA whose rows are the firstn rows ofAand whose columns are thei₁-th,...,i_n-th column ofA. We denote by A^∗_i1···in the(m−n)×(m−n)submatrix ofAwhich is obtained by deleting the rows and the columns corresponding to the submatrixA_i1···in. It is easy to verify that

ω₁·ω₂ = detA_12...n and ω^∗₁·ω₂^∗ = detA^∗_12...n and thus we have to prove that

(3.5) detA_12...n =±detA^∗_12...n, i.e.

detA_12...n = detA^∗_12...n if detA= 1 and

detA_12...n =−detA^∗_12...n if detA=−1.

Assume thatdetA= 1. Let us consider the expression

F = X

1≤i₁<i2<···<in≤m

h

(detA_i1i2···in −(−1)^1+2+···+n(−1)^{i1+i2+···+in}detA^∗_i1i2···in)i2

.

Usingkω2k= 1andkω₂^∗k= 1we get X

1≤i1<···<in≤m

(detAi1i2···in)² = X

1≤i1<···<in≤m

(detA^∗_i1i2···in)² = 1

and using the Laplace formula for decomposition of determinants, we obtain

F = X

1≤i1<···<in≤m

(detA_i1i2···in)²+ X

1≤i1<···<in≤m

(detA^∗_i

1i2···i_n)²

−2 X

1≤i1<···<in≤m

(−1)^n(n+1)/2(−1)^{i1+i2+···+in}detAi1i2···indetA^∗_i1i2···in

= 1 + 1−2·detA= 2−2 = 0.

HenceF = 0implies that

detA_i1i2···in = (−1)^n(n+1)/2(−1)^{i1+i2+···+in}detA^∗_i

1i2···i_n. In particular, fori1 = 1, . . . , in=nwe obtain

detA_i1i2···in = detA^∗_i1i2···in.

Assume thatdetA=−1. Then we consider the expression

F⁰ = X

1≤i1<i2<···<in≤m

h

(detAi1i2···in+ (−1)^1+2+···+n(−1)^{i1+i2+···+in}detA^∗_i1i2···in) i2

and analogously we obtain that

detA_i1i2···i_n =−(−1)^n(n+1)/2(−1)^{i1+i2+···+in}detA^∗_i

1i2···in. In particular, fori₁ = 1, . . . , i_n=nwe obtain

detA_i1i2···in =−detA^∗_i1i2···in.

Finally we make the following remark. The presented approach ton-inner products appears to be essential for applications in functional analysis. Since the corresponding n-norm is the same as the correspondingn-norm from the definition of Misiak, we have the same results in the normed spaces. It is an open question whether from Definition 2.1 a generalized n-inner product and n-semi-inner product with characteristic p can be introduced. It may also be of interest to research the strong convexity in the possibly introduced space with n-semi-inner product with characteristicp.

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