On Gram's determinant in $2$-inner product spaces

ON GRAM’S DETERMINANT IN 2-INNER PRODUCT SPACES

Y. J. Cho, M. Mati´ c, and J. Peˇ cari´ c

Abstract. An analogue of Gram’s inequality for 2-inner product spaces is given. Further, a number of inequalities involving Gram’s determinant are stated and proved in terms of 2-inner products.

1. Introduction

The concepts of 2-inner products and 2-inner product spaces have been intensively studied by many authors in the last three decades. A systematic presentation of the recent results related to the theory of 2- inner product spaces as well as an extensive list of the related references can be found in the book [1]. Here we give the basic definitions and the elementary properties of 2-inner product spaces.

Let X be a linear space of dimension greater than 1 over the field K = R of real numbers or the field K = C of complex numbers. Suppose that (·, ·|·) is a K-valued function defined on X × X × X satisfying the following conditions:

(2I ₁ ) (x, x|z) ≥ 0 and (x, x|z) = 0 if and only if x and z are linearly dependent,

(2I 2 ) (x, x|z) = (z, z|x), (2I ₃ ) (y, x|z) = (x, y|z),

(2I 4 ) (αx, y|z) = α(x, y|z) for any scalar α ∈ K, (2I ₅ ) (x + x ⁰ , y|z) = (x, y|z) + (x ⁰ , y|z).

Received September 4, 2000.

2000 Mathematics Subject Classification: Primary 46C05, 46C99; Secondary 26D15, 26D20.

Key words and phrases: 2-inner product, Gram’s determinant, Gram’s inequality.

The first author wishes to acknowledge the financial support of the Korea Research

Foundation Grant (KRF-2000-DP0013).

(·, ·|·) is called a 2-inner product on X and (X, (·, ·|·)) is called a 2-inner product space (or 2-pre-Hilbert space). Some basic properties of 2-inner product (·, ·|·) can be immediately obtained as follows:

(1) If K = R, then (2I ₃ ) reduces to (y, x|z) = (x, y|z).

(2) From (2I ₃ ) and (2I ₄ ), we have

(0, y|z) = 0, (x, 0|z) = 0 and also

(1.1) (x, αy|z) = ¯ α(x, y|z).

(3) Using (2I 3 )–(2I 5 ), we have

(z, z|x ± y) = (x ± y, x ± y|z) = (x, x|z) + (y, y|z) ± 2Re(x, y|z) and

(1.2) Re(x, y|z) = 1

4 [(z, z|x + y) − (z, z|x − y)].

In the real case K = R, (1.2) reduces to

(1.3) (x, y|z) = 1

4 [(z, z|x + y) − (z, z|x − y)]

and, using this formula, it is easy to see that, for any α ∈ R, (1.4) (x, y|αz) = α ² (x, y|z).

In the complex case K = C, using (1.1) and (1.2), we have Im(x, y|z) = Re[−i(x, y|z)] = 1

4 [(z, z|x + iy) − (z, z|x − iy)], which, in combination with (1.2), yields

(1.5) (x, y|z) = 1

4 [(z, z|x+y)−(z, z|x−y)]+ i

4 [(z, z|x+iy)−(z, z|x−iy)].

Using the above formula and (1.1), we have, for any α ∈ C, (1.6) (x, y|αz) = |α| ² (x, y|z).

However, for α ∈ R (1.6) reduces to (1.4). Also, from (1.6) it follows that

(x, y|0) = 0.

(4) For any three given vectors x, y, z ∈ X, consider the vector u = (y, y|z)x−(x, y|z)y. By (2I ₁ ), we know that (u, u|z) ≥ 0 with the equality if and only if u and z are linearly dependent. The inequality (u, u|z) ≥ 0 can be rewritten as

(1.7) (y, y|z)[(x, x|z)(y, y|z) − |(x, y|z)| ² ] ≥ 0.

For x = z, (1.7) becomes

−(y, y|z)|(z, y|z)| ² ≥ 0, which implies that

(1.8) (z, y|z) = (y, z|z) = 0

provided y and z are linearly independent. Obviously, when y and z are linearly dependent, (1.8) holds too. Thus (1.8) is true for any two vectors y, z ∈ X. Now, if y and z are linearly independent, then (y, y|z) > 0 and, from (1.7), it follows

(1.9) |(x, y|z)| ² ≤ (x, x|z)(y, y|z).

Using (1.8), it is easy to check that (1.9) is trivially fulfilled when y and z are linearly dependent. Therefore, the inequality (1.9) holds for any three vectors x, y, z ∈ X and is strict unless the vectors u = (y, y|z)x − (x, y|z)y and z are linearly dependent. In fact, we have the equality in (1.9) if and only if the three vectors x, y and z are linearly dependent.

In any given 2-inner product space (X, (·, · | ·)), we can define a function k · | · k on X × X by

(1.10) kx|zk = p

(x, x|z)

for all x, z ∈ X. It is easy to see that this function satisfies the following conditions:

(2N ₁ ) kx|zk ≥ 0 and kx|zk = 0 if and only if x and z are linearly dependent,

(2N ₂ ) kz|xk = kx|zk,

(2N ₃ ) kαx|zk = |α|kx|zk for any scalar α ∈ K,

(2N 4 ) kx + x ⁰ |zk ≤ kx|zk + kx ⁰ |zk.

Any function k · | · k defined on X × X and satisfying the conditions (2N ₁ )–(2N ₄ ) is called a 2-norm on X and (X, k · | · k) is called a linear 2-normed space. Whenever a 2-inner product space (X, (·, ·|·)) is given, we consider it as a linear 2-normed space (X, k · | · k) with the 2-norm defined by (1.10).

A natural extension of the Cauchy-Schwarz-Bunjakowsky inequality (1.11) |(x, y)| ² ≤ (x, x)(y, y)

in inner product space (X, (·, ·)) is the Gram’s inequality (1.12) Γ(x 1 , x 2 , · · · , x k ) ≥ 0,

which holds for any choice of vectors x ₁ , x ₂ , · · · , x _k ∈ X and is strict unless x 1 , x 2 , · · · , x k are linearly dependent. Also, there are a number of inequalities of various types related to Gram’s determinant

Γ(x ₁ , x ₂ , · · · , x _k ) =

¯ ¯

¯ ¯

¯ ¯

¯ ¯

(x 1 , x 1 ) (x 1 , x 2 ) . . . (x 1 , x k ) (x ₂ , x ₁ ) (x ₂ , x ₂ ) . . . (x ₂ , x _k )

.. . .. . . .. .. . (x _k , x ₁ ) (x _k , x ₂ ) . . . (x _k , x _k )

¯ ¯

¯ ¯

¯ ¯

¯ ¯

(see, for instance, [2, pp. 381–385] or [3, Ch. XX]).

The inequality (1.9) is an analogue of the Cauchy-Schwarz-Bunjakow- sky inequality (1.11) for 2-inner product spaces.

The aim of this paper is to give an analogue of Gram’s inequality (1.12) for 2-inner product spaces as well as the analogues for 2-inner product spaces of some classical inequalities involving Gram’s determi- nant.

In Section 2, we give a definition of Gram’s determinant in 2-inner product space and then prove a version of Gram’s inequality (1.12) for 2-inner product spaces. Also we give a versions of Parseval’s identity and of Bessel’s inequality in 2-inner product spaces.

In Section 3, we prove some further inequalities involving 2-inner product analogue of Gram’s determinant.

Section 4 is devoted to a version for 2-inner product spaces of the well known inequality which can be regarded as a generalization via Gram’s determinant of Cauchy-Schwarz inequality for sequences (see, for instance, [3, p. 599]).

In Section 5, we give a 2-inner product analogue of one well known

result which can be regarded as a generalization via Gram’s determinant

of Bessel’s inequality (see [3, pp. 396-397]). Also we give two interesting consequences of this result (Corollary 6 and Theorem 8) which are in turn 2-inner product analogues of the known classical results (see [3, pp.

603-604]).

2. Gram’s inequality

Let (X, (·, ·|·)) be 2-inner product space over the field of real numbers K = R or the field of complex numbers K = C. For any given vectors x ₁ , x ₂ , · · · , x _k ∈ X and z ∈ X, define a matrix G(x ₁ , x ₂ , · · · , x _k |z) by

G(x ₁ , x ₂ , · · · , x _k |z) =



 



(x ₁ , x ₁ |z) (x ₁ , x ₂ |z) · · · (x ₁ , x _k |z) (x 2 , x 1 |z) (x 2 , x 2 |z) · · · (x 2 , x k |z)

.. . .. . . .. .. . (x k , x 1 |z) (x k , x 2 |z) · · · (x k , x k |z)



 



and Gram’s determinant Γ(x 1 , x 2 , · · · , x k |z) of the vectors x 1 , x 2 , · · · , x k

with respect to the vector z by

(2.1)

Γ(x 1 , x 2 , · · · , x k |z) = det G(x 1 , x 2 , · · · , x k |z)

=

¯ ¯

¯ ¯

¯ ¯

¯ ¯

(x ₁ , x ₁ |z) (x ₁ , x ₂ |z) · · · (x ₁ , x _k |z) (x 2 , x 1 |z) (x 2 , x 2 |z) · · · (x 2 , x k |z)

.. . .. . . .. .. . (x k , x 1 |z) (x k , x 2 |z) · · · (x k , x k |z)

¯ ¯

¯ ¯

¯ ¯

¯ ¯ .

Theorem 1. Let x ₁ , x ₂ , · · · , x _k ∈ X and z ∈ X be given vectors in 2-inner product space X. Then we have

(2.2) Γ(x 1 , x 2 , · · · , x k |z) ≥ 0.

Moreover, the equality holds in (2.2) if and only if the vectors x ₁ , x ₂ , · · · , x _k , z are linearly dependent.

Proof. First we consider the case of the equality in (2.2). Suppose that the vectors x 1 , x 2 , · · · , x k , z are linearly dependent. Then we have (2.3) α ₁ x ₁ + · · · + α _k x _k + βz = 0

for some scalars α ₁ , · · · , α _k , β ∈ K and at least one of them is different from zero. From (2.3), it follows that

(α 1 x 1 + · · · + α k x k + βz, x j |z) = 0

for j = 1, · · · , k, that is, since (z, x _j |z) = 0,

(2.4) α ₁ (x ₁ , x _j |z) + · · · + α _k (x _k x _j |z) = 0

for j = 1, · · · , k. If α 1 = · · · = α k = 0 and β 6= 0, then, from (2.3), we have z = 0 and obviously Γ(x ₁ , x ₂ , · · · , x _k |0) = 0. If α _j 6= 0 for at least one j ∈ {1, · · · , k}, then the system (2.4) has a nontriv- ial solution (α 1 , · · · , α k ), which means that the matrix of the system G(x ₁ , x ₂ , · · · , x _k |z) must be singular and hence Γ(x ₁ , x ₂ , · · · , x _k |z) = 0.

So, if the vectors x 1 , x 2 , · · · , x k , z are linearly dependent, then Γ(x 1 , x 2 ,

· · · , x _k |z) = 0.

Conversely, suppose that Γ(x ₁ , x ₂ , · · · , x _k |z) = 0. Then the system (2.4) has a nontrivial solution (α 1 , · · · , α k ). But (2.4) can be rewritten as

(2.5) (α ₁ x ₁ + · · · + α _k x _k , x _j |z) = 0

for j = 1, · · · , k. Multiplying the j ^th equation in (2.5) by ¯ α _j and then summing over j = 1, · · · , k, we have

(α 1 x 1 + · · · + α k x k , α 1 x 1 + · · · + α k x k |z) = 0.

This means that the vectors α ₁ x ₁ + · · · + α _k x _k and z are linearly depen- dent so that there are the scalars α, β ∈ K such that α 6= 0 or β 6= 0 and

α(α ₁ x ₁ + · · · + α _k x _k ) + βz = 0.

Since α j 6= 0 for at least one j ∈ {1, · · · , k}, we conclude that the vectors x ₁ , x ₂ , · · · , x _k , z are linearly dependent.

Suppose that the vectors x ₁ , x ₂ , · · · , x _k , z are linearly independent.

Then, for r ∈ {1, · · · , k}, the vectors x 1 , x 2 , · · · , x r , z are linearly inde- pendent and

Γ(x 1 , x 2 , · · · , x r |z) 6= 0 for r = 1, · · · , k. Define the vectors y ₁ , · · · , y _k as

y ₁ = x ₁ and

(2.6) y _r =

¯ ¯

¯ ¯

¯ ¯

¯ ¯

x 1

G(x ₁ , · · · , x _r−1 |z) .. . x r−1

(x _r , x ₁ |z) · · · (x _r , x _r − 1|z) x _r

¯ ¯

¯ ¯

¯ ¯

¯ ¯

for r = 2, · · · , k. Expanding the determinant in (2.6) over the last column, we have

(2.7) y _r = α ₁ x ₁ + · · · + α _r−1 x _r−1 + Γ(x ₁ , · · · , x _r−1 |z)x _r and

(2.8)

(y r , x s |z)

= α ₁ (x ₁ , x _s |z) + · · · + α _r−1 (x _r−1 , x _s |z) + Γ(x ₁ , · · · , x _r−1 |z)(x _r , x _s |z)

=

¯ ¯

¯ ¯

¯ ¯

¯ ¯

¯

x ₁ G(x ₁ , · · · , x _r−1 |z) .. .

x r−1

(x r , x 1 |z) · · · (x r , x r − 1|z) x r

¯ ¯

¯ ¯

¯ ¯

¯ ¯

¯

for r = 2, · · · , k and 1 ≤ s ≤ r. If 1 ≤ s < r, then the determinant in (2.8) has two equal columns and hence

(y r , x s |z) = 0, 1 ≤ s < r.

For s = r, it follows from (2.8) that

(y _r , x _r |z) = Γ(x ₁ , · · · , x _r |z).

Now, using the expansion (2.7) and the above equalities, we have (y r , y r |z) = Γ(x 1 , · · · , x r−1 |z)(x r , y r |z)

= Γ(x ₁ , · · · , x _r−1 |z)Γ(x ₁ , · · · , x _r |z) 6= 0

and hence

(y _r , y _r |z) > 0 for r = 2, · · · , k. In fact, we have

(2.9) Γ(x 1 , · · · , x r |z) = (y r , y r |z)

Γ(x 1 , · · · , x r−1 |z) = (y r , y r |z) Γ(x 1 , · · · , x r−1 |z) for r = 2, · · · , k. Now

Γ(x 1 |z) = (x 1 , x 1 |z) > 0,

by the assumed independence of x ₁ and z. Using this and (2.9) with r = 2, we have further

Γ(x 1 , x 2 |z) = (y ₂ , y ₂ |z)

Γ(x ₁ |z) = (y ₂ , y ₂ |z) Γ(x 1 |z) > 0.

Continuing in this way, we conclude that Γ(x 1 , · · · , x r |z) > 0

for all r ∈ {1, · · · , k}. This completes the proof. ¤ Remark 1. The inequality (2.2) is an analogue of the Gram’s in- equality for 2-inner product spaces. In the case when k = 2, (2.2) reduces to

(x 1 , x 1 |z)(x 2 , x 2 |z) − |(x 1 , x 2 |z)| ² ≥ 0

with the equality if and only if the vectors x ₁ , x ₂ , z are linearly depen- dent. This is just the inequality (1.9) so that Gram’s inequality can be regarded as a generalization of Cauchy–Schwarz-Bunjakowsky inequal- ity.

Note that, in the case when the vectors x ₁ , x ₂ , · · · , x _k , z are linearly independent, we can define the vectors y 1 , y 2 , · · · , y k as in the proof above and, from (2.7), it follows that

L(y ₁ , y ₂ , · · · , y _r ) = L(x ₁ , x ₂ , · · · , x _r )

for r = 1, · · · , k. Moreover, from the proof above, we see that (y _r , y _s |z) = 0, 1 ≤ s < r ≤ k.

Also we have

ky ₁ |zk ² = (y ₁ , y ₁ |z) = Γ(x ₁ |z) and

ky r |zk ² = (y r , y r |z) = Γ(x 1 , · · · , x r−1 |z)Γ(x 1 , · · · , x r |z) for r = 2, · · · , k.

Now, the following result is evident:

Corollary 1. Let x ₁ , x ₂ , · · · , x _k , z ∈ X be given linearly indepen- dent vectors in 2-inner product space X. Let the vectors y ₁ , y ₂ , · · · , y _k be defined as in the proof of Theorem 1. Define the vectors e 1 , e 2 , · · · , e k

as

e 1 = y ₁

ky 1 |zk = x ₁ Γ(x ₁ |z) ^1/2 and

e _r = y _r

ky r |zk = y _r

[Γ(x 1 , · · · , x r−1 |z)Γ(x 1 , · · · , x r |z)] ^1/2 for r = 2, · · · , k. Then we have

(i) For r, s ∈ {1, 2, · · · , k},

(e _r , e _s |z) = δ _rs =

½ 0 for r 6= s, 1 for r = s.

(ii) For r ∈ {1, 2, · · · , k},

L(e ₁ , e ₂ , · · · , e _r ) = L(x ₁ , x ₂ , · · · , x _r ).

Remark 2. Note that, in the case when there is an infinite sequence x 1 , x 2 , · · · of linearly independent vectors in the space X, we can take any vector z ∈ X such that z / ∈ L(x ₁ , x ₂ , · · · ) and then construct an infinite sequence of vectors e ₁ , e ₂ , · · · such that the conclusions of Corollary 1 are valid for all r, s ∈ N.

Suppose now that Y is a finite-dimensional linear subspace of 2-inner product space (X, (·, ·|·)) and that z ∈ X is such that z / ∈ Y . If dim Y = n, then, by the Corollary 1, we can construct the base {e 1 , · · · , e n } for Y such that

(2.10) (e _i , e _j |z) = δ _ij for all i, j ∈ {1, · · · , n}.

Any vector x ∈ Y has unique representation of the form x = P _n

i=1 α _i e _i . Using (2.10), we have (x, e j |z) = α j for all j = 1, · · · , n so that

x = X n i=1

(x, e _i |z)e _i

for all x ∈ Y. Therefore, if x, y ∈ Y are two given vectors from the subspace Y, then, using (2.10), we have

(2.11)

(x, y|z) =

³ X ⁿ

i=1

(x, e _i |z)e _i , X n j=1

(y, e _j |z)e _j |z

´

= X n i=1

X n j=1

(x, e _i |z)(y, e _j |z)δ _ij

= X n i=1

(x, e i |z)(e i , y|z),

which is an analogue of Parseval’s identity for 2-inner product spaces.

Especially, for any x ∈ Y , (2.11) with y = x becomes

kx|zk ² = (x, x|z) = X n

i=1

|(x, e _i |z)| ² .

Further, for any x ∈ X, define the vectors u ∈ Y and v ∈ X as

u = X n i=1

(x, e i |z)e i , v = x − u.

For j = 1, · · · , n, we have

(v, e j |z) = µ

x − X n i=1

(x, e i |z)e i , e j |z

¶

= (x, e j |z) − X n

i=1

(x, e i |z)δ ij = 0,

which implies that (v, y|z) = 0 for every y ∈ Z.

Theorem 2. Let Y be a finite-dimensional linear subspace of 2-inner product space X and let z ∈ X is such that z / ∈ Y . Then every x ∈ X can be uniquely represented as

x = u + v,

where u ∈ Y and v ∈ X with (v, y|z) = 0 for all y ∈ Y .

Proof. The existence of the proposed representation for x ∈ X is already proved. It remains to prove the uniqueness. So, suppose that

x = u + v = u ⁰ + v ⁰ ,

where u, u ⁰ ∈ Y and (v, y|z) = (v ⁰ , y|z) = 0 for all y ∈ Y. Then we have v − v ⁰ = u ⁰ − u ∈ Y

and

(u − u ⁰ , u − u ⁰ |z) = (v − v ⁰ , u − u ⁰ |z) = (v, u − u ⁰ |z) − (v ⁰ , u − u ⁰ |z) = 0.

This implies that u − u ⁰ and z are linearly dependent, which is possible only when u−u ⁰ = 0 since z / ∈ Y. Thus, we must have v −v ⁰ = u ⁰ −u = 0, that is v = v ⁰ and u = u ⁰ . This completes the proof. ¤ Corollary 2. Let Y be a finite-dimensional linear subspace of 2- inner product space X and let z ∈ X is such that z / ∈ Y . If {e 1 , · · · , e n } is the base for Y such that (2.10) holds, then, for any x ∈ X,

(2.12)

X n i=1

|(x, e _i |z)| ² ≤ kx|zk ² .

The equality in (2.12) holds if and only if x = u + γz for some u ∈ Y and some scalar γ ∈ K.

Proof. By Theorem 2, every x ∈ X can be represented as x = u + v, where u ∈ Y and (v, y|z) = 0 for all y ∈ Y . More precisely, we have

x = u + v, u = u = X n i=1

(x, e i |z)e i , (v, u|z) = 0, which yields that

kx|zk ² = (u + v, u + v|z)

= (u, u|z) + (u, v|z) + (v, u|z) + (v, v|z)

= ku|zk ² + kv|zk ²

≥ ku|zk ²

= X n i=1

|(x, e _i |z)| ² .

Thus (2.12) is valid. Further, it is evident that we have the equality if and only if kv|zk ² = 0, which is equivalent to the requirement that v and z are linearly dependent, that is, αv + βz = 0 for some scalars α and β with α 6= 0 or β 6= 0. Now, α = 0 would imply that β 6= 0 and z = 0, which can not be true. So we have α 6= 0 and v = x − u = − ^β _α z, that is, x = u − ^β _α z. This completes the proof. ¤ Remark 3. The inequality (2.12) is an analogue of Bessel’s inequal- ity for 2-inner product spaces. It is easy to see that it is also valid for an infinite sequence of vectors. Namely, if e 1 , e 2 , · · · is an infinite sequence of vectors from X and z ∈ X such that

z / ∈ L(e ₁ , e ₂ , · · · ) and

(e _i , e _j |z) = δ _ij

for all i, j = 1, 2, · · · , then we can apply Corollary 2 to the subspace Y = L(e ₁ , · · · , e _n ) to obtain the inequality (2.12) for any fixed n ∈ N.

When n → ∞, we have that X ∞

i=1

|(x, e i |z)| ² ≤ kx|zk ² for any x ∈ X.

3. Some inequalities involving Gram’s determinant

Throughout this section, we assume the notation from the previous two sections. We prove some inequalities involving Gram’s determinant in 2-inner product spaces defined by (2.1). First we need one technical result.

Lemma 1. Let Y be any linear subspace of 2-inner product space X and let z ∈ X such that z / ∈ Y. Suppose that x ∈ X can be represented as

x = u + v,

where u ∈ Y and (v, y|z) = 0 for all y ∈ Y. Then, for arbitrarily chosen vectors x ₁ , · · · , x _m ∈ Y , we have

(3.1) Γ(x, x 1 , · · · , x m |z) = Γ(u, x 1 , · · · , x m |z)+kv|zk ² Γ(x 1 , · · · , x m |z).

Especially, if u ∈ L(x ₁ , · · · , x _m ), then

(3.2) Γ(x, x ₁ , · · · , x _m |z) = kv|zk ² Γ(x ₁ , · · · , x _m |z).

Proof. Under given assumptions, we have, for all j = 1, · · · , m, (x _j , x|z) = (x _j , u + v|z) = (x _j , u|z), (x, x _j |z) = (u + v, x _j |z) = (u, x _j |z).

Also it follows that

(x, x|z) = (u + v, u + v|z) = (u, u|z) + (v, v|z) = (u, u|z) + kv|zk ² . Using this and the elementary properties of determinant, we have

Γ(x, x 1 , · · · , x m |z) =

¯ ¯

¯ ¯

¯ ¯

¯ ¯

(u, u|z) + kv|zk ² (u, x 1 |z) · · · (u, x m |z) (x ₁ , u|z) (x ₁ , x ₁ |z) · · · (x ₁ , x _m |z)

.. . .. . . .. .. .

(x _m , u|z) (x _m , x ₁ |z) · · · (x _m , x _m |z)

¯ ¯

¯ ¯

¯ ¯

¯ ¯

=

¯ ¯

¯ ¯

¯ ¯

¯ ¯

(u, u|z) (u, x ₁ |z) · · · (u, x _m |z) (x 1 , u|z) (x 1 , x 1 |z) · · · (x 1 , x m |z)

.. . .. . . .. .. .

(x m , u|z) (x m , x 1 |z) · · · (x m , x m |z)

¯ ¯

¯ ¯

¯ ¯

¯ ¯

+

¯ ¯

¯ ¯

¯ ¯

¯ ¯

kv|zk ² (u, x 1 |z) · · · (u, x m |z) 0 (x ₁ , x ₁ |z) · · · (x ₁ , x _m |z)

.. . .. . . .. .. . 0 (x _m , x ₁ |z) · · · (x _m , x _m |z)

¯ ¯

¯ ¯

¯ ¯

¯ ¯

= Γ(u, x ₁ , · · · , x _m |z) + kv|zk ² Γ(x ₁ , · · · , x _m |z), which is just the identity (3.1). The identity (3.2) follows directly from (3.1) since u ∈ L(x ₁ , · · · , x _m ) implies that u, x ₁ , · · · , x _m , z are linearly dependent and hence Γ(u, x ₁ , · · · , x _m |z) = 0. This completes

the proof. ¤

Now, we can prove some inequalities involving Gram’s determinant.

Theorem 3. Let x ₁ , · · · , x _m ∈ X be given vectors in 2-inner product space X all different from null vector 0 and let z ∈ X such that z / ∈ L(x ₁ , · · · , x _m ). Then we have

(3.3) Γ(x ₁ , · · · , x _m |z) ≤ kx ₁ |zk ² · · · kx _m |zk ² . For m ≥ 2, the equality in (3.3) holds if and only if

(x i , x j |z) = 0, 1 ≤ i < j ≤ m.

Proof. For m = 1, we have Γ(x ₁ |z) = kx ₁ |zk ² and (3.3) is trivially satisfied. So, take m ≥ 2 and first suppose that x ₁ , · · · , x _m are linearly dependent. Then the left-hand side in (3.3) is equal to zero, while the right-hand side is strictly positive. Namely, the equality kx _j |zk ² = 0 for some j is possible only with x _j = 0, which is excluded. Thus (3.3) holds with strict inequality in this case.

Next, suppose that x ₁ , · · · , x _m are linearly independent and define Y = L(x 2 , · · · , x m ). By Theorem 2, the vector x 1 can be represented as

x ₁ = u + v, u ∈ Y, (v, y|z) = 0 for all y ∈ Y . Applying Lemma 1, we have

(3.4) Γ(x 1 , x 2 , · · · , x m |z) = kv|zk ² Γ(x 2 , · · · , x m |z).

On the other side, (v, u|z) = 0 implies that

(3.5) kx ₁ |zk ² = ku|zk ² + kv|zk ² ≥ kv|zk ² .

Since Γ(x ₂ , · · · , x _m |z) > 0, from (3.4) and (3.5), it follows that (3.6) Γ(x ₁ , x ₂ , · · · , x _m |z) ≤ kx ₁ |zk ² Γ(x ₂ , · · · , x _m |z).

Moreover, the equality in (3.6) holds if and only if ku|zk ² = 0, which is possible only with u = 0. But u = 0 is equivalent with x ₁ = v, that is, (x ₁ , y|z) = 0 for all y ∈ Y, that is,

(x 1 , x i |z) = 0

for all i = 2, · · · , m. Applying analogous observations to Γ(x ₂ , · · · , x _m |z), · · · , Γ(x _m−1 , x _m |z), we easily get the proposed conclusions. This

completes the proof. ¤

Theorem 4. Let x 1 , · · · , x m ∈ X be linearly independent vectors in 2-inner product space X and

Y := L(x 1 , · · · , x m ).

Let z ∈ X be such that z / ∈ Y. Then, for any x ∈ X, we have

y∈Y inf kx − y|zk = min

y∈Y kx − y|zk = [ Γ(x, x ₁ , · · · , x _m |z)

Γ(x 1 , · · · , x m |z) ] ^1/2 .

Proof. Let x ∈ X be given. By Theorem 2, x can be uniquely repre- sented as

x = u + v, u ∈ Y, (v, y|z) = 0

for all y ∈ Y . Now, if y ∈ Y is arbitrarily chosen, then we have x − y = u − y + v, u − y ∈ Y, (v, u − y|z) = 0.

Therefore, we have

kx − y|zk ² = ku − y|zk ² + kv|zk ² ≥ kv|zk ² and the equality occurs when y = u. We conclude that

(3.7) inf

y∈Y kx − y|zk = min

y∈Y kx − y|zk = kv|zk.

On the other hand, by Lemma 1, we have

Γ(x, x ₁ , · · · , x _m |z) = kv|zk ² Γ(x ₁ , · · · , x _m |z).

Also, Γ(x 1 , · · · , x m |z) > 0 since x 1 , · · · , x m are linearly independent and z / ∈ Y, so that

kv|zk ² = Γ(x, x ₁ , · · · , x _m |z) Γ(x 1 , · · · , x m |z) ,

which, in combination with (3.7), proves our assertion. This completes

the proof. ¤

Corollary 3. Let x 1 , · · · , x m ∈ X be linearly independent vectors in 2-inner product space X, m ≥ 2, and let z ∈ X be such that z / ∈ L(x 1 , · · · , x m ). Then we have

(3.8)

Γ(x 1 , · · · , x m |z)

Γ(x ₁ , · · · , x _k |z) ≤ Γ(x 2 , · · · , x m |z) Γ(x ₂ , · · · , x _k |z) ≤ · · ·

≤ Γ(x k , · · · , x m |z)

Γ(x _k |z) ≤ Γ(x _k+1 , · · · , x _m |z) for 1 ≤ k < m. Moreover, the equality

Γ(x _r−1 , · · · , x _m |z)

Γ(x _r−1 , · · · , x _k |z) = Γ(x _r , · · · , x _m |z) Γ(x _r , · · · , x _k |z) occurs for some r ∈ {2, · · · , k} if and only if

x _r−1 = u _r + v _r , u _r ∈ L(x _r , · · · , x _k ), (v _r , x _i |z) = 0 for all i = r, · · · , m. The equality

Γ(x k , · · · , x m |z)

Γ(x _k |z) = Γ(x k+1 , · · · , x m |z)

occurs if and only if (x k , x i |z) = 0 for all i = k + 1, · · · , m.

Proof. First take k = 1. Then (3.8) reduces to (3.9) Γ(x 1 , · · · , x m |z)

Γ(x ₁ |z) ≤ Γ(x 2 , · · · , x m |z),

which is in fact the inequality (3.6). Also, the equality in this inequality occurs if and only if (x ₁ , x _i |z) = 0 for all i = 2, · · · , m, as we proved for the inequality (3.6). Further, suppose that 1 < k < m. Replacing x ₁ , · · · , x _m in (3.9) by x _k , · · · , x _m , we obtain the last inequality in (3.8) and obviously the assertion on the equality case is true. Next, for r ∈ {2, · · · , k}, define the subspaces Y r and Y _r ⁰ as

Y r = L(x r , · · · , x m ), Y _r ⁰ = L(x r , · · · , x k ).

By Theorem 2, the vector x r−1 can be uniquely represented in the fol- lowing two forms

x _r−1 = u _r + v _r , u _r ∈ Y _r , (v _r , x _i |z) = 0 for all i = r, · · · , m and

x _r−1 = u ⁰ _r + v ⁰ _r , u ⁰ _r ∈ Y _r ⁰ , (v _r ⁰ , x _i |z) = 0 for all i = r, · · · , k. Applying Theorem 4, we have

y∈Y inf

kx r−1 − y|zk ² = kv r |zk ² = Γ(x r−1 , x r , · · · , x m |z) Γ(x r , · · · , x m |z) and

y∈Y inf

kx _r−1 − y|zk ² = kv _r ⁰ |zk ² = Γ(x _r−1 , x _r , · · · , x _k |z) Γ(x _r , · · · , x _k |z) .

But Y _r ⁰ ⊆ Y r , which implies that inf y∈Y

kx r−1 −y|zk ² ≤ inf y∈Y

kx r−1 − y|zk ² , that is,

Γ(x r−1 , x r , · · · , x m |z)

Γ(x _r , · · · , x _m |z) ≤ Γ(x r−1 , x r , · · · , x k |z) Γ(x _r , · · · , x _k |z) or, equivalently,

(3.10) Γ(x r−1 , x r , · · · , x m |z)

Γ(x _r−1 , x _r , · · · , x _k |z) ≤ Γ(x r , · · · , x m |z)

Γ(x _r , · · · , x _k |z) .

Moreover, the equality in (3.10) occurs if and only if (3.11) kv _r |zk ² = kv _r ⁰ |zk ² .

Now, from x r−1 = u r + v r = u ⁰ _r + v _r ⁰ , it follows that

v _r ⁰ = u _r − u ⁰ _r + v _r , u _r − u ⁰ _r ∈ Y _r , (v _r , u _r − u ⁰ _r |z) = 0, which implies

kv ⁰ _r |zk ² = ku r − u ⁰ _r |zk ² + kv r |zk ² .

From the above inequality and (3.11), we get ku r − u ⁰ _r |zk ² = 0 which is possible only with u _r − u ⁰ _r = 0. This means that u _r = u ⁰ _r and v _r = v ⁰ _r . In fact (3.11) is equivalent to the requirement

x r−1 = u r + v r , u r = u ⁰ _r ∈ L(x r , · · · , x k ), (v r , x i |z) = 0

for all i = r, · · · , m. This completes the proof. ¤

Corollary 4. Let x ₁ , · · · , x _m ∈ X be arbitrarily chosen vectors in 2-inner product space X, m ≥ 2, and let z ∈ X be such that z / ∈ L(x ₁ , · · · , x _m ). Then we have

(3.12) Γ(x ₁ , · · · , x _k , x _k+1 , · · · , x _m |z)

≤ Γ(x 1 , · · · , x k |z)Γ(x k+1 , · · · , x m |z)

for 1 ≤ k < m. Moreover, the equality in (3.12) can occur only in one of the following three cases:

(i) The vectors x 1 , · · · , x k are linearly dependent.

(ii) The vectors x _k+1 , · · · , x _m are linearly dependent.

(iii) The vectors x 1 , · · · , x m are linearly independent and (x i , x j |z) = 0 for all i ∈ {1, · · · , k} and j ∈ {k + 1, · · · , m}.

Proof. If x ₁ , · · · , x _m are linearly dependent, then (3.12) trivially holds since the left hand side is zero and the right hand side is nonnegative.

Also, the equality in this case occurs in (3.12) if and only if the right

hand side is zero, which is equivalent with the requirement that either the

vectors x ₁ , · · · , x _k are linearly dependent, or the vectors x _k+1 , · · · , x _m

are linearly dependent. Further, if the vectors x 1 , · · · , x m are linearly

independent, then we can apply the first and the last inequality from (3.8) to obtain the inequality

Γ(x ₁ , · · · , x _m |z)

Γ(x 1 , · · · , x k |z) ≤ Γ(x _k+1 , · · · , x _m |z),

which is equivalent to (3.12). Also, the equality occurs in (3.12) if and only if we have the equalities throughout in (3.8), that is,

(3.13) Γ(x k , · · · , x m |z)

Γ(x _k |z) = Γ(x k+1 , · · · , x m |z) and

(3.14) Γ(x r−1 , · · · , x m |z)

Γ(x _r−1 , · · · , x _k |z) = Γ(x r , · · · , x m |z) Γ(x _r , · · · , x _k |z) for all r ∈ {2, · · · , k}. Now, (3.13) is equivalent with

(3.15) (x _k , x _i |z) = 0

for all i = k + 1, · · · , m. Next, (3.14) with r = k is equivalent to (3.16) x _k−1 = u _k + v _k , u _k ∈ L(x _k ), (v _k , x _i |z) = 0

for all i = k, · · · , m. It is easy to see that (3.15) and (3.16) together are equivalent with

(x i , x j |z) = 0

for i ∈ {k − 1, k} and j ∈ {k + 1, · · · , m}. Continuing the argument in this way for r = k − 1, · · · , 2, then we have the equalities throughout in (3.8) if and only if

(x _i , x _j |z) = 0

for all i ∈ {1, · · · , k} and j ∈ {k+1, · · · , m}. This completes the proof.¤

4. A generalization of Cauchy-Schwarz inequality

Let (X, (·, ·|·)) be 2-inner product space over the field of real numbers

K = R or the field of complex numbers K = C. For given m ∈ N,

consider two sequences of vectors x ₁ , · · · , x _m ∈ X and y ₁ , · · · , y _m ∈ X.

Then, for any given z ∈ X, we can define the matrix A of order m by

(4.1) A =



 



(x ₁ , y ₁ |z) (x ₁ , y ₂ |z) · · · (x ₁ , y _m |z) (x ₂ , y ₁ |z) (x ₂ , y ₂ |z) · · · (x ₂ , y _m |z)

.. . .. . . .. .. .

(x _m , y ₁ |z) (x _m , y ₂ |z) · · · (x _m , y _m |z)



 

 .

If we define

Y = L(x ₁ , · · · , x _m , y ₁ , · · · , y _m ),

then Y is a finite-dimensional linear subspace of X of dimension dim Y = n. If z is such that z / ∈ Y, then we can choose the base {e 1 , · · · , e n } for Y such that

(e _i , e _j |z) = δ _ij , i, j ∈ {1, · · · , n}.

Using the Parseval’s identity (2.11), it is easy to see that A can be represented as

(4.2) A =



 

(x 1 , e 1 |z) · · · (x 1 , e n |z) .. . . .. .. . (x m , e 1 |z) · · · (x m , e n |z)



 



 

(e 1 , y 1 |z) · · · (e 1 , y m |z) .. . . .. .. . (e n , y 1 |z) · · · (e n , y m |z)



  .

If m > n, then obviously the vectors x ₁ , · · · , x _m must be linearly depen- dent (the same is true for the vectors (y 1 , · · · , y m ) which implies that the rows (columns) of the matrix A are linearly dependent and hence

det A = 0.

Lemma 2. If m ≤ n, then we have the identity

(4.3) det A = X

1≤j

<j

<···<j

≤n

ξ(j ₁ , j ₂ , · · · , j _m )η(j ₁ , j ₂ , · · · , j _m ), where

(4.4) ξ(j ₁ , j ₂ , · · · , j _m ) =

¯ ¯

¯ ¯

¯ ¯

¯

(x 1 , e j

|z) · · · (x 1 , e j

|z) .. . . .. .. . (x _m , e _j

|z) · · · (x _m , e _j

|z)

¯ ¯

¯ ¯

¯ ¯

¯ and

(4.5) η(j ₁ , j ₂ , · · · , j _m ) =

¯ ¯

¯ ¯

¯ ¯

¯

(e j

, y 1 |z) · · · (e j

, y m |z) .. . . .. .. . (e _j

, y ₁ |z) · · · (e _j

, y _m |z)

¯ ¯

¯ ¯

¯ ¯

¯

.

Proof. Applying Binet-Cauchy’s theorem, we get (4.3) directly from

(4.2). ¤

Theorem 5. Let x ₁ , · · · , x _m be given vectors in 2-inner product space X. Set Y = L(x ₁ , · · · , x _m ) and take any z ∈ X such that z / ∈ Y.

If {e 1 , · · · , e n }, n = dim Y ≥ m is the base for Y such that (e _i , e _j |z) = δ _ij

for all i, j ∈ {1, · · · , n}, then

(4.6) Γ(x ₁ , · · · , x _m |z) = X

1≤j

<j

<···<j

≤n

|ξ(j ₁ , j ₂ , · · · , j _m )| ² ,

where ξ(j ₁ , j ₂ , · · · , j _m ) is defined by (4.4).

Proof. Set y _j = x _j for j = 1, · · · , m. Then, for the matrix A defined by (4.1), we have

A = G(x ₁ , · · · , x _m |z), det A = Γ(x ₁ , · · · , x _m |z).

Also, for ξ(j 1 , j 2 , · · · , j m ) and η(j 1 , j 2 , · · · , j m ) respectively given by (4.4) and (4.5), we have

η(j ₁ , j ₂ , · · · , j _m ) = ξ(j ₁ , j ₂ , · · · , j _m ).

Therefore, (4.3) reduces to (4.6) in this case. ¤

Theorem 6. Let {x ₁ , · · · , x _m } and {y ₁ , · · · , y _m } be two sets of lin- early independent vectors in 2-inner product space X. Set Y = L(x 1 ,

· · · , x _m , y ₁ , · · · , y _m ) and take any z ∈ X such that z / ∈ Y. If A is defined by (4.1), then

(4.7) | det A| ² ≤ Γ(x ₁ , · · · , x _m |z)Γ(y ₁ , · · · , y _m |z).

The equality occurs in (4.7) if and only if {x 1 , · · · , x m } spans the same

subspace as {y ₁ , · · · , y _m } does, that is, if and only if L(x ₁ , · · · , x _m ) =

L(y ₁ , · · · , y _m ) = Y.

Proof. Set n = dim Y. Obviously n ≥ m under given assumptions.

Take any base {e ₁ , · · · , e _n } for Y such that (e _i , e _j |z) = δ _ij

for all i, j ∈ {1, · · · , n}. Then the identity (4.3) is valid and we can apply Cauchy’s inequality for sequences to obtain the inequality

| det A| ²

≤ X

1≤j

<j

<···<j

≤n

|ξ(j ₁ , j ₂ , · · · , j _m )| ² X

1≤j

<j

<···<j

≤n

|η(j ₁ , j ₂ , · · · , j _m )| ² .

By (4.6), the first sum on the right hand side of the above inequality is equal to Γ(x ₁ , · · ·, x _m |z), while the second sum is equal to Γ(y ₁ , · · ·, y _m |z) since, for the transpose M ^τ of a square matrix M , we have det M ^τ = det M. Thus, the above inequality is equivalent to (4.7). It remains the question on the equality case in (4.7).

The orthonormal base {e ₁ , · · · , e _n } for Y can always be chosen so that the first m vectors are obtained by applying the procedure of getting orthonormal vectors described in Corollary 1 to the vectors x ₁ , · · · , x _m . It is easy to see that, in this case, we can express the vectors x ₁ , · · · , x _m in the form

x ₁ = γ ₁ ^1/2 e ₁ , x _r = ³ γ _r

γ r−1

´ _1/2

[α _r,1 e ₁ + · · · + α _r,r−1 e _r−1 + e _r ]

for r = 2, · · · , m, where

γ r = Γ(x 1 , · · · , x r |z)

for r = 1, 2, · · · , m. Therefore, for j = 1, · · · , m, we have (x ₁ , y _j |z) = γ ₁ ^1/2 (e ₁ , y _j |z)

and

(x _r , y _j |z) = ³ γ _r γ _r−1

´ _1/2

[α _r,1 (e ₁ , y _j |z)+· · ·+α _r,r−1 (e _r−1 , y _j |z)+(e _r , y _j |z)]

for r = 2, · · · , m. Using this and the elementary properties of determi- nant, we get

det A = Γ(x ₁ , · · · , x _m |z) ^1/2 det B, where

B =



 



(e ₁ , y ₁ |z) (e ₁ , y ₂ |z) · · · (e ₁ , y _m |z) (e ₂ , y ₁ |z) (e ₂ , y ₂ |z) · · · (e ₂ , y _m |z)

.. . .. . . .. .. . (e _m , y ₁ |z) (e _m , y ₂ |z) · · · (e _m , y _m |z)



 

 .

This means that

| det A| ² = Γ(x 1 , · · · , x m |z)| det B| ² . Note that actually we have

det B = η(1, 2, · · · , m),

where η(j ₁ , j ₂ , · · · , j _m ) for 1 ≤ j ₁ < j ₂ < · · · < j _m ≤ n is given by (4.5).

Now, the equality in (4.7) is equivalent to the requirement that (4.8) Γ(y ₁ , · · · , y _m |z) = | det B| ² = |η(1, 2, · · · , m)| ² . On the other side, by Theorem 5, we have

(4.9) Γ(y 1 , · · · , y m |z) = X

1≤j

<j

<···<j

≤n

|η(j 1 , j 2 , · · · , j m )| ² .

From the equalities (4.8) and (4.9), it follows that the equality in (4.7) holds if and only if

(4.10) η(j ₁ , j ₂ , · · · , j _m ) = 0

for all (j ₁ , j ₂ , · · · , j _m ) 6= (1, 2, · · · , m). Further, consider the vectors

v _i =

¯ ¯

¯ ¯

¯ ¯

¯ ¯

¯

(e 1 , y 1 |z) · · · (e 1 , y m |z) e 1

.. . . .. .. . .. . (e _m , y ₁ |z) · · · (e _m , y _m |z) e _m

(e _i , y ₁ |z) · · · (e _i , y _m |z) e _i

¯ ¯

¯ ¯

¯ ¯

¯ ¯

¯

(i = m + 1, · · · , n).

Expanding the above determinant over the last column and using (4.10), we have

v i = η(1, 2, · · · , m)e i

for i = m + 1, · · · , n. On the other side, we have, for all j = 1, 2, · · · , m and i = m + 1, · · · , n,

(v i , y j |z) =

¯ ¯

¯ ¯

¯ ¯

¯ ¯

(e 1 , y 1 |z) · · · (e 1 , y m |z) (e 1 , y j |z)

.. . . .. .. . .. .

(e _m , y ₁ |z) · · · (e _m , y _m |z) (e _m , y _j |z) (e i , y 1 |z) · · · (e i , y m |z) (e i , y j |z)

¯ ¯

¯ ¯

¯ ¯

¯ ¯

= 0

since two columns in this determinant are identical. This implies that

(e _i , y _j |z) = (v _i , y _j |z) η(1, 2, · · · , m) = 0

for all i = m + 1, · · · , n and j = 1, 2, · · · , m. Using this and the fact that any y ∈ Y is uniquely represented as y = P _n

i=1 (y, e _i |z)e _i , we see that, for all j = 1, 2, · · · , m,

y _j = X m i=1

(y _j , e _i |z)e _i ∈ L(e ₁ , · · · , e _m ) = L(x ₁ , · · · , x _m ).

This means that y 1 , · · · , y m span the same subspace as the one spanned by x ₁ , · · · , x _m , since y ₁ , · · · , y _m are linearly independent. Therefore, (4.10) is equivalent to the requirement that L(x ₁ , · · · , x _m ) = L(y ₁ , · · · ,

y m ). This completes the proof. ¤

Corollary 5. Let x 1 , · · · , x n and y 1 be given vectors in 2-inner product space X. Suppose x ₁ , · · · , x _n are linearly independent and take any z ∈ X such that z / ∈ L(x ₁ , · · · , x _n , y ₁ ). Then

(4.11) Γ(x 1 + y 1 , x 2 , · · · , x n |z) ^1/2

≤ Γ(x ₁ , x ₂ , · · · , x _n |z) ^1/2 + Γ(y ₁ , x ₂ , · · · , x _n |z) ^1/2 . The equality occurs in (4.11) if and only if

(4.12) y ₁ = λx ₁ + u, λ ≥ 0, u ∈ L(x ₂ , · · · , x _n ).

Proof. Using the elementary properties of determinant, we easily have the following identity

Γ(x 1 + y 1 , x 2 , · · · , x n |z) = Γ(x 1 , x 2 , · · · , x n |z) + det A + det A + Γ(y ₁ , x ₂ , · · · , x _n |z),

where

A =



 



(x ₁ , y ₁ |z) (x ₁ , x ₂ |z) · · · (x ₁ , x _n |z) (x ₂ , y ₁ |z) (x ₂ , x ₂ |z) · · · (x ₂ , x _n |z)

.. . .. . . .. .. . (x _n , y ₁ |z) (x _n , x ₂ |z) · · · (x _n , x _n |z)



 

 .

Applying Theorem 6 to the sets of vectors {x 1 , x 2 , · · · , x n } and {y 1 , x 2 ,

· · · , x _n }, we have

| det A| ≤ Γ(x ₁ , x ₂ , · · · , x _n |z) ^1/2 Γ(y ₁ , x ₂ , · · · , x _n |z) ^1/2 . Therefore, we have

Γ(x ₁ + y ₁ , x ₂ , · · · , x _n |z)

= Γ(x ₁ , x ₂ , · · · , x _n |z) + det A + det A + Γ(y ₁ , x ₂ , · · · , x _n |z)

= Γ(x ₁ , x ₂ , · · · , x _n |z) + 2Re[det A] + Γ(y ₁ , x ₂ , · · · , x _n |z)

≤ Γ(x 1 , x 2 , · · · , x n |z) + 2| det A| + Γ(y 1 , x 2 , · · · , x n |z)

≤ Γ(x ₁ , x ₂ , · · · , x _n |z) + 2Γ(x ₁ , x ₂ , · · · , x _n |z) ^1/2 Γ(y ₁ , x ₂ , · · · , x _n |z) ^1/2 + Γ(y 1 , x 2 , · · · , x n |z)

= £

Γ(x 1 , x 2 , · · · , x n |z) ^1/2 + Γ(y 1 , x 2 , · · · , x n |z) ^1/2 ¤ ₂ ,

which yields (4.11). Obviously we have equality in (4.11) if and only if (4.13) Re[det A] = | det A|

= Γ(x 1 , x 2 , · · · , x n |z) ^1/2 Γ(y 1 , x 2 , · · · , x n |z) ^1/2 .

The first equality in (4.12) is equivalent with det A ≥ 0, while the second

one holds if and only if y ₁ , x ₂ , · · · , x _n are linearly dependent or L(y ₁ ,

x ₂ , · · · , x _n ) = L(x ₁ , x ₂ , · · · , x _n ). In the case when y ₁ , x ₂ , · · · , x _n are

linearly dependent, we have y 1 = u ∈ L(x 2 , · · · , x n ) and det A = 0,

while, in the case when L(y ₁ , x ₂ , · · · , x _n ) = L(x ₁ , x ₂ , · · · , x _n ), we have

y ₁ = λx ₁ + u for some λ 6= 0 and some u ∈ L(x ₂ , · · · , x _n ). In this case,

we get det A = λΓ(x ₁ , x ₂ , · · · , x _n |z) so that the condition det A ≥ 0 is

equivalent with the condition λ ≥ 0. This proves that the equality occurs

in (4.11) if and only if (4.12) holds. This completes the proof. ¤