Study on Orthogonal Sets for Birkhoff Orthogonality

Abstract

We introduce the notion of orthogonal sets for Birkhoff orthogonality, which we will call Birkhoff orthogonal sets in this paper. As a generalization of orthogonal sets in Hilbert spaces, Birkhoff orthogonal sets are not necessarily linearly independent sets in finite-dimensional real normed spaces. We prove that the Birkhoff orthogonal set $A=\{x_1,x_2,\dots ,x_n\}(n\ge 3)$ containing $n-3$ right symmetric points is linearly independent in smooth normed spaces. In particular, we obtain similar results in strictly convex normed spaces when $n=3$ and in both smooth and strictly convex normed spaces when $n=4$. These obtained results can be applied to the mutually Birkhoff orthogonal sets studied in recently.

1. Introduction

Orthogonality plays a key role in studying the geometry of Hilbert spaces. In order to study the geometry of general normed spaces, many notions of generalized orthogonality have been introduced in the literature [1,2,3,4,5]. Let H be a Hilbert space, $A\subset H$ is said to be an orthogonal set if for any $x,y\in A(x\ne y)$, there is $x\perp y$ (x is orthogonal to y). Orthogonal sets are linearly independent sets in Hilbert spaces; it is thus, natural to study the linear independence of generalized orthogonal sets in general normed spaces.

Throughout this paper, we use X to denote a finite-dimensional real normed space.

Definition 1 

([3,4,5]). For $x,y\in X$, $x\in X$ is said to be Roberts orthogonal to $y\in X$, denoted by $x{\perp }_{R}y$, if

$$\Vert x+\lambda y\Vert =\Vert x-\lambda y\Vert$$

for each $\lambda \in R$.

$x\in X$ is said to be isosceles orthogonal to $y\in X$, denoted by $x{\perp }_{I}y$, if

$$\Vert x+y\Vert =\Vert x-y\Vert .$$

$x\in X$ is said to be Birkhoff orthogonal to $y\in X$, denoted by $x{\perp }_{B}y$, if

$$\Vert x+\lambda y\Vert \ge \Vert y\Vert$$

for each $\lambda \in R$. y is called a Birkhoff orthogonal element to x.

If $x,y\in X\setminus \left\{0\right\}$ and $x{\perp }_{B}y$, then x and y are linearly independent. If $x{\perp }_{R}y$, then there must be $x{\perp }_{I}y$ and $x{\perp }_{B}y$. Inspired by the definition of an orthogonal set in Hilbert spaces, we introduce several types of generalized orthogonal set in X.

Definition 2. 

$A\subset X\setminus \left\{0\right\}$ is called a Roberts orthogonal set, if for any $x,y\in A(x\ne y)$, there is $x{\perp }_{R}y$.

$A\subset X\setminus \left\{0\right\}$ is called an isosceles orthogonal set, if for any $x,y\in A(x\ne y)$, there is $x{\perp }_{I}y$.

$A\subset X\setminus \left\{0\right\}$ is called a Birkhoff orthogonal set, if for any $x,y\in A(x\ne y)$, there is $x{\perp }_{B}y$ or $y{\perp }_{B}x$.

In [6], Alonso proposed that for $x,y,z\in X(dimX=2)$, if $x{\perp }_{I}y$, $x{\perp }_{I}z$ and $y{\perp }_{I}z$, then $\Vert x\Vert \Vert y\Vert \Vert z\Vert =0$. It is shown that isosceles orthogonal sets are linearly independent sets in two-dimensional normed spaces. Since a Roberts orthogonal set must be an isosceles orthogonal set, Roberts orthogonal sets are also linearly independent sets in two-dimensional normed spaces. However, this result does not apply to Birkhoff orthogonal sets.

Birkhoff orthogonality plays a key role in general normed spaces. A few important geometric properties of the normed space, such as smoothness, strictly convex, uniform convexity and so on, can use this orthogonality. Unlike Roberts orthogonality and isosceles orthogonality, Birkhoff orthogonality is not necessarily symmetric in a normed space. In [7], Sain et al. introduced the definition of left symmetric points and right symmetric points in Banach spaces. $x\in X$ is called a left symmetric point if $x{\perp }_{B}y$ implies that $y{\perp }_{B}x$ for all $y\in X$. Similarly, $x\in X$ is called a right symmetric point if $y{\perp }_{B}x$ implies that $x{\perp }_{B}y$ for all $y\in X$. We note that in the study of symmetry in Birkhoff orthogonality, many scholars have focused on characterizing left symmetric points and right symmetric points in different types of normed spaces [7,8,9,10,11,12,13,14]. However, there exist $x,y\in X$, which are neither left symmetric points nor right symmetric points but satisfy $x{\perp }_{B}y$ and $y{\perp }_{B}x$. For example, in $l_{\infty }^n(n>2)$, $(1,-1,0,\dots ,0)$ and $(1,1,0,\dots ,0)$ are neither left symmetric points nor right symmetric points, but $(1,1,0,\dots ,0)$ and $(1,-1,0,\dots ,0)$ are mutually Birkhoff orthogonal.

In [15], Arambašić et al. gave the precise definition of mutual Birkhoff orthogonality, which is a new notion of Birkhoff orthogonality with symmetry. Two elements $x,y\in X$ are mutually Birkhoff orthogonal (denoted by $x\perp \perp y$) if $x{\perp }_{B}y$ and $y{\perp }_{B}x$. x and y are mutually Birkhoff orthogonal elements of each other. Arambašić et al. focused on the set consisting of pairwise Birkhoff orthogonal elements, which we call the mutually Birkhoff orthogonal set in this paper.

Definition 3. 

$A\subset X\setminus \left\{0\right\}$ is called a mutually Birkhoff orthogonal set, if for any $x,y\in A(x\ne y)$, there is $x\perp \perp y$.

Arambašić et al. gave an example of the existence of the mutually Birkhoff orthogonal set with three elements in a two-dimensional linear space with a hexagonal norm. It is shown that mutually Birkhoff orthogonal sets are not necessarily linearly independent sets in two-dimensional normed spaces. Arambašić et al. proposed that if X is smooth, then mutually Birkhoff orthogonal sets in X are linearly independent sets. This is not necessarily true for Birkhoff orthogonal sets. Mutually Birkhoff orthogonal sets must be Birkhoff orthogonal sets, but Birkhoff orthogonal sets are not necessarily mutually Birkhoff orthogonal sets.

The main purpose of this paper is to study the linear independence of Birkhoff orthogonal sets, which are more widely used than mutually Birkhoff orthogonal sets. The conclusions obtained in this paper can be applied to the mutually Birkhoff orthogonal sets. Before proceeding any further, we need to recall the relevant properties of Birkhoff orthogonality.

Actually, Birkhoff orthogonality comes from an inner product. As illustrated in [16], James pointed out that in real normed spaces of dimension at least three, Birkhoff orthogonality is symmetric if and only if the norm is induced by the inner product. In [6,16,17,18,19], there are also some results that hint at this fact. In addition, Birkhoff orthogonality is right unique or right additive if and only if the space is smooth and Birkhoff orthogonality is left unique if and only if the space is strictly convex. In [20], Paul et al. put forward the concept of strongly Birkhoff orthogonal. $x\in X$ is said to be strongly Birkhoff orthogonal to $y\in X$, denoted by $x{\perp }_{SB}y$, if

$$\Vert x+\lambda y\Vert >\Vert x\Vert$$

for all scalars $\lambda \ne 0$. If $x{\perp }_{SB}y$, then $x{\perp }_{B}y$. However, the opposite is not true. In [21], Sain et al. proposed that if X is strictly convex, then for $x,y\in X\setminus \left\{0\right\}$, $x{\perp }_{B}y\iff x{\perp }_{SB}y.$

In this paper, we give some examples to show that Birkhoff orthogonal sets and mutually Birkhoff orthogonal sets are not necessarily linearly independent sets in X. We will prove that if X is strictly convex or smooth, then the Birkhoff orthogonal set $A=\{x_1,x_2,x_3\}\subset X$ is linearly independent. We also prove that if X is both strictly convex and smooth, then the Birkhoff orthogonal set $A=\{x_1,x_2,x_3,x_4\}\subset X$ that contains left (right) symmetric points is linearly independent. In the end, we conclude that if X is smooth, then the Birkhoff orthogonal set $A=\{x_1,x_2,\dots ,x_n\}\subset X(n\ge 3)$ that contains $n-3$ right symmetric points is linearly independent. Studying the properties of Birkhoff orthogonal sets, this paper can enrich Birkhoff orthogonality theorems and the study of geometric structure in normed spaces.

2. Main Results

Firstly, we consider the existence of non-zero mutually Birkhoff orthogonal elements. In [15], if $dimX\ge 3$, then for $x\in X\setminus \left\{0\right\}$, there is $y\in X\setminus \left\{0\right\}$ such that $x\perp \perp y$. Therefore, we consider the existence of non-zero mutually Birkhoff orthogonal elements in X$(dimX=2)$.

If $dimX=2$, then for $x\in X\setminus \left\{0\right\}$, there does not necessarily exist $y\in X\setminus \left\{0\right\}$ such that $x\perp \perp y$. For example, in $l_{\infty }^2$, there does not exist a non-zero element that is mutually Birkhoff orthogonal to (2,3). Furthermore, (0,1) is the non-zero mutually Birkhoff orthogonal element of (1,0).

As we all know, for $x\in X\setminus \left\{0\right\}$, there is $y\in X\setminus \left\{0\right\}$ such that $x{\perp }_{B}y$. Therefore, in a two-dimensional normed space whose Birkhoff orthogonality is symmetric, any non-zero element has a non-zero mutually Birkhoff orthogonal element.

The space $l_1^2$ is a two-dimensional sequence space normed by $\Vert x=(x_1,x_2)\Vert = \mid x_1\mid +\mid x_2\mid$. The space $l_{\infty }^2$ is a two-dimensional sequence space normed by $\Vert x=(x_1,x_2)\Vert =max\{\mid x_1\mid ,\mid x_2\mid \}$. For $l_p^2$ $(p=1,\infty )$, we have a result as follows.

Theorem 1. 

For $x,y\in l_p^2$ $(p=1,\infty )$, the following facts are equivalent:

(1) 

$x\perp \perp y$;

(2) 

$x{\perp }_{R}y$.

Proof. 

For $x,y\in l_p^2$ $(p=1,\infty )$, if one or both is zero, it is obviously true. Therefore, we just need to consider the case that x and y are all not zero.

• $\left(2\right)\Rightarrow \left(1\right)$. According to the symmetry of Roberts orthogonality, we can know that if $x{\perp }_{R}y$, then $x{\perp }_{B}y$ and $y{\perp }_{B}x$. Obviously, $x{\perp }_{R}y\Rightarrow x\perp \perp y$.
• $\left(1\right)\Rightarrow \left(2\right)$. We divide all non-zero elements in $l_1^2$ into the following four cases and discuss their Birkhoff orthogonal elements, respectively, ($x,y\in l_1^2$, $a,b,c,d\in R\setminus \left\{0\right\}$).

Case 1. $x=(a,0)$: If $x{\perp }_{B}y$, then $y=(0,d)$ or $(c,d)$ $(\mid c\mid = \mid d\mid )$;

Case 2. $x=(0,b)$: If $x{\perp }_{B}y$, then $y=(c,0)$ or $(c,d)$ $(\mid c\mid = \mid d\mid )$;

Case 3. $x=(a,b)$ $(ab>0)$: If $x{\perp }_{B}y$, then $y=(c,d)$ $(cd<0,$ $\mid c\mid = \mid d\mid )$;

Case 4. $x=(a,b)$ $(ab<0)$: If $x{\perp }_{B}y$, then $y=(c,d)$ $(cd>0,$ $\mid c\mid = \mid d\mid )$.

It can be seen that only $x=(0,a)$ is mutually Birkhoff orthogonal to $y=(b,0)(ab\ne 0)$, or $x=(a,-a)$ is mutually Birkhoff orthogonal to $y=(b,b)(ab\ne 0)$ in $l_1^2$.

In $l_1^2$, for $x=(0,a)$ and $y=(b,0)(ab\ne 0)$, for $\forall \lambda \in R$,

$$\begin{array}{cc}\hfill \Vert x+\lambda y\Vert & =\mid \lambda b\mid +\mid a\mid \hfill \\ & =\Vert x-\lambda y\Vert \hfill \\ & =\mid -\lambda b\mid +\mid a\mid \hfill \\ & \Rightarrow x{\perp }_{R}y.\hfill \end{array}$$

For $x=(a,-a)$ and $y=(b,b)(ab\ne 0)$, for $\forall \lambda \in R$,

$$\begin{array}{cc}\hfill \Vert x+\lambda y\Vert & =\mid a+\lambda b\mid +\mid -a+\lambda b\mid \hfill \\ & =\Vert x-\lambda y\Vert \hfill \\ & =\mid a-\lambda b\mid +\mid -a-\lambda b\mid \hfill \\ & \Rightarrow x{\perp }_{R}y.\hfill \end{array}$$

We divide all non-zero elements in $l_{\infty }^2$ into the following six cases and discuss their Birkhoff orthogonal elements, respectively, ($x,y\in l_{\infty }^2$, $a,b,c,d\in R\setminus \left\{0\right\}$).

Case 1. $x=(a,0)$: If $x{\perp }_{B}y$, then $y=(0,d)$;

Case 2. $x=(0,b)$: If $x{\perp }_{B}y$, then $y=(c,0)$;

Case 3. $x=(a,b)$ $(ab>0,\mid a\mid = \mid b\mid ))$: If $x{\perp }_{B}y$, then $y=(0,d)$, or $(c,0)$, or $(c,d)$ $(cd<0)$;

Case 4. $x=(a,b)$ $(ab<0,\mid a\mid = \mid b\mid ))$: If $x{\perp }_{B}y$, then $y=(0,d)$, or $(c,0)$, or $(c,d)$ $(cd>0)$;

Case 5. $x=(a,b)$ $(\mid a\mid > \mid b\mid ))$: If $x{\perp }_{B}y$, then $y=(0,d)$;

Case 6. $x=(a,b)$ $(\mid a\mid < \mid b\mid ))$: If $x{\perp }_{B}y$, then $y=(c,0)$.

It can be seen that only $x=(0,a)$ is mutually Birkhoff orthogonal to $y=(b,0)(ab\ne 0)$, or $x=(a,-a)$ is mutually Birkhoff orthogonal to $y=(b,b)(ab\ne 0)$ in $l_{\infty }^2$.

In $l_{\infty }^2$, for $x=(0,a)$ and $y=(b,0)(ab\ne 0)$, for $\forall \lambda \in R$,

$$\begin{array}{cc}\hfill \Vert x+\lambda y\Vert & =max\{\mid \lambda b\mid ,\mid a\mid \}\hfill \\ & =\Vert x-\lambda y\Vert \hfill \\ & =max\{\mid -\lambda b\mid ,\mid a\mid \}\hfill \\ & \Rightarrow x{\perp }_{R}y.\hfill \end{array}$$

For $x=(a,-a)$ and $y=(b,b)(ab\ne 0)$, for $\forall \lambda \in R$,

$$\begin{array}{cc}\hfill \Vert x+\lambda y\Vert & =max\{\mid a+\lambda b\mid ,\mid -a+\lambda b\mid \}\hfill \\ & =\Vert x-\lambda y\Vert \hfill \\ & =max\{\mid a-\lambda b\mid ,\mid -a-\lambda b\mid \}\hfill \\ & \Rightarrow x{\perp }_{R}y.\hfill \end{array}$$

In summary, mutual Birkhoff orthogonality and Roberts orthogonality are equivalent in $l_p^2$ $(p=1,\infty )$. □

The above theorem is not true in $l_p^n(p=1,\infty )(n>2)$, for example, in $l_1^3$, $(1,1,0)\perp \perp (1,0,1)$, but $(1,1,0){/\perp }_R(1,0,1)$.

If X is neither smooth nor strictly convex, then there exist linearly dependent Birkhoff orthogonal sets and linearly dependent mutually Birkhoff orthogonal sets. Arambašić et al. gave an example in a two-dimensional linear space with a hexagonal norm. Without loss of generality, we give several examples of linearly dependent Birkhoff orthogonal sets and linearly dependent mutually Birkhoff orthogonal sets in three-dimensional normed spaces.

Example 1. 

Let $A=\{x_1,x_2,x_3,x_4\}\subset l_1^3$, where $x_1=(\frac{1}{2},\frac{-1}{4},0),x_2=(\frac{1}{2},\frac{1}{2},0),x_3=(\frac{1}{4},\frac{1}{4},\frac{1}{2}),x_4=(\frac{1}{4},\frac{1}{4},\frac{-1}{2})$. $l_1^3$ is neither smooth nor strictly convex. Obviously, A is a linearly dependent Birkhoff orthogonal set.

Example 2. 

Let $X_0=(R^3,\Vert ·\Vert )$, $\forall x=(\alpha ,\beta ,\gamma )\in X_0$,

$$\Vert x\Vert =\left\{\begin{array}{cc}\hfill {({\alpha }^4+{\beta }^4+{\gamma }^4)}^{\frac{1}{4}},& \beta \ge \frac{\sqrt{2}}{2},\\ \hfill {({\alpha }^2+{\beta }^2+{\gamma }^2)}^{\frac{1}{2}},& \beta <\frac{\sqrt{2}}{2}.\end{array}\right.$$

$X_0$ is neither smooth nor strictly convex. Let $A=\{y_1,y_2,y_3,y_4\}\subset X_0$, where $y_1=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0),y_2=(\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2},0),y_3=(0,\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}),y_4=(0,\frac{\sqrt{2}}{2},\frac{-\sqrt{2}}{2})$. Obviously, A is a linearly dependent mutually Birkhoff orthogonal set.

From the above examples, we note that if $dimX=n$ and X is neither smooth nor strictly convex, then there exist Birkhoff orthogonal sets (mutually Birkhoff orthogonal sets) with $n+1$ elements. Therefore, Birkhoff orthogonal sets and mutually Birkhoff orthogonal sets are not necessarily linearly independent sets in general normed spaces.

Let $S=\{x_1,x_2\}\subset X$ be a Birkhoff orthogonal set. Obviously, S is linearly independent. Next, we consider the linear independence of the Birkhoff orthogonal set $A=\{x_1,x_2,x_3\}\subset X$.

Theorem 2. 

If X is strictly convex, then the Birkhoff orthogonal set $A=\{x_1,x_2,x_3\}\subset X$ is linearly independent.

Proof. 

Suppose X is strictly convex and A is linearly dependent.

When A is linearly dependent, then there is $a{x}_1+b{x}_2+c{x}_3=0$ (a, b and c are not all zero). Next, we prove that a, b and c are all not zero.

If $a=0$ and $bc\ne 0$, then $b{x}_2=-c{x}_3$, i.e., $x_2$ and $x_3$ are linearly dependent. However, it can be seen from the definition of the Birkhoff orthogonal set that $x_2$ and $x_3$ are linearly independent, which contradicts the above. Similarly, whether $b=0$ and $ac\ne 0$ or $c=0$ and $ab\ne 0$, a contradiction arises.

If $a=b=0$ and $c\ne 0$, then $x_3=0$. However, a contradiction arises with $x_3\ne 0$. Similarly, whether $a=c=0$ and $b\ne 0$ or $b=c=0$ and $a\ne 0$, a contradiction arises.

Therefore, if A is linearly dependent, then there is $x_1=a_2{x}_2+a_3{x}_3(a_2{a}_3\ne 0)$.

The Birkhoff orthogonal set A in a normed space totally has the following eight possible Birkhoff orthogonal cases:

$$\begin{array}{cc}& ➀x_1{\perp }_B{x}_2,x_2{\perp }_B{x}_3,x_3{\perp }_B{x}_1➁x_1{\perp }_B{x}_3,x_2{\perp }_B{x}_1,x_3{\perp }_B{x}_2\hfill \\ & ➂x_1{\perp }_B{x}_2,x_1{\perp }_B{x}_3,x_3{\perp }_B{x}_2➃x_1{\perp }_B{x}_2,x_1{\perp }_B{x}_3,x_2{\perp }_B{x}_3\hfill \\ & ➄x_1{\perp }_B{x}_2,x_3{\perp }_B{x}_1,x_3{\perp }_B{x}_2➅x_2{\perp }_B{x}_1,x_1{\perp }_B{x}_3,x_2{\perp }_B{x}_3\hfill \\ & ➆x_2{\perp }_B{x}_1,x_3{\perp }_B{x}_1,x_3{\perp }_B{x}_2➇x_2{\perp }_B{x}_1,x_3{\perp }_B{x}_1,x_2{\perp }_B{x}_3\hfill \end{array}$$

Suppose $x_1{\perp }_B{x}_2$ and $x_3{\perp }_B{x}_2$ can be true at the same time, by the homogeneity of Birkhoff orthogonality, we know that

$$x_1{\perp }_B{x}_2\Rightarrow x_1{\perp }_B{a}_2{x}_2\Rightarrow a_2{x}_2+a_3{x}_3{\perp }_B{a}_2{x}_2.$$

Because X is strictly convex, for $\forall x,y\in X\setminus \left\{0\right\}$, $x{\perp }_{B}y\Rightarrow x{\perp }_{SB}y$.

According to the above formulas, we can obtain

$$\begin{array}{cc}\hfill x_1{\perp }_B{x}_2& \Rightarrow a_2{x}_2+a_3{x}_3{\perp }_B{a}_2{x}_2\hfill \\ & \Rightarrow a_2{x}_2+a_3{x}_3{\perp }_{SB}{a}_2{x}_2\hfill \\ & \iff \Vert a_2{x}_2+a_3{x}_3+\lambda a_2{x}_2\Vert >\Vert a_2{x}_2+a_3{x}_3\Vert ,\forall \lambda \ne 0\hfill \\ & \stackrel{\lambda =-1}{⟹}\Vert a_3{x}_3\Vert >\Vert a_2{x}_2+a_3{x}_3\Vert .\hfill \end{array}$$

But

$$\begin{array}{cc}\hfill x_3{\perp }_B{x}_2& \Rightarrow a_3{x}_3{\perp }_B{a}_2{x}_2\hfill \\ & \Rightarrow a_3{x}_3{\perp }_{SB}{a}_2{x}_2\hfill \\ & \iff \Vert a_3{x}_3+\lambda a_2{x}_2\Vert >\Vert a_3{x}_3\Vert ,\forall \lambda \ne 0\hfill \\ & \stackrel{\lambda =1}{⟹}\Vert a_2{x}_2+a_3{x}_3\Vert >\Vert a_3{x}_3\Vert .\hfill \end{array}$$

Therefore, $x_1{\perp }_B{x}_2$ and $x_3{\perp }_B{x}_2$ cannot be true at the same time.

Similarly, we can obtain $x_1{\perp }_B{x}_3$ and $x_2{\perp }_B{x}_3$ cannot be true at the same time; $x_2{\perp }_B{x}_1$ and $x_3{\perp }_B{x}_1$ cannot be true at the same time.

Thus, if A is linearly dependent in X, then the cases ➂–➇ cannot hold, we just have to think about case ➀ and case ➁.

For case ➀:

$$\begin{array}{cc}\hfill x_1{\perp }_B{x}_2& \Rightarrow x_1{\perp }_B{a}_2{x}_2\hfill \\ & \Rightarrow x_1{\perp }_{SB}{a}_2{x}_2\hfill \\ & \Rightarrow a_2{x}_2+a_3{x}_3{\perp }_{SB}{a}_2{x}_2\hfill \\ & \iff \Vert a_2{x}_2+a_3{x}_3+\lambda a_2{x}_2\Vert >\Vert a_2{x}_2+a_3{x}_3\Vert ,\forall \lambda \ne 0\hfill \\ & \stackrel{\lambda =-1}{⟹}\Vert a_3{x}_3\Vert >\Vert a_2{x}_2+a_3{x}_3\Vert .\hfill \end{array}$$

$$\begin{array}{cc}\hfill x_2{\perp }_B{x}_3& \Rightarrow a_2{x}_2{\perp }_B{a}_3{x}_3\Rightarrow a_2{x}_2{\perp }_{SB}{a}_3{x}_3\hfill \\ & \iff \Vert a_2{x}_2+\lambda a_3{x}_3\Vert >\Vert a_2{x}_2\Vert ,\forall \lambda \ne 0\hfill \\ & \stackrel{\lambda =1}{⟹}\Vert a_2{x}_2+a_3{x}_3\Vert >\Vert a_2{x}_2\Vert .\hfill \end{array}$$

So

$$\begin{array}{cc}\hfill x_1{\perp }_B{x}_2,x_2{\perp }_B{x}_3& \Rightarrow \Vert a_2{x}_2\Vert <\Vert a_2{x}_2+a_3{x}_3\Vert <\Vert a_3{x}_3\Vert \hfill \\ & \Rightarrow \mid \frac{a_2}{a_3}\mid \Vert x_2\Vert <\Vert x_3\Vert .\hfill \end{array}$$

But

$$\begin{array}{cc}\hfill x_3{\perp }_B{x}_1& \Rightarrow x_3{\perp }_{SB}{x}_1\hfill \\ & \Rightarrow x_3{\perp }_{SB}{a}_2{x}_2+a_3{x}_3\hfill \\ & \iff \Vert (1+\lambda a_3)x_3+\lambda a_2{x}_2\Vert >\Vert x_3\Vert ,\forall \lambda \ne 0\hfill \\ & \stackrel{\lambda =\frac{-1}{a_3}(a_3\ne 0)}{⟹}\mid \frac{a_2}{a_3}\mid \Vert x_2\Vert >\Vert x_3\Vert .\hfill \end{array}$$

Thus, if A is linearly dependent in X, then case ➀ cannot hold. Similarly, case ➁ cannot hold.

To sum up, the Birkhoff orthogonal set A is linearly independent in a strictly convex normed space. □

Corollary 1. 

If X is strictly convex, then the mutually Birkhoff orthogonal set $A=\{x_1,x_2,x_3\}\subset X$ is linearly independent.

Theorem 3. 

If X is smooth, then the Birkhoff orthogonal set $A=\{x_1,x_2,x_3\}\subset X$ is linearly independent.

Proof. 

Suppose X is smooth and A is linearly dependent.

When A is linearly dependent, then there is $x_1=a_2{x}_2+a_3{x}_3(a_2{a}_3\ne 0)$ (as can be seen from the proof of Theorem 2).

According to the proof of Theorem 2, there are a total of eight possible Birkhoff orthogonal cases for A in a normed space.

By the homogeneity of Birkhoff orthogonality, and the right additivity of Birkhoff orthogonality in smooth normed spaces, we can obtain

$$x_1{\perp }_B{x}_2,x_1{\perp }_B{x}_3\Rightarrow x_1{\perp }_B{a}_2{x}_2,x_1{\perp }_B{a}_3{x}_3\Rightarrow x_1{\perp }_B{a}_2{x}_2+a_3{x}_3.$$

While $x_1=a_2{x}_2+a_3{x}_3 {/\perp }_B a_2{x}_2+a_3{x}_3$, this is a contradiction.

Therefore, $x_1{\perp }_B{x}_2$ and $x_1{\perp }_B{x}_3$ cannot be true at the same time. Similarly, $x_2{\perp }_B{x}_1$ and $x_2{\perp }_B{x}_3$ cannot be true at the same time; $x_3{\perp }_B{x}_1$ and $x_3{\perp }_B{x}_2$ cannot be true at the same time.

So, if A is linearly dependent in X, cases ➂–➇ cannot hold, we just think about case ➀ and case ➁.

For case ➀:

$$\begin{array}{cc}\hfill x_1{\perp }_B{x}_2& \Rightarrow a_2{x}_2+a_3{x}_3{\perp }_B{a}_2{x}_2\hfill \\ & \iff \Vert a_2{x}_2+a_3{x}_3+\lambda a_2{x}_2\Vert \ge \Vert a_2{x}_2+a_3{x}_3\Vert ,\forall \lambda \in R.\hfill \end{array}$$

$$\begin{array}{cc}\hfill x_2{\perp }_B{x}_3& \Rightarrow a_2{x}_2{\perp }_B{a}_3{x}_3\hfill \\ & \iff \Vert a_2{x}_2+\lambda a_3{x}_3\Vert \ge \Vert a_2{x}_2\Vert ,\forall \lambda \in R\hfill \\ & \stackrel{\lambda =1}{⟹}\Vert a_2{x}_2+a_3{x}_3\Vert \ge \Vert a_2{x}_2\Vert .\hfill \end{array}$$

$$\begin{array}{cc}\hfill x_3{\perp }_B{x}_1& \Rightarrow a_3{x}_3{\perp }_B{a}_2{x}_2+a_3{x}_3\hfill \\ & \iff \Vert a_3{x}_3+\lambda a_2{x}_2+\lambda a_3{x}_3\Vert \ge \Vert a_3{x}_3\Vert ,\forall \lambda \in R\hfill \\ & \stackrel{\lambda =-1}{⟹}\Vert a_2{x}_2\Vert \ge \Vert a_3{x}_3\Vert .\hfill \end{array}$$

Then we can know

$$\begin{array}{cc}& x_1{\perp }_B{x}_2,x_2{\perp }_B{x}_3,x_3{\perp }_B{x}_1\hfill \\ \hfill \Rightarrow & \Vert a_2{x}_2+a_3{x}_3+\lambda a_2{x}_2\Vert \ge \Vert a_2{x}_2+a_3{x}_3\Vert \ge \Vert a_2{x}_2\Vert \ge \Vert a_3{x}_3\Vert ,\forall \lambda \in R\hfill \\ \hfill \iff & \Vert a_3{x}_3+(\lambda +1)a_2{x}_2\Vert \ge \Vert a_3{x}_3\Vert ,\forall \lambda \in R\hfill \\ \hfill \iff & a_3{x}_3{\perp }_B{a}_2{x}_2\hfill \\ \hfill \Rightarrow & \frac{1}{a_3}{a}_3{x}_3{\perp }_B\frac{1}{a_2}{a}_2{x}_2(a_2{a}_3\ne 0)\hfill \\ \hfill \iff & x_3{\perp }_B{x}_2.\hfill \end{array}$$

However, in case ➀, there is $x_3{\perp }_B{x}_1$, thus it causes a contradiction.

Therefore, if A is linearly dependent in X, then case ➀ cannot hold. Similarly, case ➁ cannot hold.

To sum up, the Birkhoff orthogonal set A is linearly independent in a smooth normed space. □

Corollary 2. 

Let $dimX=2$ and let X is strictly convex or smooth. Then Birkhoff orthogonal sets and mutually Birkhoff orthogonal sets are linearly independent sets in X.

Theorem 4. 

Let X be smooth, $A=\{x_1,x_2,x_3,x_4\}\subset X$ is a Birkhoff orthogonal set. If there is an element $x_m\in A$ such that $x_m$ is Birkhoff orthogonal to all other elements in A, then A is linearly independent.

Proof. 

We just need to prove that if A is linearly dependent in X, then there are no elements that satisfy the case set out in the theorem.

When A is linearly dependent, then there is $a{x}_1+b{x}_2+c_3+d_4=0$ (a, b, c and d are not all zero). Next, we prove that a, b, c and d are all not zero.

If $a=0$ and $bcd\ne 0$, then the Birkhoff orthogonal set $\{x_2,x_3,x_4\}\subseteq A\subset X$ is linearly dependent. However, by Theorem 3, we can know that the Birkhoff orthogonal set $\{x_2,x_3,x_4\}\subseteq A\subset X$ is linearly independent. This is a contradiction.

Similarly, if $b=0$ and $acd\ne 0$, a contradiction arises; if $c=0$ and $abd\ne 0$, a contradiction arises; if $d=0$ and $abc\ne 0$, a contradiction arises.

If $a=b=0$ and $cd\ne 0$, then $c{x}_3=-d{x}_4$, i.e., $x_3$ and $x_4$ are linearly dependent. However, it can be seen from the definition of the Birkhoff orthogonal set that $x_3$ and $x_4$ are linearly independent, which contradicts the above.

Similarly, if $a=c=0$ and $bd\ne 0$, a contradiction arises; if $a=d=0$ and $bc\ne 0$, a contradiction arises; if $b=c=0$ and $ad\ne 0$, a contradiction arises; if $b=d=0$ and $ac\ne 0$, a contradiction arises; if $c=d=0$ and $ab\ne 0$, a contradiction arises.

If $a=b=c=0$ and $d\ne 0$, then $x_4=0$, this contradicted with $x_4\ne 0$.

Similarly, if $a=b=d=0$ and $c\ne 0$, a contradiction arises; if $a=c=d=0$ and $b\ne 0$, a contradiction arises; if $b=c=d=0$ and $a\ne 0$, a contradiction arises.

Therefore, if the Birkhoff orthogonal set $A=\{x_1,x_2,x_3,x_4\}$ in X is linearly dependent, then any element in this orthogonal set can be linearly represented by all the remaining elements in A. To facilitate the proof, let $x_1=a_2{x}_2+a_3{x}_3+a_4{x}_4(a_2{a}_3{a}_4\ne 0)$.

$$\begin{array}{cc}& x_1{\perp }_B{x}_2,x_1{\perp }_B{x}_3,x_1{\perp }_B{x}_4\hfill \\ \hfill \Rightarrow & x_1{\perp }_B{a}_2{x}_2,x_1{\perp }_B{a}_3{x}_3,x_1{\perp }_B{a}_4{x}_4\hfill \\ \hfill \Rightarrow & x_1{\perp }_B{a}_2{x}_2+a_3{x}_3+a_4{x}_4\hfill \\ \hfill \Rightarrow & x_1{\perp }_B{x}_1.\hfill \end{array}$$

This is a contradiction. Therefore, if A is linearly dependent, then $x_1{\perp }_B{x}_2$, $x_1{\perp }_B{x}_3$ and $x_1{\perp }_B{x}_4$ cannot be true at the same time in X.

Similarly, if A is linearly dependent, then $x_2{\perp }_B{x}_1$, $x_2{\perp }_B{x}_3$ and $x_2{\perp }_B{x}_4$ cannot be true at the same time in X; $x_3{\perp }_B{x}_1$, $x_3{\perp }_B{x}_2$ and $x_3{\perp }_B{x}_4$ cannot be true at the same time in X; $x_4{\perp }_B{x}_1$, $x_4{\perp }_B{x}_2$ and $x_4{\perp }_B{x}_3$ cannot be true at the same time in X.

To sum up, if there is an element $x_m\in A$ such that $x_m$ is Birkhoff orthogonal to all other elements in A, then A is linearly independent. □

Theorem 5. 

Let X be both smooth and strictly convex, $A=\{x_1,x_2,x_3,x_4\}\subset X$ is a Birkhoff orthogonal set. If there are left (right) symmetric points in A, then A is linearly independent.

Proof. 

We note that if there are left (right) symmetric points in A, then there must be an element $x_m\in A$ such that $x_m$ is Birkhoff orthogonal to all other elements in A, otherwise all other elements in A are Birkhoff orthogonal to $x_m$.

By Theorem 4, we know that if there is an element $x_m\in A$ such that $x_m$ is Birkhoff orthogonal to all other elements in A, then A is linearly independent. Therefore, we just need to prove that if there is an element $x_m\in A$ such that all other elements in A are Birkhoff orthogonal to $x_m$, then A is also linearly independent.

The possible Birkhoff orthogonal cases of $A=\{x_1,x_2,x_3,x_4\}$ in a normed space are as shown in

Table 1. All possible Birkhoff orthogonal cases of A.

(1) 12,13,14,23,24,34	(2) 12,13,14,23,24,43	(3) 12,13,14,23,42,34	(4) 12,13,14,23,42,43
(5) 12,13,14,32,24,34	(6) 12,13,14,32,24,43	(7) 12,13,14,32,42,34	(8) 12,13,14,32,42,43
(9) 12,13,41,23,24,34	(10) 12,13,41,23,24,43	(11) 12,13,41,23,42,34	(12) 12,13,41,23,42,43
(13) 12,13,41,32,24,34	(14) 12,13,41,32,24,43	(15) 12,13,41,32,42,34	(16) 12,13,41,32,42,43
(17) 12,31,14,23,24,34	(18) 12,31,14,23,24,43	(19) 12,31,14,23,42,34	(20) 12,31,14,23,42,43
(21) 12,31,14,32,24,34	(22) 12,31,14,32,24,43	(23) 12,31,14,32,42,34	(24) 12,31,14,32,42,43
(25) 12,31,41,23,24,34	(26) 12,31,41,23,24,43	(27) 12,31,41,23,42,34	(28) 12,31,41,23,42,43
(29) 12,31,41,32,24,34	(30) 12,31,41,32,24,43	(31) 12,31,41,32,42,34	(32) 12,31,41,32,42,43
(33) 21,13,14,23,24,34	(34) 21,13,14,23,24,43	(35) 21,13,14,23,42,34	(36) 21,13,14,23,42,43
(37) 21,13,14,32,24,34	(38) 21,13,14,32,24,43	(39) 21,13,14,32,42,34	(40) 21,13,14,32,42,43
(41) 21,13,41,23,24,34	(42) 21,13,41,23,24,43	(43) 21,13,41,23,42,34	(44) 21,13,41,23,42,43
(45) 21,13,41,32,24,34	(46) 21,13,41,32,24,43	(47) 21,13,41,32,42,34	(48) 21,13,41,32,42,43
(49) 21,31,14,23,24,34	(50) 21,31,14,23,24,43	(51) 21,31,14,23,42,34	(52) 21,31,14,23,42,43
(53) 21,31,14,32,24,34	(54) 21,31,14,32,24,43	(55) 21,31,14,32,42,34	(56) 21,31,14,32,42,43
(57) 21,31,41,23,24,34	(58) 21,31,41,23,24,43	(59) 21,31,41,23,42,34	(60) 21,31,41,23,42,43
(61) 21,31,41,32,24,34	(62) 21,31,41,32,24,43	(63) 21,31,41,32,42,34	(64) 21,31,41,32,42,43

, wherein, we use $ij(i,j=1,2,3,4)$ and $ji$ to denote $x_i{\perp }_B{x}_j$ and $x_j{\perp }_B{x}_i$, respectively.

We just need to prove that if A is linearly dependent in X, then the Birkhoff orthogonal cases of A can not satisfy the above condition. It can be seen from Table 1 that there are a total of 64 possible Birkhoff orthogonal cases. Suppose A is linearly dependent, we know at this time some cases that satisfy the case set out in the Theorem 4 cannot exist in X, such as case (1) and case (33). Therefore, the possible Birkhoff orthogonal cases of A can only be the remaining 32 cases.

In the remaining 32 cases, we just consider the cases that satisfy the above condition; these are case (10), case (15), case (17), case (36), case (37), case (59) and case (62).

As can be seen from the proof of Theorem 4, if A is linearly dependent in X, then there is $x_1=a_2{x}_2+a_3{x}_3+a_4{x}_4(a_2{a}_3{a}_4\ne 0).$ For case (10):

$$\begin{array}{cc}& x_2{\perp }_B{x}_4,x_2{\perp }_B{x}_3,x_1{\perp }_B{x}_2,x_1{\perp }_B{x}_3,x_4{\perp }_B{x}_1,x_4{\perp }_B{x}_3\hfill \\ \hfill \Rightarrow & a_2{x}_2{\perp }_{SB}{a}_4{x}_4+a_3{x}_3,x_1{\perp }_{SB}{a}_2{x}_2+a_3{x}_3,a_4{x}_4{\perp }_{SB}{a}_3{x}_3-x_1\hfill \\ \hfill \Rightarrow & \Vert a_2{x}_2\Vert >\Vert a_4{x}_4\Vert >\Vert x_1\Vert >\Vert a_2{x}_2\Vert .\hfill \end{array}$$

This is a contradiction.

Similarly, for case (15), there is $\Vert a_3{x}_3\Vert >\Vert a_4{x}_4\Vert >\Vert x_1\Vert >\Vert a_3{x}_3\Vert ,$ which is a contradiction; for case (17), there is $\Vert a_2{x}_2\Vert >\Vert a_3{x}_3\Vert >\Vert x_1\Vert >\Vert a_2{x}_2\Vert$, which is a contradiction; for case (36), there is $\Vert a_2{x}_2\Vert >\Vert x_1\Vert >\Vert a_4{x}_4\Vert >\Vert a_2{x}_2\Vert ,$ which is a contradiction; for case (37), there is $\Vert a_3{x}_3\Vert >\Vert a_2{x}_2\Vert >\Vert x_1\Vert >\Vert a_3{x}_3\Vert ,$ which is a contradiction; for case (59), there is $\Vert a_4{x}_4\Vert >\Vert a_2{x}_2\Vert >\Vert a_3{x}_3\Vert >\Vert a_4{x}_4\Vert ,$ which is a contradiction; for case (62), there is $\Vert a_3{x}_3\Vert >\Vert a_2{x}_2\Vert >\Vert a_4{x}_4\Vert >\Vert a_3{x}_3\Vert ,$ which is a contradiction.

Therefore, if A is linearly dependent in X, then there is no element $x_m\in A$ such that all other elements in A are Birkhoff orthogonal to $x_m$.

To sum up, if there are left (right) symmetric points in A, then the Birkhoff orthogonal set A must be linearly independent. □

Corollary 3. 

Let X be both smooth and strictly convex, $dimX=3$. If a Birkhoff orthogonal set $A\subset X$ contains left (right) symmetric points, then A is linearly independent.

Obviously, in a smooth normed space X, if all elements in the Birkhoff orthogonal set $A=\{x_1,x_2,\dots ,x_n\}\subset X$ are right symmetric points, then A is linearly independent. Next, we further draw the following result.

Theorem 6. 

Let X be smooth and $A=\{x_1,x_2,\dots ,x_n\}\subset X(n\ge 3,n\in N)$ is a Birkhoff orthogonal set. If A contains $n-3$ right symmetric points, then A is linearly independent.

Proof. 

When $n=3$, according to Theorem 3, we can know that the conclusion is true.

When $n>3$, suppose A is linearly dependent and A contains $n-3$ right symmetric points, then there is $a_1{x}_1+a_2{x}_2+\cdots +a_n{x}_n=0$ ($a_1$, ..., $a_{n-1}$ and $a_n$ are not all zero).

We note that if a Birkhoff orthogonal set $S\subset X$ contains right symmetric points and the elements in S can be linearly represented by all the remaining elements, then there must exist an element $x_m\in S$ such that $x_m$ is Birkhoff orthogonal to all other elements in S. By the proof of Theorem 4, this is a contradiction.

If $a_1{a}_2\cdots a_n\ne 0$, according to the above result, this is a contradiction. If $a_1=0$ and $a_2\cdots a_n\ne 0$, according to the above result, this is a contradiction. Similarly, if $a_1=a_2=0$ and $a_3\cdots a_n\ne 0$, this is also a contradiction, and so on.

This proves that $a_1$, ..., $a_{n-1}$ and $a_n$ are all zero. Therefore, when $n>3$, the conclusion is true. □

Remark 1. 

In [22], Bhunia et al. proposed a new generalization of Birkhoff orthogonality over semi-Hilbert spaces.We think it would be valuable to extend the results obtained in this paper to semi-Hilbert spaces. Therefore, it is interesting to introduce and study the orthogonal sets in the sense of orthogonality with this new form. We hypothesize that this may produce similar conclusions to those in this paper, and will further study the details in the future.

3. Conclusions

In this manuscript, we proposed the notion of Birkhoff orthogonal sets in finite-dimensional real normed spaces to study Birkhoff orthogonality from a new aspect. We studied the linear independence of Birkhoff orthogonal sets, and the obtained results can also be applied to the study of symmetry in Birkhoff orthogonality. In addition, there are few related studies on generalized orthogonal sets in general normed spaces. In the future, we will continue studying the properties of other generalized orthogonal sets to enrich the knowledge of generalized orthogonal sets.