Existence and Construction of Tangential and Anisotropic Bases in Finite-Dimensional Quadratic Spaces

Abstract

This paper studies the existence and construction of bases consisting of tangential and anisotropic vectors in finite-dimensional quadratic spaces over fields of characteristic different from two. While classical theory guarantees the existence of orthogonal bases in regular quadratic spaces, the existence of bases governed by alternative geometric constraints such as tangency or isotropy has remained largely unexplored. We introduce determinant-based constructive methods extending the Gram–Schmidt process to arbitrary quadratic spaces, yielding systematic criteria for generating orthogonal, tangential, and isotropic families of vectors. Our main results establish necessary and sufficient conditions for the existence of tangential bases, including a characterization of regular spaces of positive index and strong algebraic obstructions in the hyperbolic case. In addition, we prove a general constructive existence theorem for isotropic bases in real regular quadratic spaces.

1. Introduction

The structure of finite-dimensional quadratic spaces has been a central topic in the theory of quadratic forms since the foundational works of O’Meara [1] and Lam [2]. A cornerstone of the theory is Witt decomposition, which expresses every regular quadratic space as an orthogonal sum of hyperbolic planes and an anisotropic component. In particular, it is well known that every finite-dimensional regular quadratic space admits an orthogonal basis [1].

Beyond orthogonality, however, other geometric configurations have received comparatively little attention. One such configuration is given by the tangency condition

$${(u·v)}^2=(u·u)(v·v),$$

which is equivalent to the vanishing of the Gram determinant of $\{u,v\}$. Geometrically, this condition corresponds to a rank-one degeneracy and arises naturally in classical geometry, for instance, in the theory of conics and tangent lines. From the algebraic point of view, it describes a precise form of quadratic dependence between two nonzero vectors.

While isotropic vectors and hyperbolic planes play a fundamental role in the classical structure theory, the systematic study of bases formed by mutually tangential vectors appears to be largely unexplored. Recent algorithmic approaches have investigated the anisotropic part of quadratic forms [3,4], but the existence and explicit construction of tangential or isotropic bases have not been treated in a unified and constructive manner.

The purpose of this paper is to investigate the following problems:

• Under what structural conditions does a finite-dimensional regular quadratic space admit a basis consisting of mutually tangential vectors?
• When can such a basis be chosen to be normalized or anti-normalized?
• Under what circumstances can a basis be composed entirely of isotropic vectors?

Our approach is constructive. Using vectorial determinant techniques, we develop explicit formulas for generating vectors that are orthogonal or tangent to a prescribed family. Although reminiscent of the classical Gram-Schmidt process [5], the method operates in arbitrary regular quadratic spaces over fields of characteristic different from 2, and it relies fundamentally on the behavior of Gram determinants.

The main results show that the existence of tangential bases is governed by the presence of hyperbolic components in the Witt decomposition. In particular, we prove that regular quadratic spaces of positive index admit normalized or anti-normalized tangential bases and establish criteria that link their existence to the structure of the anisotropic complement. We also analyze the hyperbolic case and formulate conjectures concerning the non-existence of certain tangential configurations.

These results provide a new geometric perspective on quadratic spaces, complementing both the classical structure theory [1,2] and recent algorithmic developments [3,4,6].

2. Preliminary Facts

For the sake of simplicity and convenience to the reader, we will begin this section with some basic and necessary concepts for understanding this work.

Definition 1

(See [5], 6.2). Let V be a vector space over a field $\mathbb{F}$. A symmetric bilinear form on V is a bilinear map

$$\phi :V\times V\to \mathbb{F}$$

satisfying, for all $u,v,w\in V$ and $\lambda \in \mathbb{F}$:

1. 

$\phi (u+v,w)=\phi (u,w)+\phi (v,w)$;

2. 

$\phi (\lambda u,v)=\lambda \phi (u,v)$;

3. 

$\phi (u,v)=\phi (v,u)$.

In what follows, we write $u·v$ instead of $\phi (u,v)$. A quadratic space is a vector space V equipped with a symmetric bilinear form. We say that V is regular if the associated bilinear form is non-degenerate.

Definition 2.

Let $u\ne 0$ be a vector in a quadratic space V. Then, the following apply:

• u isotropic if $u·u=0$.
• u anisotropic if $u·u\ne 0$.

A subspace S of V is totally isotropic if $u·v=0$ for all $u,v\in S$.

Definition 3.

Let V be a finite-dimensional quadratic space. The index of V, denoted $indV$, is the maximal dimension of a totally isotropic subspace of V. Thus, $indV=0$ if and only if V is anisotropic.

Given a basis $\mathcal{B}=\{v_1,v_2,\dots ,v_n\}$ of V, the Gram matrix of is the $\mathcal{B}$ is the matrix $[v_i·v_j]$, and its determinant is denoted by $Gram\mathcal{B}=det\left({[v_i·v_j]}_{1\le i,j\le n}\right)$.

Two vectors $u,v\in V$ are called orthogonal if $u·v=0$ and tangent if:

$${(u·v)}^2=(u·u)(v·v).$$

(1)

Algebraically, this condition characterizes the degeneracy of the Gram matrix of $\{u,v\}$, and thus encodes a rank-one interaction between the vectors.

The following classical theorems will be used throughout this paper. Proofs may be found in [1].

Theorem 1

(See [1], Theorem 42.1). Every finite-dimensional quadratic space admits an orthogonal basis.

Theorem 2

(See [1], Theorem 42.2–4). Let V be a finite-dimensional quadratic space. The following statements are equivalent:

1. 

V is regular;

2. 

V has an orthogonal anisotropic basis;

3. 

For every basis $\mathcal{B}$ of V, $Gram\mathcal{B}\ne 0$.

4. 

There exists a basis $\mathcal{B}$ of V such that $Gram\mathcal{B}\ne 0$.

For any subset $A\subset V$, define the orthogonal complement of A as

$$A^{\perp }:=\{u\in V:a·u=0\forall a\in A\}.$$

(2)

It is well known that $A^{\perp }$ is a subspace of V.

Theorem 3

(See [1], Theorem 42.6). If $U\subseteq V$ is any subspace, then:

$$dim\left(U\right)+dim\left(U^{\perp }\right)=dim\left(V\right)+dim(U\cap V^{\perp }).$$

Corollary 1.

If V is regular, then for every subspace $U\subseteq V$, we have

$$dim\left(U\right)+dim\left(U^{\perp }\right)=dim\left(V\right)\mathrm{and}U={\left(U^{\perp }\right)}^{\perp }.$$

Definition 4.

A quadratic space V is a hyperbolic plane if the following apply:

(i) 

$dim\left(V\right)=2$;

(ii) 

V is regular;

(iii) 

V contains a nonzero isotropic vector.

Theorem 4

(See [1], Theorems 42.7–42.9). If V is a hyperbolic plane, then there exist bases ${\mathcal{B}}_1=\{v_1,v_2\}$, ${\mathcal{B}}_2=\{w_1,w_2\}$ such that

(i) 

$v_1·v_1=1$, $v_2·v_2=-1$, $v_1·v_2=0$.

(ii) 

$w_1·w_1=0=w_2·w_2$, $w_1·w_2=1$.

Definition 5.

A quadratic space V is a  hyperbolic space if there exist hyperbolic planes $V_1,V_2,\dots ,V_k$ such that

$$V=V_1\perp V_2\perp \cdots \perp V_k.$$

In this case, V is regular and $dim\left(V\right)$ is even.

Theorem 5

(See [1], Theorems 42.10–11). Every regular quadratic space V of dimension n and index s admits an orthogonal decomposition:

$$V=H\perp U,$$

where H is a hyperbolic space of dimension $2s$ and U is an anisotropic subspace of dimension $n-2s$.

3. Tangency and Orthogonality

The concepts of orthogonality and tangency are traditionally studied in the context of inner product spaces, yet the extension of tangency to general quadratic spaces is not well documented in the literature. This section provides determinant-based constructions for vectors that are either orthogonal or tangent to a given family.

The vectorial determinant construction used here generalizes ideas from multilinear algebra ([5], Ch. 8) to the setting of quadratic forms.

Let V be a vector space over a field $\mathbb{F}$. A vectorial determinant is an expression of the form

$$v=\left|\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\\ a_{11}& a_{12}& \cdots & a_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n-1,1}& a_{n-1,2}& \cdots & a_{n-1,n}\end{array}\right|$$

(3)

where $v_1,v_2,\dots ,v_n\in V$ and each $a_{ij}\in \mathbb{F}$, for $1\le i\le n-1$, $1\le j\le n$. The determinant is developed along the first row, so we can write:

$$v=\sum _{i=1}^n{\lambda }_i{v}_i,$$

where ${\lambda }_i$ is the cofactor of the i-th entry in the top row of the associated scalar matrix

$$A=\left(\begin{array}{cccc}{a}_{01}& a_{02}& \cdots & a_{0n}\\ a_{11}& a_{12}& \cdots & a_{1n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n-1,1}& a_{n-1,2}& \cdots & a_{n-1,n}\end{array}\right)$$

where $a_{0i}=1$ for each $i=1,\dots ,n$.

Theorem 6

(Orthogonality via Vectorial Determinant). Let V be a regular, $\mathcal{B}=\{w_1,w_2\dots ,$$w_n\}$ a basis, and $\{v_1,v_2\dots ,v_{n-1}\}\subset V$ linearly independent. Define

$$v=\left|\begin{array}{cccc}{w}_1& w_2& \cdots & w_n\\ w_1·v_1& w_2·v_1& \cdots & w_n·v_1\\ ⋮& ⋮& \ddots & ⋮\\ w_1·v_{n-1}& w_2·v_{n-1}& \cdots & w_n·v_{n-1}\end{array}\right|$$

Then

1. 

$v\ne 0$;

2. 

$v·v_i=0$ for $1\le i\le n-1$;

3. 

$v·v=det\left({[w_i·w_j]}_{1\le i,j\le n}\right)det\left({[v_i·v_j]}_{1\le i,j\le n-1}\right)$.

Proof. 

Since $v_1,\dots ,v_{n-1}$ are linearly independent and V is regular, their Gram matrix

$$G_v={\left[v_i·v_j\right]}_{1\le i,j\le n-1}$$

is nonsingular; hence $det\left(G_v\right)\ne 0$.

(1) 

Non-vanishing of  $\mathit{v}$.

By construction, v is defined as a vectorial determinant whose coefficients are minors of the matrix formed by the scalar products $w_k·v_i$. If $v=0$, then all these minors would vanish, which would imply that the rows corresponding to the vectors $v_1,\dots ,v_{n-1}$ are linearly dependent. This contradicts the nonsingularity of $G_v$. Hence $v\ne 0$.

(2) 

Orthogonality.

Fix $i\in \{1,\dots ,n-1\}$. Taking the scalar product of v with $v_i$ amounts to replacing the first row of the defining determinant by

$$(w_1·v_i,\dots ,w_n·v_i).$$

But this row already appears among the remaining rows of the matrix. Therefore, the determinant has two identical rows, and by alternating multilinearity, it vanishes. Thus

$$v·v_i=0\mathrm{for}\mathrm{all}i.$$

(3) 

Computation of  $\mathit{v}·\mathit{v}$.

To compute $v·v$, we expand the vectorial determinant in the basis $\{w_1,\dots ,w_n\}$. Using bilinearity of the scalar product and the Cauchy-Binet formula, one obtains

$$v·v=det\left({[w_i·w_j]}_{1\le i,j\le n}\right)det\left({[v_i·v_j]}_{1\le i,j\le n-1}\right).$$

This follows from the fact that the construction of v is precisely the classical adjugate-type expression for a vector orthogonal to $v_1,\dots ,v_{n-1}$, and the quadratic norm is governed by the product of the two Gram determinants.

This completes the proof.

□

Corollary 2.

The family $\{v_1,v_2\dots ,v_{n-1}\}$ is orthogonal to an anisotropic vector $v\in V$ if and only if $det\left({[v_i·v_j]}_{1\le i,j\le n}\right)\ne 0$.

Theorem 7

(Construction of Tangent Vectors). Let V be a finite-dimensional regular quadratic space over a field $\mathbb{F}$, and Let $\mathcal{B}=\{v_1,v_2\dots ,v_n\}$ be a basis of V. Let ${\alpha }_1,{\alpha }_2\dots ,{\alpha }_{n-1}\in \mathbb{F}$ be scalars and define

$$v=\left|\begin{array}{ccccc}{v}_1& v_2& \cdots & v_n& 0\\ v_1·v_1& v_1·v_2& \cdots & v_1·v_n& {\alpha }_1\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ v_{n-1}·v_1& v_{n-1}·v_2& \cdots & v_{n-1}·v_n& {\alpha }_{n-1}\\ v_n·v_1& v_n·v_2& \cdots & v_n·v_n& \mu \end{array}\right|$$

where $\mu \in \mathbb{F}$ is a root of the polynomial

$$p\left(X\right)=\left|\begin{array}{cccccc}{\alpha }_1& {\alpha }_2& \cdots & {\alpha }_{n-1}& X& 1\\ v_1·v_1& v_1·v_2& \cdots & v_1·v_{n-1}& v_1·v_n& {\alpha }_1\\ ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ v_{n-1}·v_1& v_{n-1}·v_2& \cdots & v_{n-1}·v_{n-1}& v_{n-1}·v_n& {\alpha }_{n-1}\\ v_n·v_1& v_n·v_2& \cdots & v_n·v_{n-1}& v_n·v_n& X\end{array}\right|.$$

Let $\mathrm{\Delta }=det(v_i·v_j)$. Then, the following apply:

• $v·v_i={(-1)}^{n-1}{\alpha }_i\mathrm{\Delta }$ for each $i=1,\dots ,n-1$;
• $v·v_n={(-1)}^{n-1}\mu \mathrm{\Delta }$;
• $v·v={\mathrm{\Delta }}^2$.

In particular, if ${\alpha }_i^2=v_i·v_i$, then the vector v is tangent to each of the vectors $v_1,\dots ,v_{n-1}$.

Proof. 

Let

$$\mathrm{\Delta }=det\left({[v_i·v_j]}_{1\le i,j\le n}\right).$$

(1) 

Computation of  $\mathit{v}·{\mathit{v}}_{\mathit{i}}$  for $\mathbf{1}\le \mathbf{i}\le \mathbf{n}-\mathbf{1}$.

By bilinearity of the scalar product and the definition of the vectorial determinant, taking the scalar product of v with $v_i$ amounts to replacing the first row of the defining determinant by

$$(v·v_1,\dots ,v·v_n,0).$$

Expanding along the last column (the column containing ${\alpha }_1,\dots ,{\alpha }_{n-1},\mu$), all entries vanish except the one corresponding to ${\alpha }_i$. Hence the Laplace expansion yields

$$v·v_i={(-1)}^{n-1}{\alpha }_{i}det\left({[v_k·v_{\ell }]}_{1\le k,\ell \le n}\right)={(-1)}^{n-1}{\alpha }_i\mathrm{\Delta }.$$

(2) 

Computation of  $\mathit{v}·{\mathit{v}}_{\mathit{n}}$.

The same argument applies when taking the scalar product with $v_n$. In this case the only nonzero contribution in the Laplace expansion comes from the entry $\mu$ in the last row, giving

$$v·v_n={(-1)}^{n-1}\mu \mathrm{\Delta }.$$

(3) 

Computation of  $\mathit{v}·\mathit{v}$.

To compute $v·v$, consider the determinant

$$\left|\begin{array}{cccc}v·v_1& \cdots & v·v_n& 0\\ v_1·v_1& \cdots & v_1·v_n& {\alpha }_1\\ ⋮& \ddots & ⋮& ⋮\\ v_{n-1}·v_1& \cdots & v_{n-1}·v_n& {\alpha }_{n-1}\\ v_n·v_1& \cdots & v_n·v_n& \mu \end{array}\right|.$$

Substituting the expressions obtained in Equations (1) and (2), the first row becomes

$$({(-1)}^{n-1}{\alpha }_1\mathrm{\Delta },\dots ,{(-1)}^{n-1}{\alpha }_{n-1}\mathrm{\Delta },{(-1)}^{n-1}\mu \mathrm{\Delta },0).$$

Factoring ${(-1)}^{n-1}\mathrm{\Delta }$ from that row, the determinant becomes

$${(-1)}^{n-1}\mathrm{\Delta }\left|\begin{array}{ccccc}{\alpha }_1& \cdots & {\alpha }_{n-1}& \mu & 0\\ v_1·v_1& \cdots & v_1·v_n& {\alpha }_1\\ ⋮& \ddots & ⋮& ⋮\\ v_{n-1}·v_1& \cdots & v_{n-1}·v_n& {\alpha }_{n-1}\\ v_n·v_1& \cdots & v_n·v_n& \mu \end{array}\right|.$$

Expanding again along the last column and using the defining condition $p\left(\mu \right)=0$, we obtain

$$v·v={(-1)}^{n-1}\mathrm{\Delta }·{(-1)}^{n-1}\mathrm{\Delta }={\mathrm{\Delta }}^2.$$

This completes the proof.

□

From the previous theorem, we can obtain the following two corollaries.

Corollary 3

(Roots of the Tangency Polynomial). Let $v_n\in V$ be a vector such that $v_n·v_i=0$ for all $1,2,\dots ,n-1$, and define the symmetric matrix:

$$D=\left(\begin{array}{ccc}{v}_1·v_1& \cdots & v_1·v_{n-1}\\ ⋮& \ddots & ⋮\\ v_{n-1}·v_1& \cdots & v_{n-1}·v_{n-1}\end{array}\right),\mathrm{with}det\left(D\right)\ne 0,$$

(4)

and the row vector $B=({\alpha }_1,\dots ,{\alpha }_{n-1})\in {\mathbb{F}}^{n-1}$. Then the roots of the polynomial $p\left(X\right)$ from Theorem 7 are

$$\mu =\pm \sqrt{(1-B{D}^{-1}{B}^T)(v_n·v_n)}.$$

These roots lie in $\mathbb{F}$ if and only if there exists $\lambda \in \mathbb{F}$ such that ${\lambda }^2=(1-B{D}^{-1}{B}^T)(v_n·v_n)$.

Proof. 

We compute the determinant defining $p\left(X\right)$ by expanding along the last row. The expansion yields

$$p\left(X\right)={(-1)}^{n-1}{X}^2·det\left(D\right)-{(-1)}^{n-1}(v_n·v_n)S$$

where S is the determinant of

$$S=\left|\begin{array}{ccccc}{\alpha }_1& {\alpha }_2& \cdots & {\alpha }_{n-1}& 1\\ v_1·v_1& v_1·v_2& \cdots & v_1·v_{n-1}& {\alpha }_1\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ v_{n-1}·v_1& v_{n-1}·v_2& \cdots & v_{n-1}·v_{n-1}& {\alpha }_{n-1}\end{array}\right|.$$

Let $\mathrm{\Lambda }:=B{D}^{-1}$. Subtracting from the first row the linear combination $\sum {\alpha }_i·(\mathrm{row}i+1)$, we obtain

$$S=det\left(D\right)·(1-B{D}^{-1}{B}^T).$$

Substituting into the expression for $p\left(X\right)$, we get

$$p\left(X\right)=det\left(D\right)(X^2-(1-B{D}^{-1}{B}^T)(v_n·v_n)).$$

Solving $p\left(X\right)=0$, we find

$$\mu =\pm \sqrt{(1-B{D}^{-1}{B}^T)(v_n·v_n)}.$$

□

Corollary 4

(Number of Tangent Vectors). Assume that for each $i=1,2,\dots ,n-1$, the scalar ${\alpha }_i\in \mathbb{F}$ satisfies ${\alpha }_i^2=v_i·v_i$. Then there exist at most $2^{n-1}$ choices for the sequence $({\alpha }_1,\dots ,{\alpha }_{n-1})$, and consequently, at most $2^{n-1}$ distinct vectors tangent to the family $\{v_1,\dots ,v_{n-1}\}$.

Furthermore, if $\mathbb{F}=\mathbb{R}$ and the vector $v_n$ is such that $v_n·v_i=0$ for all $i=1,\dots ,n-1$, and $B=({\alpha }_1,\dots ,{\alpha }_{n-1})$ with ${\alpha }_i^2=v_i·v_i$, then the number of linearly independent tangent vectors to $\{v_1,\dots ,v_{n-1}\}$ is as follows:

• Two, if $1-B{D}^{-1}{B}^T>0$;
• One, if $1-B{D}^{-1}{B}^T=0$;
• None, if $1-B{D}^{-1}{B}^T<0$.

In this case, the tangent vector v constructed in Theorem 7 has the form:

$$v=\pm det\left(D\right)\sqrt{1-B{D}^{-1}{B}^T}·{\displaystyle \frac{v_n}{v_n·v_n}}\left|\begin{array}{cccc}{v}_1& \cdots & v_{n-1}& 0\\ v_1·v_1& \cdots & v_1·v_{n-1}& {\alpha }_1\\ ⋮& \ddots & ⋮& ⋮\\ v_{n-1}·v_1& \cdots & v_{n-1}·v_{n-1}& {\alpha }_{n-1}\end{array}\right|.$$

Proof. 

Each ${\alpha }_i\in \mathbb{F}$ must satisfy ${\alpha }_i^2=v_i·v_i$, which admits exactly two square roots in an algebraically closed field (or at most two in $\mathbb{R}$, one if zero). Thus, there are $2^{n-1}$ possible sequences for $({\alpha }_1,\dots ,{\alpha }_{n-1})$.

In the real case $\mathbb{F}=\mathbb{R}$, the number of real roots of the tangency polynomial $p\left(X\right)$ depends on the sign of the discriminant:

• If $1-B{D}^{-1}{B}^T>0$, $p\left(X\right)$ has two real roots.
• If the expression equals zero, we have a double root (one solution).
• If the expression is negative, $p\left(X\right)$ has no real roots, and hence no tangent vector exists.

This completes the proof. □

4. Tangential and Anisotropic Bases

We begin by introducing terminology regarding the normalization of bases in quadratic spaces.

Definition 6.

Let V be a quadratic space of dimension n. A basis $\mathcal{B}=\{v_1,v_2,\dots ,v_n\}$ is said to be as follows:

• normalized if $v_i·v_i=1$ for all i;
• anti-normalized if $v_i·v_i=-1$ for all i.

The following result provides sufficient conditions for the existence of a normalized (or anti-normalized) tangential basis in certain regular spaces.

Theorem 8.

Let V be a regular quadratic space over $\mathbb{R}$ with $dim\left(V\right)=2ind\left(V\right)+1$. Then V admits a tangential basis that is either normalized or anti-normalized.

Proof. 

By Theorem 5, V decomposes as

$$V=\langle v_1,v_1^{\prime }\rangle \perp \langle v_2,v_2^{\prime }\rangle \perp \cdots \perp \langle v_s,v_s^{\prime }\rangle \perp \langle v\rangle ,$$

where $ind\left(V\right)=s$, and each $v_i·v_i=v_i^{\prime }·v_i^{\prime }=1$, while $v·v=\pm 1$ (we assume $\pm 1$ since V is regular over $\mathbb{R}$).

Define the following:

• $w_i:=v_i+v$.
• $w_i^{\prime }:=2v^{\prime }-{\displaystyle \frac{v}{v·v}}$.
• $w_0:=v$.

It is straightforward to verify the following:

• $w_i·w_i=v·v$.
• $w_i^{\prime }·w_i^{\prime }=v·v$.
• $w_i·w_i^{\prime }=1$.
• $w_i·v=v·v$, and $w_i^{\prime }·v=-1$.

Thus, the set

$$\{w_1,w_1^{\prime },\dots ,w_s,w_s^{\prime },v\}$$

is a tangential basis of V, normalized if $v·v=1$ and anti-normalized if $v·v=-1$. □

Before generalizing this theorem, we need the next lemma.

Lemma 1

(Auxiliary). Let $b>1$ be a real number, and let $V=\langle u_1\rangle \perp \cdots \perp \langle v_n\rangle$ be an anisotropic quadratic space over $\mathbb{R}$, $u_i·u_i=1$ for each $i\in \{1,\dots ,n\}$ (positive case) or $u_i·u_i=-1$ (negative case) for all i. Suppose $u_1^{\prime },u_2^{\prime },\dots ,u_{n-1}^{\prime }$ are linearly independent vectors such that the following apply:

• $u_i^{\prime }.·u_j^{\prime }=b$ for $i\ne j$;
• $u_i^{\prime }·u_i^{\prime }=b^2$ (positive case), or $u_i^{\prime }·u_i^{\prime }=-b^2$ (negative case).

Then there exists a vector $u^{\prime }\in V$ such that the family $\{u_1^{\prime },\dots ,u_{n-1}^{\prime },u^{\prime }\}$ is linearly independent and the following apply:

• $u^{\prime }.·u_i^{\prime }=-b$, $u^{\prime }·u^{\prime }=b^2$, in the positive case;
• $u^{\prime }·u_i^{\prime }=-b$, $u^{\prime }·u^{\prime }=-b^2$, in the negative case.

Proof. 

Let $w\in V$ such that $w·u_i^{\prime }=0$ for each $i\in \{1,2,\dots ,n-1\}$ and $w·w=\pm 1$, ($+1$ in the positive case and $-1$ in the negative case). Let

$$u_0^{\prime }=\lambda \sum _{i=1}^{n-1}{u}_i^{\prime },\mathrm{where}\lambda ={\displaystyle \frac{1}{n-2+b}}.$$

We see that

$$(n-1)\lambda ={\displaystyle \frac{n-1}{(n-1)+(b-1)}}<1,\mathrm{and}\mathrm{then}{(n-1)}^2{\lambda }^2<1.$$

Hence

$$\begin{array}{ccc}\hfill u_0^{\prime }·u_0^{\prime }& =& {\lambda }^2\left(\sum _{i=1}^{n-1}{u}_i^{\prime }·u_i^{\prime }+2\sum _{i<j}{u}_i^{\prime }·u_j^{\prime }\right)\hfill \\ & =& {\lambda }^2((n-1)b^2+(n-1)(n-2)b)\hfill \\ & =& {\lambda }^2(n-1)b(b+n-2)\hfill \\ & =& \lambda (n-1)b.\hfill \end{array}$$

Since $\lambda (n-1)={\displaystyle \frac{n-1}{(b-1)+(n-1)}}<1$, we conclude that $u_0^{\prime }·u_0^{\prime }<b<b^2$. Then, if $\mu =\sqrt{b^2-u_0^{\prime }·u_0^{\prime }},=u_0^{\prime }+\mu w$ is the vector we were looking for. In the negative case we have $u_0^{\prime }·u_i^{\prime }=-b$ for each $i\in \{1,2,\dots ,n-1\}$ and

$$u_0^{\prime }·u_0^{\prime }={\lambda }^2[-(n-1)b^2-(n-1)(n-2)b]=-{\lambda }^2(n-1)b{\lambda }^{-1}=-\lambda (n-1)b.$$

Clearly $-\lambda (n-1)b>-b^2$ since $\lambda (n-1)<1<b$. Therefore, $u_0^{\prime }·u_0^{\prime }>-b^2$. Defining ${\mu }^{\prime }=\sqrt{b^2+u_0^{\prime }·u_0^{\prime }}$, the vector $u^{\prime }=u_0^{\prime }+{\mu }^{\prime }w$ satisfies $u^{\prime }·u^{\prime }=-b^2$ and $u^{\prime }·u_i^{\prime }=-b$ for each $i\in \{1,2,\dots ,n-1\}$. □

The next theorem is the most important result of this paper.

The idea of the proof is to construct a tangential basis on a maximal regular subspace and then extend it to the full space by carefully controlling the interaction with the anisotropic component.

Theorem 9.

Let V be a regular quadratic space over $\mathbb{R}$ with $dim\left(V\right)=n>2$ and $ind\left(V\right)>0$. Then V admits a tangential basis which is either normalized or anti-normalized.

Proof. 

Let $s=ind\left(V\right)>0$. By Witt decomposition, V splits orthogonally as

$$V=\langle v_1,v_1^{\prime }\rangle \perp \cdots \perp \langle v_s,v_s^{\prime }\rangle \perp \langle u_1\rangle \perp \dots \perp \langle u_t\rangle ,$$

where each $\langle v_i,v_i^{\prime }\rangle$ is a hyperbolic plane and $u_j·u_j=\pm 1$ according to the signature.

• Step 1: Construction on a core subspace.

Consider the non-degenerate subspace

$$V_0=\langle v_1,v_1^{\prime }\rangle \perp \cdots \perp \langle v_s,v_s^{\prime }\rangle \perp \langle u_1\rangle .$$

Since $ind\left(V_0\right)=s>0$ and $dim\left(V_0\right)>2$, Theorem 8 ensures that $V_0$ admits a normalized (or anti-normalized) tangential basis.

• Step 2: Extension to the anisotropic complement.

The remaining summand

$$W=\langle u_2\rangle \perp \cdots \perp \langle u_t\rangle$$

is anisotropic. By Lemma 1, we may construct vectors $u_2^{\prime },\dots ,u_t^{\prime }$ in W whose scalar products satisfy the required tangency relations relative to a fixed vector in $V_0$.

• Step 3: Global assembly.

Choose a vector $u\in V_0$ compatible with the tangential structure constructed in Step 1. Then the family

$$\{w_1,w_1^{\prime },\dots ,w_s,w_s^{\prime },u_1,u+u_2^{\prime },\dots ,u+u_t^{\prime }\}$$

forms a basis of V.

Orthogonality of the Witt decomposition guarantees that the only nontrivial scalar products arise within $V_0$ and within the constructed combinations $u+u_j^{\prime }$. By construction and by Lemma 1, all required Gram determinants of pairs vanish, so the vectors are mutually tangent. Moreover, the normalization (or anti-normalization) property is preserved.

Hence, V admits a tangential basis that is normalized or anti-normalized. □

We conclude this section by presenting an explicit example in dimension 4, corresponding to the minimal case of positive index, which concretely illustrates the existence of a tangential basis predicted by Theorem 9.

Example 1

(Dimension 4). Consider the quadratic space $({\mathbb{R}}^4,q)$ with

$$q\left(x\right)=x_1^2+x_2^2-x_3^2-x_4^2,$$

whose Gram matrix is $\mathrm{diag}(1,1,-1,-1)$. Then $dim\left(V\right)=4$ and $ind\left(V\right)=2$.

Define

$$e_1=(1,0,1,0),e_2=(1,0,-1,0),e_3=(0,1,0,1),e_4=(0,1,0,-1).$$

Each vector is isotropic since $e_i·e_i=0$. Moreover, the family $\{e_1,e_2,e_3,e_4\}$ forms a basis of ${\mathbb{R}}^4$.

Because $e_i·e_i=0$ for all i, the tangency condition

$${(e_i·e_j)}^2=(e_i·e_i)(e_j·e_j)$$

holds for every pair. Hence this is a tangential basis in dimension 4.

The following theorem gives us the necessary and sufficient conditions for a quadratic space to have a tangential base.

Theorem 10

(Tangency Inheritance in Direct Sums). Let $V=U\oplus V^{\perp }$, where $U\subset V$ is a regular subspace. Then V admits a tangential basis if and only if U does.

Proof. 

Suppose $dimU=n$ and $dim{V}^{\perp }=m$. Suppose V has a tangential basis $\{v_1,v_2,\cdots ,v_{n+m}\}$, every $v_i$ can be written in the form $v_i=u_i+{\overline{u}}_i$, $u_i\in U$ and ${\overline{u}}_i\in V^{\perp }$. We claim the vectors $\{u_1,u_2,\dots ,u_{n+m}\}$ generate U. Indeed, if $w\in U$ is arbitrary, we have scalars ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{n+m}$ such that

$$w={\lambda }_1{v}_1+{\lambda }_2{v}_2+\dots +{\lambda }_{n+m}{v}_{n+m}.$$

Therefore, $w-({\lambda }_1{u}_1+{\lambda }_2{u}_2+\cdots +{\lambda }_{n+m}{u}_{n+m})={\lambda }_1{\overline{u}}_1+{\lambda }_2{\overline{u}}_2+\cdots +{\lambda }_{n+m}{\overline{u}}_{n+m}$. Since $w-({\lambda }_1{u}_1+{\lambda }_2{u}_2+\cdots +{\lambda }_{n+m}{u}_{n+m})\in U\cap V^{\perp }$, we have $w={\lambda }_1{u}_1+{\lambda }_2{u}_2+\cdots +{\lambda }_{n+m}{u}_{n+m}$ as was to be proved. Therefore, a subfamily of $\{u_1,u_2,\dots ,u_{n+m}\}$ with n elements is a basis for U, say $U=\langle u_1,u_2,\dots ,u_n\rangle$. Observe now,

$${(u_i·u_j)}^2={((u_i+{\overline{u}}_i)·(u_j+{\overline{u}}_j))}^2={(v_i·v_j)}^2=(v_i·v_i)(v_j·v_j)=(u_i·u_i)(u_j·u_j).$$

Therefore, $\{u_1,\dots ,u_n\}$ is a tangential basis U.

Suppose now that U has a tangential basis $\{u_1,\dots ,u_n\}$. Let $\{{\overline{u}}_1,\dots ,{\overline{u}}_m\}$ be a basis for $V^{\perp }$. Denote $w_i=u_1+{\overline{u}}_i$, for each $i\in \{1,\dots ,m\}$. We calculate ${(u_i·w_j)}^2$ and ${(w_i·w_j)}^2$:

$$\begin{array}{ccc}\hfill {(w_i·w_j)}^2& =& {(u_i·(u_1+{\overline{u}}_j))}^2={(u_i·u_1)}^2=(u_i·u_i)(u_1·u_1)=(u_i·u_i)(w_j·w_j),\hfill \\ \hfill {(w_i·w_j)}^2& =& {((u_1+{\overline{u}}_i)·(u_1+{\overline{u}}_j))}^2=(u_1·u_1)(u_1·u_1)=(w_i·w_i)(w_j·w_j).\hfill \end{array}$$

Since $V=U\oplus V^{\perp }$, $\{u_1,u_2,\dots ,u_n,w_1,\dots ,w_m\}$ is a tangential basis for V. □

Example 2.

Let $\mathbb{K}$ be a field of characteristic $\ne 2$ and let

$$V=\{a(x^2+y^2)+2gxz+2fyz+c{z}^2|a,g,f,c\in \mathbb{K}\};$$

$[a(x^2+y^2)+2gxz+2fyz+c{z}^2]·[a^{\prime }(x^2+y^2)+2g^{\prime }xz+2f^{\prime }yz+c^{\prime }{z}^2]=2g{g}^{\prime }+2f{f}^{\prime }-a^{\prime }c$.

V is a regular space of dimension 4. If $\mathbb{K}=\mathbb{R}$, then V has index 1, but if $\mathbb{K}=\mathbb{C}$, V has index 2. If $\mathbb{K}=\mathbb{R}$, we can establish a one-to-one correspondence between subspaces of V of the form $\langle x^2+y^2+2gxz+2fyz+c{z}^2\rangle$, $g^2+f^2-c\ge 0$ and circles in the plane ${\mathbb{R}}^2$. $v\in V$ represents a real circle if and only if $v·v>0$ and $v·z^2\ne 0$. Orthogonality and tangency between vectors correspond here to orthogonality and tangency between circles as usually defined in Euclidean geometry.

We will finish this work with the definition of isometries.

Definition 7.

Two quadratic spaces $V_1,V_2$ over the same field $\mathbb{F}$ are isometric if there exists an isomorphism $\psi :V_1\to V_2$ such that $v·v^{\prime }=\psi \left(v\right)·\psi \left(v^{\prime }\right)$ for each pair of vectors $v,v^{\prime }\in V_1$. Such mapping ψ is called an isometry between $V_1$ and $V_2$.

The following theorem will be very useful for our purposes; its proof can be consulted in detail in ([1], Theorem 42.16).

Theorem 11.

Let U and V be isometric regular subspaces of a quadratic space W. Then $U^{\perp }$ and $V^{\perp }$ are isometric too.

The problem of the existence of an isotropic basis of a regular space V over the field of real numbers is solved with the following theorem.

Theorem 12

(Existence of Isotropic Bases). Let V be a regular quadratic space over $\mathbb{R}$, with index $s=ind\left(V\right)>0$. Then V admits an isotropic basis.

Proof. 

If V is a hyperbolic space and we keep the notation of Theorem 8, it is clear that $\{v_1,v_1^{\prime },\dots ,v_s,v_s^{\prime }\}$ is the desired basis. Suppose now that $t=dimV-2indV>0$. Let us consider the extension $V^{\ast }=V\perp \langle u_{t+1}\rangle$ where $u_{t+1}·u_{t+1}=u_i·u_i$ for each $i\in \{1,2,\dots ,t\}$. By Theorem 10, $V^{\ast }$ has a tangential basis $\{v_1,v_2,\dots ,v_{2s+t+1}\}$ where $v_i·v_i=1$ or $v_i·v_i=-1$ for each $i\in \{1,2,\dots ,2s+t+1\}$. We define the vectors $w_i=v_i\pm v_{2s+t+1}$, where $i\in \{1,2,\dots ,2s+t\}$ and the $sing+$ is taken if $v_i·v_{2s+t+1}\ne v_{2s+t+1}·v_{2s+t+1}$ and the $sing-$ is taken if $v_i·v_{2s+t+1}=v_{2s+t+1}·v_{2s+t+1}$. It is clear that every $w_i$ is isotropic and ${\langle v_{2s+t+1}\rangle }^{\perp }=\langle w_1,w_2,\dots ,w_{2s+t}\rangle$.

Since the subspaces $\langle v_{2s+t+1}\rangle$ and $\langle v_{t+1}\rangle$ of $V^{\ast }$ are isometric, Theorem 11 implies that their orthogonal complements are isometric too. Therefore, $V={\langle u_{t+1}\rangle }^{\perp }$ is isometric to ${\langle u_{2s+t+1}\rangle }^{\perp }$ and V has an isotropic basis. □

Corollary 5.

If $V=U\oplus V^{\perp }$, U is regular and $indU>0$, then V has an isotropic basis.

Proof. 

By Theorem 12, U has an isotropic basis $\mathcal{B}$. If $\overline{\mathcal{B}}$ any basis of $V^{\perp }$, then $\mathcal{B}\cup \overline{\mathcal{B}}$ is an isotropic basis of V. □

5. Examples and Applications

The following are some representative examples that illustrate the construction and use of tangential and anisotropic bases in different analytical and geometric contexts.

Example 3

(Tangential bases in a two-dimensional quadratic space). Let $V={\mathbb{R}}^2$ endowed with the quadratic form

$$q(x,y)=x^2-y^2,$$

and associated bilinear form

$$(x,y)·(x^{\prime },y^{\prime })=x{x}^{\prime }-y{y}^{\prime }.$$

Then V is a regular quadratic space of dimension 2 and index 1.

Consider first the standard orthogonal basis

$$\begin{array}{ccc}\hfill e_1=(1,0),& & e_2=(0,1),\hfill \end{array}$$

for which

$$\begin{array}{ccccc}\hfill e_1·e_1=1,& \hfill & \hfill e_2·e_2=-1,& \hfill & \hfill e_1·e_2=0.\end{array}$$

This basis reflects the classical orthogonal decomposition of V, but it does not exhibit any tangential interaction.

Now define the vectors

$$\begin{array}{ccc}\hfill v_1=(1,1),& & v_2=(2,3),\hfill \end{array}$$

A direct computation shows that

$$\begin{array}{ccccc}\hfill v_1·v_1=0,& \hfill & \hfill v_2·v_2=-5,& \hfill & \hfill v_1·v_2=-1.\end{array}$$

Although the vectors are neither orthogonal nor both anisotropic, they satisfy the tangency condition

$${(v_1·v_2)}^2=(v_1·v_1)(v_2·v_2).$$

Hence, $v_1$ and $v_2$ are tangent vectors in the sense of Definition (1).

Since $\{v_1,v_2\}$ is linearly independent, it forms a tangential basis of V. This example illustrates that, even in the lowest nontrivial dimension, tangential bases differ fundamentally from orthogonal ones and may involve isotropic vectors.

Geometrically, the vectors $v_1$ and $v_2$ lie on the light cone defined by $q(x,y)=0$ and on a time-like branch of the hyperbola $q(x,y)=-5$, respectively, meeting the tangency condition algebraically encoded by the quadratic form. See Figure 1.

The next example illustrates that, even in the presence of an orthogonal basis, one may construct a tangential basis exhibiting a radically different geometric structure, involving both isotropic and anisotropic vectors.

Example 4

(Tangential bases in a three-dimensional quadratic space). Let $V={\mathbb{R}}^3$ endowed with the quadratic form

$$q(x,y)=x^2+y^2-z^2,$$

and associated bilinear form

$$(x,y,z)·(x^{\prime },y^{\prime },z^{\prime })=x{x}^{\prime }+y{y}^{\prime }+z{z}^{\prime }.$$

Then V is a regular quadratic space of dimension 3 and index 1.

The standard basis

$$\begin{array}{ccccc}\hfill e_1=(1,0,0),& \hfill & \hfill e_2=(0,1,0),& \hfill & \hfill e_3=(0,0,1),\end{array}$$

is orthogonal, with

$$\begin{array}{ccc}\hfill e_1·e_1=1=e_2·e_2,& \hfill & \hfill e_3·e_3=-1.\end{array}$$

This basis reflects the classical orthogonal decomposition $V={\mathbb{R}}^2\oplus \mathbb{R}$, but does not capture any tangential interaction.

Define now the vectors

$$\begin{array}{ccccc}\hfill v_1=(1,0,1),& \hfill & \hfill v_2=(0,1,1),& \hfill & \hfill v_3=(1,1,1),\end{array}$$

A direct computation yields

$$\begin{array}{ccccc}\hfill v_1·v_1=0,& \hfill & \hfill v_2·v_2=0,& \hfill & \hfill v_3·v_3=1,\end{array}$$

and

$$\begin{array}{ccccc}\hfill v_1·v_2=-1,& \hfill & \hfill v_1·v_3=0,& \hfill & \hfill v_2·v_3=0.\end{array}$$

Therefore, the vectors, $v_1$ and $v_2$ are isotropic and tangent to each other, while $v_3$ is anisotropic and tangent to both $v_1$ and $v_2$ (See Figure 2).

Since $\{v_1,v_2,v_3\}$ is linearly independent, it forms a tangential basis of V.

Figure 2. Projection of a tangential basis in the quadratic space $({\mathbb{R}}^3,x^2+y^2-z^2)$ onto the plane $z=1$. The circle represents the isotropic cone, while $v_1$ and $v_2$ are isotropic and tangent to the anisotropic vector $v_3$.

Figure 2. Projection of a tangential basis in the quadratic space $({\mathbb{R}}^3,x^2+y^2-z^2)$ onto the plane $z=1$. The circle represents the isotropic cone, while $v_1$ and $v_2$ are isotropic and tangent to the anisotropic vector $v_3$.

The following examples are intended to illustrate potential geometric interpretations of tangency in applied settings, rather than to provide a complete application framework.

Example 5

(Reproducing kernel Hilbert spaces and anisotropic geometry). Let $X\subset {\mathbb{R}}^d$ be a nonempty set and let

$$K:X\times X\to \mathbb{R}$$

be a positive definite kernel. Denote by ${\mathcal{H}}_K$ the reproducing kernel Hilbert space (RKHS) associated with K.

For each $x\in X$, define the feature vector

$$\phi \left(x\right):=K(·,x)\in {\mathcal{H}}_K.$$

By the reproducing property, the inner product in ${\mathcal{H}}_K$ satisfies

$${\langle \phi \left(x\right),\phi \left(y\right)\rangle }_{{\mathcal{H}}_K}=K(x,y).$$

Consider the anisotropic Gaussian kernel

$$K_A(x,y)=exp(-{(x-y)}^{T}A(x-y)),$$

where $A\in {\mathbb{R}}^{d\times d}$ is a symmetric positive definite matrix.

The matrix A induces different scales and preferred directions in the feature space ${\mathcal{H}}_{K_A}$.

Let $x_1,x_2\in X$ be two data points such that

$$K_A{(x_1,x_2)}^2=K_A(x_1,x_1)K_A(x_2,x_2).$$

Then the corresponding feature vectors $\phi \left(x_1\right)$ and $\phi \left(x_2\right)$ satisfy

$${\langle \phi \left(x_1\right),\phi \left(x_2\right)\rangle }^2=\langle \phi \left(x_1\right),\phi \left(x_1\right)\rangle \langle \phi \left(x_2\right),\phi \left(x_2\right)\rangle ,$$

and are therefore tangent vectors in the quadratic space $({\mathcal{H}}_{K_A},\langle ·,·\rangle )$.

This condition expresses a perfect alignment between feature representations, reflecting a strong geometric dependence between the data points along the anisotropic directions determined by A.

Problem 1.

Let $X\subset {\mathbb{R}}^2$ be a dataset consisting of two classes

$$\begin{array}{ccc}\hfill {\mathcal{C}}_{+}=\{x_1,x_2\},& \hfill & \hfill {\mathcal{C}}_{-}=\left\{x_3\right\},\end{array}$$

and consider the anisotropic Gaussian kernel

$$K_A(x,y)=exp(-{(x-y)}^{T}A(x-y)),$$

where

$$\begin{array}{ccc}\hfill A=\left(\begin{array}{cc}\alpha & 0\\ 0& \beta \end{array}\right),& \hfill & \hfill \alpha \gg \beta >0.\end{array}$$

Assume that $x_1,x_2\in {\mathcal{C}}_{+}$ differ mainly in the second coordinate, while $x_3\in {\mathcal{C}}_{-}$ differs primarily in the first coordinate:

(a) 

Show that for sufficiently large $\alpha /\beta$,

$$K_A{(x_1,x_2)}^2\approx K_A(x_1,x_1)K_A(x_2,x_2).$$

(b) 

Interpret this result in terms of tangential vectors in the reproducing kernel Hilbert space ${\mathcal{H}}_{K_A}$.

(c) 

Explain the effect of this phenomenon on the margin of a support vector machine trained with the kernel $K_A$.

Similar effects of anisotropic kernel design on the geometry of feature spaces have been observed in margin-based learning algorithms; see, for instance [7].

5.1. Applications to Machine Learning

The theory of reproducing kernel Hilbert spaces provides the mathematical foundation of many kernel-based learning algorithms, including support vector machines and kernel principal component analysis [7]. In this framework, kernels induce implicit feature maps into (possibly infinite-dimensional) Hilbert spaces, where geometric properties of the data become accessible through inner products.

Anisotropic kernels, which assign different weights to distinct directions in the data space, have been extensively studied in the context of metric learning and adaptive kernel design. The present example shows that such anisotropy may also be interpreted geometrically in terms of tangential phenomena in the associated reproducing kernel Hilbert space.

Anisotropic Kernels and Tangential Effects in RKHS

Let $X\subset {\mathbb{R}}^d$ be a dataset and let

$$K_A(x,y)=exp(-{(x-y)}^{T}A(x-y)),$$

be an anisotropic Gaussian kernel, where A is a symmetric positive definite matrix. Denote by ${\mathcal{H}}_{K_A}$ the associated reproducing kernel Hilbert space and by $\phi :X\to {\mathcal{H}}_{K_A}$ the canonical feature map.

Problem 2.

Consider two data points $x_1,x_2\in X$ such that their difference lies predominantly in a direction weakly weighted by A. Show that, as the anisotropy ratio of A increases,

$$K_A{(x_1,x_2)}^2\to K_A(x_1,x_1)K_A(x_2,x_2).$$

and interpret this limit in terms of tangential vectors in ${\mathcal{H}}_{K_A}$.

Proposition 1

(Geometric interpretation). Under the assumptions of Problem 2, the feature vectors $\phi \left(x_1\right)$ and $\phi \left(x_2\right)$ become asymptotically tangential in the quadratic space $({\mathcal{H}}_{K_A},\langle ·,·\rangle )$.

Consequently, the effective dimension of the span

$$span\{\phi \left(x_1\right),\phi \left(x_2\right)\},\mathit{collapses}.$$

Proof. 

By the reproducing property,

$$\langle \phi \left(x_1\right),\phi \left(x_2\right)\rangle =K_A(x_1,x_2).$$

Since $K_A(x_i,x_i)=1$ for all i, the claim follows directly from the limiting equality

$${\langle \phi \left(x_1\right),\phi \left(x_2\right)\rangle }^2=\left|\right|\phi \left(x_1\right){\left|\right|}^2\left|\right|\phi \left(x_2\right){\left|\right|}^2,$$

which is precisely the tangency condition. □

Remark 1

(Implications for learning algorithms). From the viewpoint of kernel methods [8], tangential collapse corresponds to a degeneration of the Gram matrix and reflects a loss of effective dimensionality in the feature space.

6. Conclusions

In this paper, we have established new theoretical results concerning the existence and construction of tangential and anisotropic bases in finite-dimensional quadratic spaces over fields of characteristic different from 2. Our work builds upon the classical foundations laid out by O’Meara [1] and Lam [2], as well as recent algorithmic developments in the theory of quadratic forms over number fields [3,4].

Specifically, we performed the following:

• Introduced and formalized the concept of tangency in quadratic spaces via the condition ${(u·v)}^2=(u·u)(v·v)$, offering a geometric generalization of orthogonality with links to conic and circle geometry (cf. [1,2]).
• Developed two determinant-based constructions (Theorems 6 and 7) that allow the computation of vectors orthogonal or tangent to a given family of vectors. These constructions echo the computational spirit of works by Koprowski and collaborators [3,4,6], while addressing a distinct structural problem.
• Proved that certain regular quadratic spaces over $\mathbb{R}$ of dimension greater than $2ind\left(V\right)$ admit normalized or anti-normalized tangential bases (Theorems 8 and 9), extending the scope of decomposition theorems known in the literature ([1], Ch. 42).
• Gave necessary and sufficient conditions for the existence of tangential and isotropic bases via orthogonal decompositions and isometries (Theorems 10 and 12), contributing to the structural classification of quadratic spaces.

These results provide new tools for understanding how geometric properties such as tangency and isotropy manifest within the algebraic structure of quadratic forms. The methods introduced are constructive and may be applicable in algorithmic settings, complementing existing approaches in symbolic computation and algebraic geometry [3,4].

Although we have provided affirmative answers in several cases, important questions remain open. In particular, we pose the following:

Problem 3.

If V is a regular quadratic space over $\mathbb{R}$ of arbitrary finite dimension, then it does not admit a tangential basis unless its index is strictly positive.

Problem 4.

No hyperbolic space over $\mathbb{R}$, admits a tangential basis.

These conjectures align with the structural rigidity observed in the decomposition of regular spaces into hyperbolic and anisotropic parts, as characterized in ([1], Thm. 42.10–42.11). If true, they imply the following refinement:

• The only real regular quadratic spaces that admit tangential bases are those with positive index and non-hyperbolic structure.

In the context of geometric representation, our results have a concrete interpretation. For instance, in Example 2, the correspondence between quadratic forms and circles in ${\mathbb{R}}^2$ suggests that tangency between vectors corresponds to Euclidean tangency between circles. This reinforces the idea also hinted at in [2] that certain algebraic conditions reflect deeper geometric relationships.

Therefore, the concept of tangency in quadratic spaces is not merely formal; it has rich interpretative value in both pure mathematics and applications, such as geometry, physics, and even machine learning (where inner product structures appear in kernel methods). Future investigations might explore these connections more deeply, possibly using the computational techniques proposed in [3,4,5,6].