# A characterization of inner product spaces via norming vectors

Guillaume Aubrun, Mathis Cavichioli

► To cite this version:

Guillaume Aubrun, Mathis Cavichioli. A characterization of inner product spaces via norming vectors. Canadian Mathematical Bulletin, 2025, pp.1-5. 10.4153/S0008439524000985. hal-04951651

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# A CHARACTERIZATION OF INNER PRODUCT SPACES VIA NORMING VECTORS

GUILLAUME AUBRUN AND MATHIS CAVICHIOLI

**ABSTRACT.** A finite-dimensional normed space is an inner product space if and only if the set of norming vectors of any endomorphism is a linear subspace. This theorem was proved by Sain and Paul for real scalars. In this paper, we give a different proof which also extends to the case of complex scalars.

## 1. INTRODUCTION

The characterization of Euclidean spaces among normed spaces, or Hilbert spaces among Banach spaces, is a classical theme in functional analysis. It can be traced back to the Jordan-von Neumann theorem [3], which states that a normed space is Euclidean if and only if it satisfies the parallelogram identity. This line of research has been very fruitful: for example, the monograph [1] compiles about 350 characterizations of inner product spaces.

We consider vector spaces over a field $K$ which is either $R$ or $C$. Recall that an inner product on a vector space $X$ is a map $\langle \cdot, \cdot \rangle : X \times X \to K$ that satisfies the usual axioms of conjugate symmetry, linearity in one variable and positive-definiteness. If $\| \cdot \|$ is a norm on $X$, we say that the normed space $(X, \| \cdot \|)$ is an inner product space if there exists an inner product $\langle \cdot, \cdot \rangle$ on $X$ such that the identity $\|x\|^2 = \langle x, x \rangle$ holds for every $x \in X$. A finite-dimensional inner product space over the real field is also called a Euclidean space.

In this note, we consider a characterization of finite-dimensional inner product spaces by a special property of their endomorphisms. Recall that the operator norm of a linear operator $u : X \to X$, denoted $\|u\|_{op}$, is defined as the smallest $C \ge 0$ such that the inequality $\|u(x)\| \le C\|x\|$ holds for every $x \in X$. We consider the set $N(u)$ of norming vectors, defined as

$$N(u) = \{x \in X : \|u(x)\| = \|u\|_{op} \cdot \|x\|\}.$$

The following theorem has been proved by Sain and Paul (see [5, Theorem 2.2]) in the real case only. Our goal is to provide a completely different and self-contained proof, which extends naturally to the complex case.

**Theorem.** Let $X$ be a finite-dimensional normed space over the real or complex field. The following are equivalent.

(1) The space $X$ is an inner product space.

(2) For every linear operator $u : X \to X$, the set $N(u)$ is a linear subspace.

<footnote n="2020">2020 Mathematics Subject Classification. 46C15.</footnote>

<footnote>Key words and phrases. Characterization of inner-product spaces, norming vectors, positive John position.</footnote>
The implication (1) ⇒ (2) can be shown as follows. If X is an inner product space, we may consider the adjoint operator $u^*: X \to X$. Let $\lambda_{\max}(u^*u)$ denote the largest eigenvalue of the self-adjoint operator $u^*u$ and E be the associated eigenspace. It is elementary to check then for every $x \in X$,

$$ \|ux\|^2 = \langle u^*ux, x \rangle \le \lambda_{\max}(u^*u)\|x\|^2 $$

with equality if and only if $x \in E$. Since $\lambda_{\max}(u^*u) = \|u\|_{\text{op}}^2$, it follows that the set $\mathcal{N}(u)$ coincides with E. In particular, $\mathcal{N}(u)$ is a linear subspace.

In the following sections we prove the harder implication (2) ⇒ (1). Our arguments are inherently finite-dimensional; we leave open the question of whether the theorem extends to infinite-dimensional normed spaces.

## 2. THE MAIN PROPOSITION

In this section we introduce our main tool. It is strongly related to the concept of positive John position of convex bodies which is studied in [2]. We use a slightly different approach that allows us to cover also the complex case in a natural way.

Consider a norm $\| \cdot \|$ on $\mathbf{K}^n$ and equip the algebra $\mathbf{M}_n(\mathbf{K})$ of $n \times n$ matrices with the corresponding operator norm $\| \cdot \|_{\text{op}}$. We denote by $\text{GL}_n(\mathbf{K})$ the group of invertible matrices. Given $Q \in \text{GL}_n(\mathbf{K})$, we consider the set

$$ C_Q = \{A \in \mathbf{M}_n(\mathbf{K}) : \|AQ\|_{\text{op}} \le 1\}. $$

Let $\mathbf{M}_n^+(\mathbf{K})$ be the cone of positive semi-definite matrices. The set $\mathbf{M}_n^+(\mathbf{K}) \cap C_Q$ is compact and therefore contains an element of maximal determinant. (This element is unique but we do not need this information.) Such an element A satisfies $\|AQ\|_{\text{op}} = 1$.

**Proposition.** Let $Q \in \text{GL}_n(\mathbf{K})$ and A of maximal determinant in $\mathbf{M}_n^+(\mathbf{K}) \cap C_Q$. Then the set

$$ \mathcal{N}(AQ) = \{x \in \mathbf{K}^n : \|AQx\| = \|x\|\} $$

spans $\mathbf{K}^n$ as a vector space.

We show in the next section how the Proposition implies our Theorem. In the real case, the Proposition follows easily from [2, Theorem 1.2]. For completeness, we include a self-contained proof.

**Proof of the Proposition.** Introduce the unit ball $\mathcal{B} = \{x \in \mathbf{K}^n : \|x\| \le 1\}$. If we identify the dual space with $\mathbf{K}^n$, the unit ball for the dual norm is

$$ \mathcal{B}^* = \{y \in \mathbf{K}^n : |\langle x, y \rangle| \le 1 \text{ for every } x \in \mathcal{B}\}. $$

By the Hahn-Banach theorem, we have for $x \in \mathbf{K}^n$

$$ \|x\| = \max_{y \in \mathcal{B}^*} |\langle x, y \rangle|. $$

Consider the compact set $T = \mathcal{B} \times \mathcal{B}^*$ and let $C(T)$ be the Banach space of continuous functions from $T$ to $\mathbf{K}$. We define a map $\alpha : \mathbf{M}_n(\mathbf{K}) \to C(T)$ as follows: if $M \in \mathbf{M}_n(\mathbf{K})$, $x \in \mathcal{B}$ and $y \in \mathcal{B}^*$, then

$$ \alpha(M)(x, y) = \langle Mx, y \rangle. $$

The map $\alpha$ is linear and satisfies $\|\alpha(M)\| = \|M\|_{\text{op}}$ for every $M \in \mathbf{M}_n(\mathbf{K})$.

Denote $\mathcal{N} = \mathcal{N}(AQ)$. Assume by contradiction that the set $\mathcal{N}$ does not span $\mathbf{K}^n$. Its linear image $(A^{1/2}Q)(\mathcal{N})$ does not span $\mathbf{K}^n$ either, and therefore there exists a nonzero orthogonal projection $P$ such that $PA^{1/2}Qx = 0$ for every $x \in \mathcal{N}$. Let $H$
be the matrix defined as $H = \lambda P - \text{Id}$, where $\lambda$ is a positive number chosen so that $\text{Tr} H > 0$.

Introduce functions $f, g \in C(T)$ defined as $f = \alpha(AQ)$ and $g = \alpha(A^{1/2}HA^{1/2}Q)$. Observe that $\|f\| = \|AQ\|_{\text{op}} = 1$. For $t = (x, y) \in T$, we have a chain of implications

$$|f(t)| = 1 \Rightarrow \|AQ(x)\| = 1 \Rightarrow x \in \mathcal{N} \Rightarrow HA^{1/2}Qx = -A^{1/2}Qx \Rightarrow g(t) = -f(t).$$

Consequently, the functions $f$ and $g$ satisfy the hypothesis of the following lemma, whose proof is postponed.

**Lemma.** Let $T$ be a nonempty compact topological space and $f, g \in C(T)$. Assume that for every $t \in T$ such that $|f(t)| = \|f\|$, we have $\text{Re} f(t)g(t) < 0$. Then, for $\delta > 0$ small enough, we have $\|f + \delta g\| < \|f\|$.

The lemma implies that for $\delta > 0$ small enough, we have

$$\|A^{1/2}(\text{Id} + \delta H)A^{1/2}Q\|_{\text{op}} = \|f + \delta g\| \le 1$$

and thus $A^{1/2}(\text{Id} + \delta H)A^{1/2} \in C_Q$. As $\delta$ goes to zero, we have $\det(\text{Id} + \delta H) = 1 + \delta \text{Tr} H + o(\delta)$ and therefore $\det(\text{Id} + \delta H) > 1$ for $\delta > 0$ small enough. The inequality

$$\det(A^{1/2}(\text{Id} + \delta H)A^{1/2}) = \det(A) \det(\text{Id} + \delta H) > \det(A)$$

contradicts the maximality of $A$.

□

It remains to prove the lemma.

**Proof of the Lemma.** Both $f$ and $g$ are nonzero (otherwise the hypothesis fails) and we may assume by rescaling that $\|f\| = \|g\| = 1$. Let $T_1$ be the nonempty closed subset of $T$ defined as $T_1 = \{t \in T : |f(t)| = 1\}$. Since the function $t \mapsto \text{Re} f(t)g(t)$ is continuous, it achieves its maximum on $T_1$ and therefore there exists $\varepsilon > 0$ such that $\text{Re} f(t)g(t) \le -\varepsilon$ for every $t \in T_1$. Denote by $T_2$ the closed subset of $T$ defined as $T_2 = \{t \in T : \text{Re} f(t)g(t) \ge -\varepsilon\}$. Since $f$ is continuous, there exists $\eta > 0$ such that $|f(t)| \le 1 - \eta$ for every $t \in T_2$. For $t \in T$ and $\delta > 0$, we compute

$$|(f + \delta g)(t)|^2 &= |f(t)|^2 + 2\delta \text{Re} f(t)g(t) + \delta^2|g(t)|^2 \\ &\le \begin{cases} (1 - \eta)^2 + 2\delta + \delta^2 & \text{if } t \in T_2 \\ 1 - 2\delta\varepsilon + \delta^2 & \text{if } t \notin T_2 \end{cases}$$

It follows that $\|f + \delta g\|^2 \le \max((1 - \eta)^2 + 2\delta + \delta^2, 1 - 2\delta\varepsilon + \delta^2)$ and therefore $\|f + \delta g\| < 1$ for $\delta > 0$ small enough. □

### 3. PROOF OF THE THEOREM

The implication (1) $\implies$ (2) has been proved in the introduction. Conversely, let $X$ be a finite-dimensional normed space satisfying condition (2) from the Theorem. Let $\text{lso}(X)$ be the group of isometries of $X$, defined as

$$\text{lso}(X) = \{u : X \to X \text{ linear : } \|u(x)\| = \|x\| \text{ for every } x \in X\}.$$

There is an inner product on $X$ which is invariant with respect to $\text{lso}(X)$ (see for example [4, p. 131, Theorem 2]) and therefore, without loss of generality, we may assume that $X = (\mathbb{R}^n, \| \cdot \|)$ and that $\text{lso}(X)$ is a subgroup of the orthogonal group $O_n$. (From now on we consider only the real case, the complex case is similar using the unitary group $U_n$.)
Fix $Q \in \mathcal{O}_n$ and let $A$ be an element of maximal determinant in $M_n^+(K) \cap \mathcal{C}_Q$. The set $\mathcal{N}(AQ)$ is a linear subspace of $\mathbf{R}^n$ (by hypothesis (2) from the Theorem) which spans $\mathbf{R}^n$ (by the conclusion of the Proposition). It follows that $\mathcal{N}(AQ) = \mathbf{R}^n$ and therefore that $AQ \in \text{Iso}(X) \subset \mathcal{O}_n$. The matrix $A = (AQ)Q^{-1}$ is both orthogonal and positive semidefinite; it follows that $A$ is the identity matrix and therefore $Q \in \text{Iso}(X)$.

The previous paragraph shows that $\text{Iso}(X) = \mathcal{O}_n$. As a consequence, for every $x$ and $y$ in the Euclidean unit sphere $S^{n-1}$, there exists $u \in \text{Iso}(X)$ such that $u(x) = y$. It follows that the norm $\|\cdot\|$ is constant on $S^{n-1}$, hence is a multiple of the Euclidean norm. This shows that $X$ is an inner product space.

## ACKNOWLEDGEMENT

The first author was supported in part by ANR under the grant ESQuisses (ANR-20-CE47-0014-01). We thank Miguel Martín for pointing to us the reference [5].

## REFERENCES

[1] D. Amir. *Characterizations of inner product spaces*, volume 20 of Operator Theory: Advances and Applications. Birkhäuser Verlag, Basel, 1986.

[2] S. Artstein-Avidan and E. Putterman. Some new positions of maximal volume of convex bodies. *Matematica*, 1(4):765–808, 2022.

[3] P. Jordan and J. Von Neumann. On inner products in linear, metric spaces. *Ann. of Math. (2)*, 36(3):719–723, 1935.

[4] A. L. Onishchik and È. B. Vinberg. *Lie groups and algebraic groups*. Springer Series in Soviet Mathematics. Springer-Verlag, Berlin, 1990. Translated from the Russian and with a preface by D. A. Leites.

[5] D. Sain and K. Paul. Operator norm attainment and inner product spaces. *Linear Algebra Appl.*, 439(8):2448–2452, 2013.

UNIVERSITÉ LYON 1, CNRS, INRIA, INSTITUT CAMILLE JORDAN, 43, BOULEVARD DU 11 NOVEMBRE 1918, 69100 VILLEURBANNE, FRANCE