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 [Reference Artstein-Avidan and Putterman2]. 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 $\mathsf {M}_n(\mathbf {K})$ of $n \times n$ matrices with the corresponding operator norm $\| \cdot \|_{\mathrm {op}}$ . We denote by $\mathsf {GL}_n(\mathbf {K})$ the group of invertible matrices. Given $Q \in \mathsf {GL}_n(\mathbf {K})$ , we consider the set

$$\begin{align*}\mathcal{C}_Q = \{ A \in \mathsf{M}_n(\mathbf{K}) \ : \ \| AQ \|_{\mathrm{op}} \leq 1\}. \end{align*}$$

Let $\mathsf {M}_n^+(\mathbf {K})$ be the cone of positive semi-definite matrices. The set $\mathsf {M}_n^+(\mathbf {K}) \cap \mathcal {C}_Q$ 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 $\| AQ \|_{\mathrm {op}} = 1$ .

Proposition. Let $Q \in \mathsf {GL}_n(\mathbf {K})$ and A of maximal determinant in $\mathsf {M}_n^+(\mathbf {K}) \cap \mathcal {C}_Q$ . Then the set

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

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 [Reference Artstein-Avidan and Putterman2, Theorem 1.2]. For completeness, we include a self-contained proof.

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

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

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

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

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 $\alpha : \mathsf {M}_n(\mathbf {K}) \to C(T)$ as follows: if $M \in \mathsf {M}_n(\mathbf {K})$ , $x \in \mathcal {B}$ and $y \in \mathcal {B}^*$ , then

$$\begin{align*}\alpha(M)(x,y) = \langle Mx , y\rangle. \end{align*}$$

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

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 $H= \lambda P- \mathrm {Id}$ , where $\lambda $ is a positive number chosen so that $\operatorname {\mathrm {Tr}} H>0$ .

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

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

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

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

$$\begin{align*}\| A^{1/2}(\mathrm{Id} + \delta H)A^{1/2}Q \|_{\mathrm{op}} = \|f + \delta g\| \leq 1 \end{align*}$$

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

$$\begin{align*}\det(A^{1/2}(\mathrm{Id} + \delta H)A^{1/2} ) = \det(A)\det( \mathrm{Id} + \delta H)> \det(A)\end{align*}$$

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 ${t \mapsto \Re f(t)\overline {g(t)}}$ is continuous, it achieves its maximum on $T_1$ and therefore there exists $\varepsilon>0$ such that $\Re f(t)\overline {g(t)} \leq -\varepsilon $ for every $t \in T_1$ . Denote by $T_2$ the closed subset of T defined as $T_2 = \{t \in T \ : \ \Re f(t)\overline {g(t)} \geq -\varepsilon \}$ . Since f is continuous, there exists $\eta>0$ such that $|f(t)| \leq 1-\eta $ for every $t \in T_2$ . For $t \in T$ and $\delta>0$ , we compute

$$ \begin{align*} |(f+\delta g)(t)|^2 &= |f(t)|^2 + 2 \delta \Re f(t) \overline{g(t)} + \delta^2 |g(t)|^2 \\ & \leq \begin{cases} (1-\eta)^2 + 2 \delta + \delta^2 & \text{if } t \in T_2\\ 1 - 2 \delta \varepsilon + \delta^2 & \text{if } t \not \in T_2. \end{cases} \end{align*} $$

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

3 Proof of the Theorem

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

$$\begin{align*}\mathsf{Iso}(X) = \{ u : X \to X \text{ linear } \ : \ \|u(x)\|=\|x\| \text{ for every }x \in X \}. \end{align*}$$

There is an inner product on X which is invariant with respect to $\mathsf {Iso}(X)$ (see for example [Reference Onishchik and Vinberg4, p. 131, Theorem 2]) and therefore, without loss of generality, we may assume that $X=(\mathbf {R}^n,\|\cdot \|)$ and that $\mathsf {Iso}(X)$ is a subgroup of the orthogonal group $\mathsf {O}_n$ . (From now on we consider only the real case, the complex case is similar using the unitary group $\mathsf {U}_n$ .)

Fix $Q \in \mathsf {O}_n$ and let A be an element of maximal determinant in $\mathsf {M}_n^+(\mathbf {K}) \cap \mathcal {C}_Q$ . 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 $AQ \in \mathsf {Iso}(X) \subset \mathsf {O}_n$ . The matrix $A = (AQ)Q^{-1}$ is both orthogonal and positive semidefinite; it follows that A is the identity matrix and therefore $Q \in \mathsf {Iso}(X)$ .

The previous paragraph shows that $\mathsf {Iso}(X) = \mathsf {O}_n$ . As a consequence, for every x and y in the Euclidean unit sphere $S^{n-1}$ , there exists $u \in \mathsf {Iso}(X)$ 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.