For positive semidefinite unit trace $\Gamma$, what is $\{ x \in \mathbb R^N: x^T A x = \text{tr}(A \Gamma) \ \forall A \in\mathbb R^{N \times N} \}$?

Question

My question. Let $N > 1$ and $\Gamma \in \mathbb R^{N \times N}$ be a positive semidefinite matrix with unit trace. Can we describe in a simple manner the following set? $$S(\Gamma) := \{ x \in \mathbb R^N: x^T A x = \text{tr}(A \Gamma) \ \forall A \in \mathbb R^{N \times N} \}.$$

Some ideas.

1. We have $S(0) = \{ 0 \}$.
2. For $N = 2$, the set $S(\text{id}_N)$ is empty: for $A = \scriptsize\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ we get that the second component of $x \in \mathbb R^2$ must be either $1$ or $-1$. Analogously for $A = \scriptsize\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ we get that the first component of $x \in \mathbb R^2$ must be either $1$ or $-1$. For $A = \scriptsize\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ we get that $x_1 x_2 = 0$, which is a contradiction.
3. If $x \in S(\Gamma)$, then $c x \not\in S(\Gamma)$ for all $c \in \mathbb R \setminus \{ \pm 1 \}$, so we can think of the $x \in S(\Gamma)$ as directions, the length being fixed.
4. The condition $x^T A x = \text{tr}(A \Gamma)$ is invariant under transposition of $A$, so we have $$ S(\Gamma) = \{ x \in \mathbb R^N: x^T A x = \text{tr}(A \Gamma) \ \forall A = A^\mathsf T \in \mathbb R^{N \times N} \}. $$
5. We can explicitly write the condition (where $\Gamma = (\gamma_{j, k})_{j, k = 1}^{N}$) as $$ \sum_{j, k = 1}^{N} a_{j k} x_j x_k = \sum_{j, k = 1}^{N} a_{j k} \gamma_{j k} \qquad \forall (a_{j, k})_{j, k = 1}^{N} \in \mathbb R^{N \times N}, $$ so if $\Gamma$ has rank 1 and we can thus write it as $\Gamma = x_0 x_0^{\mathsf T}$, then $x_0 \in S(\Gamma)$. As $\text{tr}(\Gamma) = 1$, this implies $\| x_0 \| = 1$.

Background. I am reading the first section of An extension of Kakutani’s theorem on infinite product measures to the tensor product of semifinite *-algebras by Donald Bures. I want to understand definition 1 for the real full matrix von Neumann algebra $\mathcal A := \mathbb R^{N \times N}$.

Definition 1. Let $\mathcal A$ be a von Neumann algebra and $\Sigma$ the set of normal states on $\mathcal A$. A representation $\phi$ of $\mathcal A$ on $H$ we mean an isomorphism of $\mathcal A$ onto a von Neumann algebra acting on $H$. For $\mu \in \Sigma$ let $$ S(\phi, \mu) := \left\{ x \in H: ( \phi(A) x \mid x)_H = \mu(A) \ \forall A \in \mathcal A \right\} $$ the set of vectors such that $\mu$ induces a vector state of $\mathcal A$ with respect to the representation $\phi$.

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I chose $H := \mathbb R^N$ and $\phi \colon \mathcal A \to B(H)$, $A \mapsto (x \mapsto A x)$. Since $\mathcal A$ is finite-dimensional, all states are normal and the state space of $\mathcal A$ is $$ \mathcal S(\mathcal A) := \left\{ A \mapsto \text{tr}(A \Gamma): \Gamma \in S_{+, 1}^N \right\} \cong S_{+, 1}^N := \left\{ X \in \mathcal A: \text{tr}(A) = 1, A = A^{\mathsf{T}} \right\}. $$

Answer 1

Let $x\in S(\Gamma)$. Then for $A=\Gamma-xx^T$, we have $$ \|\Gamma-xx^T\|_F^2 =\langle A,\Gamma-xx^T\rangle_F =\operatorname{tr}\left(A(\Gamma-xx^T)\right) =\operatorname{tr}(A\Gamma)-x^TAx =0. $$ Therefore $\Gamma$ is necessarily equal to $xx^T$.

In other words, $S(\Gamma)$ is empty if and only if $\operatorname{rank}(\Gamma)>1$. When $\operatorname{rank}(\Gamma)\le1$, we can express $\Gamma$ as $xx^T$ for some vector $x$. Hence $S(\Gamma)=\{x,-x\}$ in this case.