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Let $\{ G_k \}$ be a set of $n \times n$ (generally complex) matrices satisfying the completeness relation

\begin{equation} \sum_k G_k G_k^* = \mathbf{1}, \end{equation}

where $*$ denotes the conjugate-transpose operation. Suppose that one of these matrices, say $G_0$, has the specific form

\begin{equation} G_0 = e^{-i K}, \end{equation}

where $K$ is a non-Hermitian matrix. Given this setup, can one determine whether the operator

\begin{equation} B_X=Q^2 \big( G_0 X G_0^* - G_k X G_k^* \big), \quad \text{for } k \neq 0, \tag 1 \end{equation}

is positive? Here, $Q$ is a real symmetric matrix, and $X$ is a Hermitian matrix with $\operatorname{Tr}[X] = 1$. A particular case of interest is when

\begin{equation} X = \frac{1}{n} \mathbf{1}. \end{equation}

In this scenario, the problem reduces to analyzing the positivity of

\begin{equation} B_{\frac{1}{n}\mathbf{1}} =\frac{1}{n}Q^2 \big( G_0 G_0^* - G_k G_k^* \big), \quad \text{for } k \neq 0. \end{equation}

It is not immediately clear to me whether this expression is positive. In summary, under what conditions can the operator in $(1)$ be guaranteed to be positive?

Rodrigo de Azevedo
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User1729173
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  • Are you sure it's $Q^2 ( G_0 G_0^* - G_k G_k^* )$ and not something like $Q ( G_0 G_0^* - G_k G_k^* )Q$? Or how do you ensure that the operator you care about is even Hermitian (which is of course necessary for it to be self-adjoint)? – Frederik vom Ende Mar 16 '25 at 09:27
  • Thanks @FrederikvomEnde, I would be interested in knowing whether there is any hope for $Q(G_0G_0^* - G_k G_K^*)Q$ as well. – User1729173 Mar 16 '25 at 15:30
  • For the symmetric case Douglas' lemma may be of interest to you. A simple sufficient condition motivated by this approach is that is that if $Q$ "annihilates the negative part" of $G_0(\cdot)G_0^-G_k(\cdot)G_k^$—in the sense that $Q(G_k-cG_0)=0$ for some $c\in\mathbb C$, $|c|\leq 1$—then $QG_0XG_0^Q-QG_kXG_k^Q\geq 0$ for all $X\geq 0$. So there is certainly hope, but I fear the general situation is too messy, and I do not see how the extra conditions you provided narrow down the possibilities in any way. – Frederik vom Ende Mar 16 '25 at 17:38

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