Let $ \{G_i\}_{i=1}^N\in\mathbb{R}^{n\times m}$ be a set of full column rank matrices (i.e., $ \mathrm{rank}(G_i)=m$ for all $ i$ ) and $ \{P_i\}_{i=1}^N\in\mathbb{R}^{m\times m}$ be a set of positive definite matrices. Let $ A\in\mathbb{R}^{m\times n}$ , $ B\in\mathbb{R}^{m\times n}$ and consider the following equation $ $ \tag{1}\label{eq:1} \sum_{i=1}^N G_i G_i^\top \Delta_i G_i G_i^\top=0_{n\times n} $ $ where $ $ \Delta_i = A^\top P_i B+B^\top P_i A, $ $ and $ 0_{n\times n}$ denotes the $ n\times n$ zero matrix.

Q: Does \eqref{eq:1} imply $ G_i^\top \Delta_i G_i=0_{m\times m}$ for all $ i=1,2,\dots,N$ ?

The answer is in the affirmative if $ P_i=p_i M$ , for all $ i$ , with $ M\in\mathbb{R}^{m\times m}$ being a positive definite matrix and $ \{p_i\}_{i=1}^N$ being a set of positive real numbers. Indeed, this readily follows, for instance, from the fact that \eqref{eq:1} is equivalent to $ $ \tag{2}\label{eq:2} \sum_{i=1}^N \mathrm{tr}(G_i^\top XG_i G_i^\top \Delta_i G_i) = 0, \ \ \text{for all symmetric } X\in\mathbb{R}^{n\times n}, $ $ where $ \mathrm{tr}(\cdot)$ denotes the trace operator. (More precisely, after picking $ X=A^\top MB+B^\top M A$ in the previous expression.)

The above question can be thought of as a “discretized” version of this OP. My hope is that in this (simplified) setting either a proof or counterexample (that so far I couldn’t find) can more easily emerge.