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

My question. Suppose that either $A$ or $B$ is of full (row) rank. Does \eqref{eq:1} imply $Y_i^\top \Delta_i Y_i=0_{m}$ for all $i=1,2,\dots,N$?

Two remarks.

1. 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. This follows from the fact that \eqref{eq:1} is equivalent to $$\tag{2}\label{eq:2} \sum_{i=1}^N \mathrm{tr}(Y_i^\top XY_i Y_i^\top \Delta_i Y_i) = 0, \ \ \text{for all symmetric } X\in\mathbb{R}^{n\times n}, $$ where $\mathrm{tr}(\cdot)$ denotes the trace operator. More precisely, if we pick $X=A^\top MB+B^\top M A$ in the previous expression, the LHS of \eqref{eq:2} turns into a sum positive numbers which implies that $Y_i^\top \Delta_i Y_i=0_{m}$ for all $i=1,2,\dots,N$, as desired.
2. The assumption hat either $A$ or $B$ is of full (row) rank seems to be crucial. Indeed, if we drop this assumption, then the answer is in the negative. For a counterexample pick $N=2$ and $$ Y_1=Y_2=I_2,\ P_1=\begin{bmatrix} 2&1\\1&2\end{bmatrix},\ P_2=\begin{bmatrix} 2&-1\\-1&2\end{bmatrix},\ A=\begin{bmatrix} 0&0\\1&0 \end{bmatrix},\ B=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$

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

My question. Suppose that either $A$ or $B$ is of full (row) rank. Does \eqref{eq:1} imply $Y_i^\top \Delta_i Y_i=0_{m}$ for all $i=1,2,\dots,N$?

Two remarks.

1. 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. This follows from the fact that \eqref{eq:1} is equivalent to $$\tag{2}\label{eq:2} \sum_{i=1}^N \mathrm{tr}(Y_i^\top XY_i Y_i^\top \Delta_i Y_i) = 0, \ \ \text{for all symmetric } X\in\mathbb{R}^{n\times n}, $$ where $\mathrm{tr}(\cdot)$ denotes the trace operator. More precisely, if we pick $X=A^\top MB+B^\top M A$ in the previous expression, the LHS of \eqref{eq:2} turns into a sum positive numbers which implies that $Y_i^\top \Delta_i Y_i=0_{m}$ for all $i=1,2,\dots,N$, as desired.
2. The assumption that either $A$ or $B$ is of full (row) rank seems to be crucial. Indeed, if we drop this assumption, then the answer is in the negative. For a counterexample pick $N=2$ and $$ Y_1=Y_2=I_2,\ P_1=\begin{bmatrix} 2&1\\1&2\end{bmatrix},\ P_2=\begin{bmatrix} 2&-1\\-1&2\end{bmatrix},\ A=\begin{bmatrix} 0&0\\1&0 \end{bmatrix},\ B=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$