Return to Question

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 $$ \Delta_i := A^\top P_i B+B^\top P_i A, $$ $0_{n\times n}$ denotes the $n\times n$ zero matrix and $(\cdot)^\top$ denotes transposition.

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, 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 $G_i^\top \Delta_i G_i=0_{m\times m}$ for all $i=1,2,\dots,N$, as desired.

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.

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 $$ \Delta_i := A^\top P_i B+B^\top P_i A, $$ $0_{n\times n}$ denotes the $n\times n$ zero matrix and $(\cdot)^\top$ denotes transposition.

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 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, 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 $G_i^\top \Delta_i G_i=0_{m\times m}$ for all $i=1,2,\dots,N$, as desired.

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.

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 $$ \Delta_i := A^\top P_i B+B^\top P_i A, $$ $0_{n\times n}$ denotes the $n\times n$ zero matrix and $(\cdot)^\top$ denotes transposition.

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 desired result follows.)

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.

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 $$ \Delta_i := A^\top P_i B+B^\top P_i A, $$ $0_{n\times n}$ denotes the $n\times n$ zero matrix and $(\cdot)^\top$ denotes transposition.

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, 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 $G_i^\top \Delta_i G_i=0_{m\times m}$ for all $i=1,2,\dots,N$, as desired.

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.

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 $$ \Delta_i := A^\top P_i B+B^\top P_i A, $$ $0_{n\times n}$ denotes the $n\times n$ zero matrix and $(\cdot)^\top$ denotes transposition.

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.

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 $$ \Delta_i := A^\top P_i B+B^\top P_i A, $$ $0_{n\times n}$ denotes the $n\times n$ zero matrix and $(\cdot)^\top$ denotes transposition.

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 desired result follows.)

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.