# On a vanishing integral inner product

Let $G(z)$ be an $n\times m$ rational matrix-valued function of full column rank on the unit circle. Further, let $P(z)$ be an $m\times m$ rational matrix-valued function positive definite on the unit circle and let $A$, $B$ be $m\times n$ complex (constant) matrices. Consider the following inner-product condition (on $\mathcal{L}_2^{m\times m}[-\pi,\pi]$) $$\tag{1}\label{eq:cond} \langle G^*XG,G^* \Delta G \rangle_2 = \mathrm{tr}\int_{-\pi}^{\pi} G(e^{i\theta})^* X G(e^{i\theta}) G(e^{i\theta})^* \Delta(e^{i\theta}) G(e^{i\theta}) \frac{\mathrm{d}\theta}{2\pi}=0,\ \, \forall X\in\mathcal{H}_n,$$ where $(\cdot)^*$ denotes Hermitian transposition, $\mathcal{H}_n$ denotes the space of Hermitian $n\times n$ matrices, and $$\Delta(e^{i\theta}) := A^* P(e^{i\theta}) B + B^* P(e^{i\theta}) A.$$

My question. Does condition \eqref{eq:cond} necessarily imply $G(e^{i\theta})^*\Delta(e^{i\theta})G(e^{i\theta})\equiv 0$?

1. A simple observation is that the answer is in the affirmative if $P(e^{i\theta})=p(e^{i\theta})M$ where $p(e^{i\theta})$ is scalar and $M$ is a constant $m\times m$ positive definite matrix. Indeed, in this case, the thesis readily follows from the fact that, picking $X=A^* M B + B^* M A$ in \eqref{eq:cond}, yields $$\left\|p^{1/2} G^*(A^* M B + B^* M A)G\right\|_2=0.$$
2. Condition \eqref{eq:cond} can be equivalently rewritten as $$\tag{2}\label{eq:cond2} \int_{-\pi}^{\pi} G(e^{i\theta}) G(e^{i\theta})^* \Delta(e^{i\theta}) G(e^{i\theta}) G(e^{i\theta})^* \frac{\mathrm{d}\theta}{2\pi}=0,$$ since $\mathrm{tr}(XY)=0$, $\forall X \in\mathcal{H}_n$ and a fixed $Y\in\mathcal{H}_n$, is equivalent to $Y=0$. Condition \eqref{eq:cond2} asserts that $G(e^{i\theta})^* \Delta(e^{i\theta}) G(e^{i\theta})$ is in the kernel of the linear operator $\Sigma\colon T \mapsto \int_{-\pi}^{\pi}G T G^*\frac{\mathrm{d}\theta}{2\pi}$, mapping rational Hermitian $m\times m$ matrix-valued functions to Hermitian (constant) $n\times n$ matrices.

Based on my previous comments, a point that I didn't manage to properly understand (and might be useful to provide an answer to my question above) is the following one.

A (perhaps) simpler question. In the special case $P(e^{i\theta})=p(e^{i\theta})M$ where $p(e^{i\theta})$ is scalar and $M$ is a constant $m\times m$ positive definite matrix, how to prove directly (i.e., without exploiting \eqref{eq:cond}) that condition \eqref{eq:cond2} necessarily implies $G(e^{i\theta})^*\Delta(e^{i\theta})G(e^{i\theta})\equiv 0$?

Any (even partial) comment and/or suggestion is very welcome. Thanks in advance.

Let $P(e^{i\theta})=Mp(e^{i\theta})$. (In what follows, I will drop the dependence on $\theta$ to lighten notation.) Consider the linear operator: \begin{align} \Sigma\colon\ &\mathcal{L}_2^{m\times m}[-\pi,\pi]\to \mathcal{H}_n\\ &\ T \mapsto \int_{-\pi}^\pi p G T G^* \frac{\mathrm{d}\theta}{2\pi}. \end{align} (Here $\mathcal{L}_2^{m\times m}[-\pi,\pi]$ denotes the space of $m\times m$ matrix-valued functions that are square integrable on $[-\pi,\pi]$). Notice that the adjoint of $\Sigma$ has the form \begin{align} \Sigma^\dagger\colon &\ \mathcal{H}_n \to \mathcal{L}_2^{m\times m}[-\pi,\pi]\\ &\ Z \mapsto p G^* Z G. \end{align} Now the well-known fact $\ker(\Sigma^\dagger)=\ker(\Sigma\Sigma^\dagger)$ immediately implies that $$G^*\Delta G \equiv 0 \iff \int_{-\pi}^\pi G G^* \Delta G G^* \frac{\mathrm{d}\theta}{2\pi}=0.$$