4
$\begingroup$

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 $G(e^{i\theta})^*\Delta(e^{i\theta})G(e^{i\theta})\equiv 0$?


Some comments.

  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.

$\endgroup$

1 Answer 1

1
$\begingroup$

I think I've managed to provide an answer to the second (simpler) question. I'll post it here.

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 $$ G^*\Delta G \equiv 0 \iff \int_{-\pi}^\pi G G^* \Delta G G^* \frac{\mathrm{d}\theta}{2\pi}=0. $$

At this point, I don't know if this result could be helpful for providing an answer to the main question. Perhaps some of you might find it so. (Frankly, I couldn't get any insight from this approach.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.