0
$\begingroup$

Let $$ Q = \begin{bmatrix} Q_0 & Q_1 & Q_2 & \cdots\\ Q_{-1} & Q_{0} & Q_1 & \cdots\\ Q_{-2} & Q_{-1} & Q_0 & \cdots\\ \vdots & \vdots & \vdots & \ddots \end{bmatrix} $$ be a block Toeplitz matrix with $Q_n\in \mathbb{R}^{p\times p}$ and $Q_{-n} = Q_{n}^{\sf{T}}$, i.e., $Q$ is symmetric. Additionally, let $Q_0$ be positive-definite.

Questions:

  1. When is a finite such $Q$ invertible?
  2. When does the inverse represent a bounded linear operator for infinite $Q$?
Improve this question
$\endgroup$
1
  • $\begingroup$ @Derek - $Q_0$ being positive definite (and hence invertible) does not imply that $Q$ is invertible. For example, suppose $Q_0$ is $1 \times 1$ and equal to the scalar $1$, and $Q$ is the $2\times 2$ all-ones matrix. The inequality Q > Q0+Q0+...Q0 is not true in general. $\endgroup$
    – Nathaniel Johnston
    Commented Mar 29 at 17:20

0

Reset to default

You must log in to answer this question.