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We have $p \in \mathbb{N}$ and $\mathcal{H}$ is a Hilbert space.

  • let's consider a matrix $\boldsymbol{\Gamma}_p := (C_{i-j})_{i,j=1, ..., p} \in \mathcal{S_H}^{p\times p}\!\!\,,$ that is a $p\times p$-matrix of Hilbert-Schmidt operators $C_{1-p}, ..., C_{p-1}\colon \mathcal{H} \rightarrow \mathcal{H}$,
  • $C_0$ is selfadjoint and positive semidefinite and $\forall l \in \{1, ..., p-1\}\colon$ $C^*_l = C_{-l}$ where $"^*"$ denotes the adjoint of an operator and $||C_0||_{\mathcal{S}_H} \geq ||C_i||_{\mathcal{S_H}}, \forall i \in \{1-p, ..., p-1\}.$

With that given, the matrix

$$ \boldsymbol{\Gamma}_p = \begin{bmatrix} C_{0} & C_{-1} & C_{-2} & \ldots & \ldots &C_{1-p} \\ C_1 & C_0 & C_{-1} & \ddots & & \vdots \\ C_2 & C_1 & \ddots & \ddots & \ddots& \vdots \\ \vdots & \ddots & \ddots & \ddots & C_{-1} & C_{-2}\\ \vdots & & \ddots & C_1 & C_0& C_{-1} \\ C_{p-1}& \ldots & \ldots & C_2 & C_1 &C_0 \end{bmatrix} $$

is selfadjoint, diagonal-dominant (respective $||\cdot||_{\mathcal{S_H}}$) with positive semidefinite diagonal elements.

Questions:

  1. Is $\boldsymbol{\Gamma}_p$ positive semidefinit?
  2. Could it be even strictly positive definite (but I guess it doesn't hold, since $C_0$ is not necessarily strictly positive definite) ?

Thank you very much!

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  • $\begingroup$ How can you hope for invertibility if $C_0$ is not invertible? $\endgroup$ Feb 23, 2016 at 11:08
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    $\begingroup$ The $3\times 3$-matrix $\begin{pmatrix}2&-a&-a\\-a&2&-a\\-a&-a&2\end{pmatrix}$ is a counterexample for 2 and 3 if $1<a\le 2$. Maybe, this question is better for math.stackexchange. $\endgroup$ Feb 23, 2016 at 11:10
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    $\begingroup$ @ Jochen: $C_0$ is invertible, @ Sebastian: I don't see why this should be a counterexample, since all "leading main-minors" (führende Hauptminoren) are positive with $a \in (1,2].$ Am I right? $\endgroup$
    – Obriareos
    Feb 23, 2016 at 11:57
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    $\begingroup$ No, the vector $(1,1,1)^t$ is an eigenvector with eigenvalue $2-2a$, which is negative for $a>1$. We need $a\le 2$ only because of your hypothesis that the off-diagonal terms should not have larger norm than the diagonal terms. For $a=1$, the matrix is therefore not invertible. $\endgroup$ Feb 23, 2016 at 14:29
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    $\begingroup$ The compact operator $C_0$ can be invertible only if your Hilbert space is finite dimensional. $\endgroup$ Feb 23, 2016 at 15:57

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