Assume $Q$ is a general Toeplitz matrix. Under what conditions can we make sure $$\mathrm{det}(I-Q^2)\neq 0?$$

Let's denote the determinant by $|\cdot|$. We can show that $$|I-Q^2| = |I-Q||I+Q|\neq 0 .$$ Thus,
$$|I-Q|\neq 0 \text{ and }|I+Q|\neq 0 .$$ What does it mean?

  • 3
    $\begingroup$ It means that $1$ and $-1$ are not eigenvalues of $Q$. $\endgroup$ – Robert Israel Apr 11 '19 at 20:25

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