I want to prove that the sequence:
$$ ( 2 + \psi^2 ; \psi ) $$ is positive semi definite for $|\psi | < 1$, and ideally find conditions on $\psi$ about when it isn't.
Recall that for a sequence, the definition of being positive semi definite is: $\forall t_1, \cdots, t_n \in \mathbb T, \forall a_1, \cdots, a_n \in \mathbb R^*\colon$ $$ \sum_1^n \sum_k^n s_{t_j-t_k} a_j a_k \geq 0 $$
My attempt:
I reduced the problem to finding a condition on $\psi$ such that this inequality hold (in that case, the sequence is not positive semi definite).
$$ 2A + 2B - 2 \psi A B + \psi^2 [ A + B ] < 0 $$
my problem is that $A,B$ can be any number and I don't know how to tackle that problem. I computed the roots (potentially complex) but it isn't instructive.
Ok actually I had a brilliant idea. Instead of looking at the sequence, I looked at the derived matrix.
In other words, I want to prove that $M$ is semi positive definite where: $$ M = \begin{bmatrix} 2+ \psi^2 & -\psi \\ -\psi & 2+\psi^2 \end{bmatrix} $$
I computed the eigenvalues with the characteristic polynomial, and in fact, there is no real roots thus the matrix is positive definite.
Then, this results in having my original sequence positive definite $\forall \psi$!
EDIT
OFC there are real roots because the matrix is symetric (spectral theorem). I did a small mistake computing the determinant ( the formula is ad-bc ) and then I found those two roots :
$$ r_1 = \psi^2 + \psi + 2, \qquad r_2 = \psi^2 - \psi + 2 $$
which polynomials in $\psi$ don't have any root. So, the roots never get negative, then the matrix has 2 positive eigenvalues.