I would like to work out some simple condition on the entries of a particular sort of matrix for the matrix to be positive semidefinite. This matrix has the following form
$$ {\bf Q} = \begin{bmatrix}r_0 & r_1 & r_2 & \dots & r_n \\r_1^* & r_0 & r_1 &\dots &r_{n-1}\\r_2^* & r_1^* & r_0 &\dots &r_{n-2}\\\vdots & \vdots &\vdots &\ddots &\vdots\end{bmatrix} $$
where $*$ denotes complex conjugate; in other words it is a Hermitian Toeplitz matrix. In addition, it has a sparsity constraint, only a small number, $q$, of the $n$ coefficients are non-zero; so there are a large number ($2(n-q)$) of diagonal stripes of zeros. ($2(n-q)$ since the diagonal is never zero).
In general, I think it is hard to determine whether Hermitian Toeplitz matrices are positive (semi)definite without looking at the particular example matrix and explicitly checking its eigenvalues, see this question:
However when the matrix has a particular form more progress can be made:
https://mathoverflow.net/questions/68099/diagonalizing-a-certain-real-and-symmetric-toeplitz-matrix
How to prove the positive-definiteness of a symmetric Toeplitz matrix like this?
Is this matrix obviously positive definite?
I'm hopeful that the sparsity constraint might provide some foothold for progress, but I haven't managed to make this work myself.
Therefore my question is as stated before: is there some simple condition on the values $r_0, r_1, r_2, \dots, r_n$, of which only $q$ are non-zero, that is synonymous with ${\bf Q}$ being positive (semi)definite? A relaxation would be to consider only real numbers $r_0, r_1, r_2, \dots, r_n$ giving instead a sparse symmetric Toeplitz matrix to study. If you have something that works only for positive-definiteness not positive semidefiniteness I would be equally excited!