I am trying to prove that a matrix following a certain structure based on its index is always positive definite. Each entry in the matrix has the following value:
$$ \text{e}^{- \alpha \cdot |i-j|} ~~~~,\alpha \in \mathbb{R}_{>0}$$
Where $i$ and $j$ represent the indices of the rows and columns of the matrix. Thereby, a $N \times N$ matrix $M$ would look like the following:
$$ M = \begin{bmatrix} 1 & \text{e}^{- \alpha} & \cdots & \text{e}^{- \alpha \cdot (N-1)} \\ \text{e}^{- \alpha} & 1 & \ldots & \text{e}^{- \alpha \cdot (N-2)} \\ \vdots & \vdots & \ldots &\vdots\\ \text{e}^{- \alpha \cdot (N-1)} & \text{e}^{- \alpha \cdot (N-2)} & \ldots & 1 \end{bmatrix}$$
I have tried to prove that the quadratic form of the matrix is positive:
$$ \forall x \in \mathbb{R}^N\smallsetminus\{0\}\quad x^T M x > 0 $$
But I don't know how to prove that the following summation is always positive:
$$ \sum_{i=1}^N \sum_{j=1}^N x_i x_j \, \text{e}^{- \alpha \cdot |i-j|} $$
I do realize that setting $\alpha=0$ results in $\left( \sum_{i=1}^N x_i\right)^2$, but I assume that this is not proof enough.
If anyone could give a hint I would be very gratefull!
