Is this matrix positive-semidefinite in general? for the matrix written below I was wondering if one can show that it is positive-semidefinite for $n>3$ and $0< \alpha<1$. (Or not. For $n=2, 3$ it works by showing that all principal minors are non-negative.)
$$
C_{n,n} =
 \begin{pmatrix}
  1 & \alpha^1& \alpha^2 & \cdots & \alpha^{n-1} \\
  \alpha^1 & 1 &  \alpha^1&\cdots & \alpha^{n-2} \\
  \vdots  & \vdots & \vdots   & \ddots & \vdots  \\
   \alpha^{n-1} & \alpha^{n-2} & \alpha^{n-3}& \cdots & 1 \end{pmatrix}
 =\begin{pmatrix}
  \alpha^0 & \alpha^1& \alpha^2 & \cdots & \alpha^{n-1} \\
  \alpha^1 & \alpha^0 &  \alpha^1&\cdots & \alpha^{n-2} \\
  \vdots  & \vdots & \vdots   & \ddots & \vdots  \\
   \alpha^{n-1} & \alpha^{n-2} & \alpha^{n-3}& \cdots & \alpha^0
 \end{pmatrix}
$$

Taking the first column, and substracting to it $\alpha$ times the column 2, we get $\det C_{n,n}=(1-\alpha^2)\det C_{n-1,n-1}$, hence we can conclude by Sylvester's criterion. 
