Is this matrix positive-semidefinite in general?

Question

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.)

$$ C_{n,n} = \begin{pmatrix} 1 & \alpha^1& \alpha^2 & \cdots & \alpha^{n-1} \\ \alpha^1 & 1 & \alpha^1&\cdots & \alpha^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \alpha^{n-1} & \alpha^{n-2} & \alpha^{n-3}& \cdots & 1 \end{pmatrix} =\begin{pmatrix} \alpha^0 & \alpha^1& \alpha^2 & \cdots & \alpha^{n-1} \\ \alpha^1 & \alpha^0 & \alpha^1&\cdots & \alpha^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \alpha^{n-1} & \alpha^{n-2} & \alpha^{n-3}& \cdots & \alpha^0 \end{pmatrix} $$

Answer 1

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.