linear algebra - Inverse of nonnegative Toeplitz matrice

Consider a right-hand circulant matrice of size $n$ (called also Toeplitz matrice)
\begin{equation}
T=
\left( \begin{array}{ccccc}
a_1 & a_2 & a_3 & \ldots & a_n \\
a_n & a_1 & a_2 & \ldots & a_{n-1}\\
\vdots\\
a_2 & a_3 & a_4 & \ldots & a_1
\end{array} \right)

\end{equation}

$\{a_i\}$ are nonnegative and not all equal.

Is $T$ alwayls invertible?

I proved that:

$\bullet$ for n=2 the $Det(T) = (a_1+a_2)(a_1-a_2) \neq 0$ $

$\bullet$ for n=3 the $Det(T) = \frac{1}{2}(a_1+a_2+a_3)\left( (a_1-a_2)^2 + (a_1-a_3)^2(a_2-a_3)^2 \right) \neq 0$

What about $n>3$?

Answer

a long hint: does writing $T$ as a linear combination of the powers of $P $ where $p_{i i-1} = 1$ and zero every where else. $P$ has eigenvalues the $n$th roots of unity. so $T$ thaw $a_1+a_2\omega+a_3\omega^2+\cdots$ and the determinant of $T$ is the product of
$(a_1+a_2\omega+a_3\omega^2+\cdots)(a_1+a_2\omega+a_3\omega^2+\cdots))\cdots$

showing that the determinant is nonzero reduces to showing that all the factors are nonzero. that is if $\omega$ in an $n$th root of unity and $a_1, a_2, \cdots$ are non negative and not all equal, then $$(a_1+a_2\omega+a_3\omega^2+\cdots )\neq 0$$

it is more likely to be true if $n$ is a prime number. i am not sure if it is true for nonprime numbers. e.g., take $n = 4, \omega = i, a_1 = a2 = -a_3 = -a_4 = 1 $